1. Let the system be one mole of argon gas at room temperature and
atmospheric pressure. Compute the total energy (kinetic only, neglecting
atomic rest energies), entropy, enthalpy, Helmholtz free energy, and
Gibbs free energy. Express all answers in SI units. Get solution
2. Consider the production of ammonia from nitrogen and hydrogen,N2 + 3H2 → 2NH3,at 298 K and 1 bar. From the values of ΔH and S tabulated at the back of this book, compute ΔG for this reaction and chock that it is consistent with the value given in the table. Get solution
3. Use the data at the back of this book to verify the values of ΔH and ΔG quoted above for the lead-acid below reaction.Reaction:... Get solution
4. In a hydrogen fuel cell, the steps of the chemical reaction areat – electrode: H2 + 2OH– → 2H2O + 2e–;at + electrode: ½O2 + H2O + 2e– → 2OH–.Calculate the voltage of the cell. What is the minimum voltage required for electrolysis of water? Explain briefly. Get solution
5. Consider a fuel cell that uses methane (“natural gas”) as fuel. The reaction isCH4 + 2O2 → 2H2O + CO2(a) Use the data at the back of this book to determine the values of ΔH and ΔG for this reaction, for one mole of methane. Assume that the reaction takes place at room temperature and atmospheric pressure.(b) Assuming ideal performance, how much electrical work can you get out of the cell, for each mole of methane fuel?(c) How much waste heat is produced, for each mole of methane fuel?(d) The steps of this reaction areat – electrode: CH4 + 2H2O → CO2 + 8H+ + 8e–;at + electrode: 2O2 + 8H+ + 8e– → 4H2O.What is the voltage of the cell? Get solution
6. A muscle can be thought of as a fuel cell, producing work from the metabolism of glucose:C6H12O6 + 6O2 → 6CO2 + 6H2O.(a) Use the data at the back of this book to determine the values of ΔH and ΔG for this reaction, for one mole of glucose. Assume that the reaction takes place at room temperature and atmospheric pressure.(b) What is the maximum amount of work that a muscle can perform, for each mole of glucose consumed, assuming ideal operation?(c) Still assuming ideal operation, how much heat is absorbed or expelled by the chemicals during the metabolism of a mole of glucose? (Be sure to say which direction the heat flows.)(d) Use the concept of entropy to explain why the heat flows in the direction it does.(e) How would your answers to parts (b) and (c) change, if the operation of the muscle is not ideal? Get solution
7. The metabolism of a glucose molecule (see previous problem) occurs in many steps, resulting in the synthesis of 38 molecules of ATP (adenosine triphosphate) out of ADP (adenosine diphosphate) and phosphate ions. When the ATP splits back into ADP and phosphate, it liberates energy that is used in a host of important processes including protein synthesis, active transport of molecules across cell membranes, and muscle contraction. In a muscle, the reaction ATP → ADP + phosphate is catalyzed by an enzyme called myosin that is attached to a muscle filament. As the reaction takes place, the myosin molecule pulls on an adjacent filament, causing the muscle to contract. The force it exerts averages about 4 piconewtons and acts over a distance of about 11 nm. From this data and the results of the previous problem, compute the “efficiency” of a muscle, that is, the ratio of the actual work done to the maximum work that the laws of thermodynamics would allow.Problem:A muscle can be thought of as a fuel cell, producing work from the metabolism of glucose:C6H12O6 + 6O2 → 6CO2 + 6H2O.(a) Use the data at the back of this book to determine the values of ΔH and ΔG for this reaction, for one mole of glucose. Assume that the reaction takes place at room temperature and atmospheric pressure.(b) What is the maximum amount of work that a muscle can perform, for each mole of glucose consumed, assuming ideal operation?(c) Still assuming ideal operation, how much heat is absorbed or expelled by the chemicals during the metabolism of a mole of glucose? (Be sure to say which direction the heat flows.)(d) Use the concept of entropy to explain why the heat flows in the direction it does.(e) How would your answers to parts (b) and (c) change, if the operation of the muscle is not ideal? Get solution
8. Derive the thermodynamic identity for G (below equation), and from it the three partial derivative relations below.EQUATION:dG = – S dT+ V dP+ μ dNPartial Derivative Relations:... Get solution
9. Sketch a qualitatively accurate graph of G vs. T for a pure substance as it changes from solid to liquid to gas at fixed pressure. Think carefully about the slope of the graph. Mark the points of the phase transformations and discuss the features of the graph briefly. Get solution
10. Suppose you have a mole of water at 25°C and atmospheric pressure. Use the data at the back of this book to determine what happens to its Gibbs free energy if you raise the temperature to 30°C. To compensate for this change, you could increase the pressure on the water. How much pressure would be required? Get solution
11. Suppose that a hydrogen fuel cell, as described in the text, is to be operated at 75° C and atmospheric pressure. We wish to estimate the maximum electrical work done by the cell, using only the room-temperature data at the back of this book. It is convenient to first establish a zero-point for each of the three substances, H2, O2, and H2O. Let us take G for both H2 and O2 to be zero at 25° C, so that G for a mole of H2O is −237 kJ at 25°C.(a) Using these conventions, estimate the Gibbs free energy of a mole of H2 at 75°C. Repeat for O2 and H2O.(b) Using the results of part (a), calculate the maximum electrical work done by the cell at 75°C, for one mole of hydrogen fuel. Compare to the ideal performance of the cell at 25°C. Get solution
12. Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,...where each ∂/∂V is taken with S fixed, each ∂/∂S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U) you can evaluate the partial derivatives in parentheses to obtain...a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H, F, and G). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you’ve done four of them the novelty begins to wear off. For applications of these Maxwell relations, see the next four problems. Get solution
13. Use a Maxwell relation from the previous problem 1 and the third law of thermodynamics to prove that the thermal expansion coefficient β (defined in Problem 2) must be zero at T = 0.Problem 1:Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,...where each ∂/∂V is taken with S fixed, each∂/∂S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U) you can evaluate the partial derivatives in parentheses to obtain...a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H, F, and G). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you’ve done four of them the novelty begins to wear off. For applications of these Maxwell relations, see the next four problems.Problem 2:When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:...(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive. Get solution
14. The partial-derivative relations derived in Problems 1, 4, and 5, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between Cp and CV. (a) With the heat capacity expressions from Problem 4 in mind, first consider S to be a function of T and V. Expand dS in terms of the partial derivatives (∂S/∂T)V and (∂S/∂V)T. Note that one of these derivatives is related to CV.(b) To bring in CP, consider V to be a function of T and P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set dP = 0 and note that you have derived a nontrivial expression for (∂S/∂T)P. This derivative is related to CP, so you now have a formula for the difference CP – CV.(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1. Your final result should be...(d) Check that this formula gives the correct value of CP – CV for an ideal gas.(e) Use this formula to argue that CP cannot be less than CV.(f) Use the data in Problem 1 to evaluate CP – CV for water and for mercury at room temperature. By what percentage do the two heat capacities differ?(g) Figure 1 shows measured values of CP for three elemental solids, compared to predicted values of CV. It turns out that a graph of β vs. T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why CP and CV agree at low temperatures but diverge in the way they do at higher temperatures.Problem 1:Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep V fixed as T increases, as follows. (a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume, dV1, in terms of dT and the thermal expansion coefficient β introduced in Problem 2.(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, dV2, in terms of dP and the isothermal compressibility κT defined as...(This is the reciprocal of the isothermal bulk modulus defined in Problem 3.)(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dP to dT is equal to ... since there is no net change in volume. Express this partial derivative in terms of β and κT. Then express it more abstractly in terms of the partial derivatives used to define β and κT. For the second expression you should obtain...This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.(d) Compute β, κT and ... for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).(e) For water at 25°C, β = 2.57 × 10−4 K−1 and κT = 4.52 × 10−10 Pa−1. Suppose you increase the temperature of some water from 20°C to 30°C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at 25°C) β = 1.81 × 10−4 K−1 and κT = 4.04 × 10−11 Pa−l. Given the choice, would you rather measure the heat capacities of these substances at constant V or at constant P?Problem 2:When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:...(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.Problem 3:By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by...where ρ is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness. More precisely, if we imagine applying an increase in pressure ΔP to a chunk of the material, and this increase results in a (negative) change in volume ΔV, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:...This definition is still ambiguous, however, because I haven’t said whether the compression is to take place isothermally or adiabatically (or in some other way).(a) Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.(b) Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.(c) Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the rms speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.(d) When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not?Problem 4:Use the thermodynamic identity to derive the heat capacity formula...which is occasionally more convenient than the more familial’ expression in terms of U. Then derive a similar formula for CP, by first writing dH in terms of dS and dP.Problem 5:Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,...where each ∂/∂V is taken with S fixed, each ∂/∂S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U) you can evaluate the partial derivatives in parentheses to obtain...a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H, F, and G). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you’ve done four of them the novelty begins to wear off.Figure 1: Measured heat capacities at constant pressure (data points) forone mole each of three different elemental solids. The solid curves show the heatcapacity at constant volume predicted by the model used in Section 7.5, with thehorizontal scale chosen to best fit the data for each substance. At sufficiently hightemperatures, CV for each material approaches the value 3R predicted by theequipartition theorem. The discrepancies between the data and the solid curvesat high T are mostly due to the differences between CP and CV. At T = 0 alldegrees of freedom are frozen out, so both CP and CV go to zero. Data from Y. S.Touloukian, ed., Thermophysical Properties of Matter (Plenum, New York, 1970).... Get solution
15. The formula for CP – CV derived in the previous problem 1 can also be derived starting with the definitions of these quantities in terms of U and H. Do so. Most of the derivation is very similar, but at one point you need to use the relation P = – (∂F/∂V)T. Problem 1: The partial-derivative relations derived in Problems 2, 5, and 6, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between Cp and CV. (a) With the heat capacity expressions from Problem 5 in mind, first consider S to be a function of T and V. Expand dS in terms of the partial derivatives (∂S/∂T)V and (∂S/∂V)T. Note that one of these derivatives is related to CV. (b) To bring in CP, consider V to be a function of T and P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set dP = 0 and note that you have derived a nontrivial expression for (∂S/∂T)P. This derivative is related to CP, so you now have a formula for the difference CP – CV.(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 2. Your final result should be...(d) Check that this formula gives the correct value of CP – CV for an ideal gas.(e) Use this formula to argue that CP cannot be less than CV.(f) Use the data in Problem 2 to evaluate CP – CV for water and for mercury at room temperature. By what percentage do the two heat capacities differ?(g) Figure 1 shows measured values of CP for three elemental solids, compared to predicted values of CV. It turns out that a graph of β vs. T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why CP and CV agree at low temperatures but diverge in the way they do at higher temperatures.Problem 2:Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep V fixed as T increases, as follows. (a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume, dV1, in terms of dT and the thermal expansion coefficient β introduced in Problem 3. (b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, dV2, in terms of dP and the isothermal compressibility κT defined as...(This is the reciprocal of the isothermal bulk modulus defined in Problem 4.)(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dP to dT is equal to ... since there is no net change in volume. Express this partial derivative in terms of β and κT. Then express it more abstractly in terms of the partial derivatives used to define β and κT. For the second expression you should obtain...This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.(d) Compute β, κT and ... for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).(e) For water at 25°C, β = 2.57 × 10−4 K−1 and κT = 4.52 × 10−10 Pa−1. Suppose you increase the temperature of some water from 20°C to 30°C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at 25°C) β = 1.81 × 10−4 K−1 and κT = 4.04 × 10−11 Pa−l. Given the choice, would you rather measure the heat capacities of these substances at constant V or at constant P?Problem 3:When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:...(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.Problem 4:By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by...where ρ is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness. More precisely, if we imagine applying an increase in pressure ΔP to a chunk of the material, and this increase results in a (negative) change in volume ΔV, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:...This definition is still ambiguous, however, because I haven’t said whether the compression is to take place isothermally or adiabatically (or in some other way).(a) Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.(b) Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.(c) Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the rms speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.(d) When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not?Problem 5:Use the thermodynamic identity to derive the heat capacity formula...which is occasionally more convenient than the more familial’ expression in terms of U. Then derive a similar formula for CP, by first writing dH in terms of dS and dP.Problem 6:Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,...where each ∂/∂V is taken with S fixed, each ∂/∂S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U) you can evaluate the partial derivatives in parentheses to obtain...a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H, F, and G). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you’ve done four of them the novelty begins to wear off.Figure 1: Measured heat capacities at constant pressure (data points) forone mole each of three different elemental solids. The solid curves show the heatcapacity at constant volume predicted by the model used in Section 7.5, with thehorizontal scale chosen to best fit the data for each substance. At sufficiently hightemperatures, CV for each material approaches the value 3R predicted by theequipartition theorem. The discrepancies between the data and the solid curvesat high T are mostly due to the differences between CP and CV. At T = 0 alldegrees of freedom are frozen out, so both CP and CV go to zero. Data from Y. S.Touloukian, ed., Thermophysical Properties of Matter (Plenum, New York, 1970).... Get solution
16. A formula analogous to that for CP – CV relates the isothermal and isentropic compressibilities of a material:... (Here κS = – (1/V)(∂V/∂P)S is the reciprocal of the adiabatic bulk modulus considered in Problem.) Derive this formula. Also check that it is true for an ideal gas.Problem:By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by...where ρ is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness. More precisely, if we imagine applying an increase in pressure ΔP to a chunk of the material, and this increase results in a (negative) change in volume ΔV, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:...This definition is still ambiguous, however, because I haven’t said whether the compression is to take place isothermally or adiabatically (or in some other way).(a) Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.(b) Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.(c) Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the rms speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.(d) When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not? Get solution
17. The enthalpy and Gibbs free energy, as defined in this section, give Special treatment to mechanical (compression-expansion) work, – P dV. Analogous quantities can be defined for other kinds of work, for instance, magnetic work. Consider the situation shown in below Figure, where a long solenoid (N turns, total length L) surrounds a magnetic specimen (perhaps a paramagnetic solid). If the magnetic field inside the specimen is ... and its total magnetic moment is ..., then we define an auxiliary field ... (often called simply the magnetic field) by the relation...where μo is the “permeability of free space,” 4π × 10–7 N/A2. Assuming cylindrical symmetry, all vectors must point either left of right, so we can drop the → symbols and agree that rightward is positive, leftward negative. From Ampere’s law, one can also show that when the current in the wire is I, the ... field inside the solenoidis NI/L, whether or not the specimen is present. (a) Imagine making an infinitesimal change in the current in the wire, resulting in infinitesimal changes in B, M, and .... Use Faraday’s law to show that the work required (from the power supply) to accomplish this change is Wtotal = V...dB. (Neglect the resistance of the wire.)(b) Rewrite the result of part (a) in terms of ... and M, then subtract off the work that would be required even if the specimen wore not present. If we define W, the work done on the system to be what’s left, show that W = μo...dM.(c) What is the thermodynamic identity for this system? (Include magnetic work but not, mechanical work or particle flow.)(d) How would you define analogues of the enthalpy and Gibbs free energy for a magnetic system? (The Helmholtz free energy is defined in the same way as for a mechanical system.) Derive the thermodynamic identities for each of these quantities, and discuss their interpretations.Figure: A long solenoid, surrounding a magnetic specimen, connected to a power supply that can change the current, performing magnetic work.... Get solution
18. Imagine that you drop a brick on the ground and it lands with a thud. Apparently the energy of this system tends to spontaneously decrease. Explain why. Get solution
19. In the previous section I derived the formula (∂F/∂V)T = –P. Explain why this formula makes intuitive sense, by discussing graphs of F vs. V with different slopes. Get solution
20. The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = k ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.) Get solution
21. Is heat capacity (C) extensive or intensive? What about specific heat (c)? Explain briefly. Get solution
22. Show that below equation is in agreement with the explicit formula for the chemical potential of a monatomic ideal gas derived in Section 3.5. Show how to calculate μ° for a monatomic ideal gas.Equation:μ(T, P) = μ°(T) + kT ln(P/P°). Get solution
24. Go through the arithmetic to verify that diamond becomes more stable than graphite at approximately 15 kbar. Get solution
25. In working high-pressure geochemistry problems it is usually more convenient to express volumes in units of kJ/kbar. Work out the conversion factor between this unit and m3. Get solution
26. How can diamond ever be more stable than graphite, when it has less entropy? Explain how at high pressures the conversion of graphite to diamond can increase the total entropy of the carbon plus its environment. Get solution
27. Graphite is more compressible than diamond.(a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text?(b) The isothermal compressibility of graphite is about 3 × 10–6 bar–1 , while that of diamond is more than ten times less and hence negligible in comparison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem 1.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).Problem 1:Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep V fixed as T increases, as follows. (a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume, dV1, in terms of dT and the thermal expansion coefficient β introduced in Problem 2.(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, dV2, in terms of dP and the isothermal compressibility κT defined as...(This is the reciprocal of the isothermal bulk modulus defined in Problem 3.)(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dP to dT is equal to ... since there is no net change in volume. Express this partial derivative in terms of β and κT. Then express it more abstractly in terms of the partial derivatives used to define β and κT. For the second expression you should obtain...This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.(d) Compute β, κT and ... for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).(e) For water at 25°C, β = 2.57 × 10−4 K−1 and κT = 4.52 × 10−10 Pa−1. Suppose you increase the temperature of some water from 20°C to 30°C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at 25°C) β = 1.81 × 10−4 K−1 and κT = 4.04 × 10−11 Pa−l. Given the choice, would you rather measure the heat capacities of these substances at constant V or at constant P?Problem 2:When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:...(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.Problem 3:By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by...where ρ is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness. More precisely, if we imagine applying an increase in pressure ΔP to a chunk of the material, and this increase results in a (negative) change in volume ΔV, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:...This definition is still ambiguous, however, because I haven’t said whether the compression is to take place isothermally or adiabatically (or in some other way).(a) Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.(b) Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.(c) Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the rms speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.(d) When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not? Get solution
28. Calcium carbonate, CaCO3, has two common crystalline forms, calcite and aragonite. Thermodynamic data for these phases can be found at the back of this book.(a) Which is stable at earth’s surface, calcite or aragonite?(b) Calculate the pressure (still at room temperature) at which the other phase should become stable. Get solution
29. Aluminum silicate. Al2SiO5, has three different crystalline forms: kyanite, andalusite, and sillimanite. Because each is stable under a different, set of temperature-pressure conditions, and all are commonly found in metamorphic rocks, these minerals are important indicators of the geologic history of rock bodies.(a) Referring to the thermodynamic data at the back of this book, argue that at 298 K the stable phase should be kyanite, regardless of pressure.(b) Now consider what happens at fixed pressure as we vary the temperature. Let ΔG be the difference in Gibbs free energies between any two phases, and similarly for ΔS. Show that the T dependence of ΔG is given by...Although the entropy of any given phase will increase significantly as the temperature increases, above room temperature it is often a good approximation to take ΔS. the difference in entropies between two phases, to be independent of T. This is because the temperature dependence of S is a function of the heat capacity (as we saw in Chapter 3), and the heat capacity of a solid at high temperature depends, to a good approximation, only on the number of atoms it contains.(c) Taking ΔS to be independent of T, find the range of temperatures over which kyanite, andalusite, and sillimanite should be stable (at 1 bar).(d) Referring to the room-temperature heat capacities of the three forms of Al2SiO5, discuss the accuracy the approximation ΔS = constant. Get solution
30. Sketch qualitatively accurate graphs of G vs. T for the three phases of H2O (ice, water, and steam) at atmospheric pressure. Put all three graphs on the same set of axes, and label the temperatures 0°C and 100°C. How would the graphs differ at a pressure of 0.001 bar? Get solution
31. Sketch qualitatively accurate graphs of G vs. P for the three phases of H2O (ice, water, and steam) at 0°C. Put all three graphs on the same set of axes, and label the point corresponding to atmospheric pressure. How would the graphs differ at slightly higher temperatures? Get solution
32. The density of ice is 917 kg/m3.(a) Use the Clausius-Clapeyron relation to explain why the slope of the phase boundary between water and ice is negative.(b) How much pressure would you have to put on an ice cube to make it melt at –1°C?(c) Approximately how deep under a glacier would you have to be before the weight of the ice above gives the pressure you found in part (b)? (Note that the pressure can be greater at some locations, as where the glacier flows over a protruding rock.)(d) Make a rough estimate of the pressure under the blade of an ice skate, and calculate the melting temperature of ice at this pressure. Some authors have claimed that skaters glide with very little friction because the increased pressure under the blade melts the ice to create a thin layer of water. What do you think of this explanation? Get solution
33. An inventor proposes to make a heat engine using water/ice as the working substance, taking advantage of the fact that water expands as it freezes. A weight to be lifted is placed on top of a piston over a cylinder of water at 1°C. The system is then placed in thermal contact with a low-temperature reservoir at –1°C until the water freezes into ice, lifting the weight. The weight is then removed and the ice is melted by putting it in contact with a high-temperature reservoir at 1°C. The inventor is pleased with this device because it can seemingly perform an unlimited amount of work while absorbing only a finite amount of heat. Explain the flaw in the inventor’s reasoning, and use the Clausius-Clapeyron relation to prove that the maximum efficiency of this engine is still given by the Carnot formula, 1 – Tc/Th. Get solution
34. Below 0.3 K the slope of the 3He solid-liquid phase boundary is negative (see Figure 5.13).(a) Which phase, solid or liquid, is more dense? Which phase has more entropy (per mole)? Explain your reasoning carefully.(b) Use the third law of thermodynamics to argue that the slope of the phase boundary must go to zero at T = 0. (Note that the 4He solid-liquid phase boundary is essentially horizontal below 1 K.)(c) Suppose that you compress liquid 3He adiabatically until it becomes a solid. If the temperature just before the phase change is 0.1 K, will the temperature after the phase change be higher or lower? Explain your reasoning carefully. Get solution
35. The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔV depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔV to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:... Get solution
36. Effect of altitude on boiling water.(a) Use the result of the previous problem 1 and the data in below Figure to plot a graph of the vapor pressure of water between 50°C and 100°C. How well can you match the data at the two endpoints?(b) Reading the graph backwards, estimate the boiling temperature of water at each of the locations for which you determined the pressure in Problem 2. Explain why it takes longer to cook noodles when you’re camping in the mountains.(c) Show that the dependence of boiling temperature on altitude is very nearly (though not exactly) a linear function, and calculate the slope in degrees Celsius per thousand feet (or in degrees Celsius per kilometer).Figure: Phase diagram for H2O (not to scale). The table gives the vaporpressure and molar latent heat for the solid-gas transformation (first three entries)and the liquid-gas transformation (remaining entries). Data from Keenan et al.(1978) and Lide (1994)....Problem 1:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔV depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔV to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Problem 2:The exponential atmosphere. (a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest, the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air.(b) Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass m of the air molecules. (The information needed to calculate m is given in Problem.) Show, then, that the pressure obeys the differential equation...called the barometric equation.(c) Assuming that the temperature of the atmosphere is independent of height (not a great assumption but not terrible either), solve the barometric equation to obtain the pressure as a function of height: P(z) = P(0)e−mgz/kT. Show also that the density obeys a similar equation.(d) Estimate the pressure, in atmospheres, at the following locations: Ogden, Utah (4700 ft or 1430 m above sea level); Leadville, Colorado (10,150 ft, 3090 m) ; Mt. Whitney, California (14,500 ft, 4420 m); Mt. Everest, Nepal/Tibet (29,000 ft, 8850 m). (Assume that the pressure at sea level is 1 atm.)Problem 3:Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), O2 (21%), and argon (1%). Get solution
37. Use the data at the back of this book to calculate the slope of the calcite-aragonite phase boundary (at 298 K). You located one point on this phase boundary in below Problem; use this information to sketch the phase diagram of calcium carbonate.Problem:Calcium carbonate, CaCO3, has two common crystalline forms, calcite and aragonite. Thermodynamic data for these phases can be found at the back of this book.(a) Which is stable at earth’s surface, calcite or aragonite?(b) Calculate the pressure (still at room temperature) at which the other phase should become stable. Get solution
38. In Problems 1 and 2, you calculated the entropies of diamond and graphite at 500 K. Use these values to predict the slope of the graphite-diamond phase boundary at 500 K, and compare to below Figure 1. Why is the slope almost constant at still higher temperatures? Why is the slope zero at T = 0?Figure 1: The experimentalphase diagram of carbon.The stability region of the gasphase is not visible on this scale;the graphite-liquid-gas triplepoint is at the bottom of thegraphite-liquid phase boundary,at 110 bars pressure. FromDavid A. Young, Phase Diagramsof the Elements (Universityof California Press, Berkeley,1991)....Problem 1:As shown in below Figure 2, the heat capacity of diamond near room temperature is approximately linear in T. Extrapolate this function up to 500 K, and estimate the change in entropy of a mole of diamond as its temperature is raised from 298 K to 500 K. Add on the tabulated value at 298 K (from the backof this book) to obtain S(500 K).Figure 2: Measured heat capacities at constant pressure (data points) forone mole each of three different elemental solids. The solid curves show the heatcapacity at constant volume predicted by the model used in Section 7.5, with thehorizontal scale chosen to best fit the data for each substance. At sufficiently hightemperatures, CV for each material approaches the value 3R predicted by theequipartition theorem. The discrepancies between the data and the solid curvesat high T are mostly due to the differences between CP and CV. At T = 0 alldegrees of freedom are frozen out, so both CP and CV go to zero. Data from Y. S.Touloukian, ed., Thermophysical Properties of Matter (Plenum, New York, 1970)....Problem 2:Experimental measurements of heat capacities are often represented in reference works as empirical formulas. For graphite, a formula that works well over a fairly wide range of temperatures is (for one mole)...where a = 16.86 J/K, b = 4.77 × 10–3 J/K2, and c = 8.54 × 105 J∙K. Suppose, then, that a mole of graphite is heated at constant pressure from 298 K to 500 K. Calculate the increase in its entropy during this process. Add on the tabulated value of S(298 K) (from the back of this book) to obtain S(500 K). Get solution
39. Consider again the aluminosilicate system treated in below Problem. Calculate the slopes of all three phase boundaries for this system: kyanite-andalusitc, kyanite-sillimanite, and andalusite-sillimanite. Sketch the phase diagram, and calculate the temperature and pressure of the triple point.Problem:Aluminum silicate. Al2SiO5, has three different crystalline forms: kyanite, andalusite, and sillimanite. Because each is stable under a different, set of temperature-pressure conditions, and all are commonly found in metamorphic rocks, these minerals are important indicators of the geologic history of rock bodies.(a) Referring to the thermodynamic data at the back of this book, argue that at 298 K the stable phase should be kyanite, regardless of pressure.(b) Now consider what happens at fixed pressure as we vary the temperature. Let ΔG be the difference in Gibbs free energies between any two phases, and similarly for ΔS. Show that the T dependence of ΔG is given by...Although the entropy of any given phase will increase significantly as the temperature increases, above room temperature it is often a good approximation to take ΔS. the difference in entropies between two phases, to be independent of T. This is because the temperature dependence of S is a function of the heat capacity (as we saw in Chapter 3), and the heat capacity of a solid at high temperature depends, to a good approximation, only on the number of atoms it contains.(c) Taking ΔS to be independent of T, find the range of temperatures over which kyanite, andalusite, and sillimanite should be stable (at 1 bar).(d) Referring to the room-temperature heat capacities of the three forms of Al2SiO5, discuss the accuracy the approximation ΔS = constant. Get solution
40. The methods of this section can also be applied to reactions in which one set of solids converts to another. A geologically important example is the transformation of albite into jadeite + quartz:...Use the data at the back of this book to determine the temperatures and pressures under which a combination of jadeite and quartz is more stable than albite. Sketch the phase diagram of this system. For simplicity, neglect the temperature and pressure dependence of both ΔS and ΔV. Get solution
41. Suppose you have a liquid (say, water) in equilibrium with its gas phase, inside some closed container. You then pump in an inert gas (say, air), thus raising the pressure exerted on the liquid. What happens?(a) For the liquid to remain in diffusive equilibrium with its gas phase, the chemical potentials of each must change by the same amount: dμl = dμg. Use this fact and below equation to derive a differential equation for the equilibrium vapor pressure, Pυ, as a function of the total pressure P. (Treat the gases as ideal, and assume that none of the inert gas dissolves in the liquid.)(b) Solve the differential equation to obtain...where the ratio V/N in the exponent is that of the liquid. (The quantity ...) is just the vapour pressure in the absence of the inert gas.) Thus, the presence of the inert gas leads to a slight increase in the vapour pressure: It causes more of the liquid to evaporate.(c) Calculate the percent increase in vapour pressure when air at atmospheric pressure is added to a system of water and water vapour in equilibrium at 25°C. Argue more generally that the increase in vapour pressure due to the presence of an inert gas will be negligible except under extreme conditions.Equation... Get solution
42. Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour.(a) Use the vapor pressure equation (below Problem) and the data in below Figure to plot a graph of the vapor pressure of water from 0°C to 40° C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature.(b) The temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?Problem:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔU depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔU to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Figure: Phase diagram for H2O (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries). Data from Keenan et al (1978) and Lide (1994).... Get solution
43. Assume that the air you exhale is at 35°C, with a relative humidity of 90%. This air immediately mixes with environmental air at 10°C and unknown relative humidity; during the mixing, a variety of intermediate temperatures and water vapour percentages temporarily occur, If you are able to “see your breath” due to the formation of cloud droplets during this mixing, what can you conclude about the relative humidity of your environment? (Refer to the vapour pressure graph drawn below Problem 1.)Problem 1:Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour. (a) Use the vapor pressure equation (below Problem 2) and the data in below Figure to plot a graph of the vapor pressure of water from 0°C to 40° C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature. (b) The temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?Problem 2:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔU depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔU to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Figure: Phase diagram for H2O (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries). Data from Keenan et al (1978) and Lide (1994).... Get solution
44. Suppose that an unsaturated air mass is rising and cooling at the dry adiabatic lapse rate found in Problem 1. If the temperature at ground level is 25° C and the relative humidity there is 50%, at what altitude will this air mass become saturated so that condensation begins and a cloud forms (see below Figure 1)? (Refer to the vapour pressure graph drawn shown below Problem 5.)Figure 1: Cumulus clouds form when rising air expands adiabaticallyand cools to the dew point the onset of condensation slowsthe cooling, increasing the tendency of the air to rise further.These clouds began to form in late morning, in a sky that was clear onlyan hour before the photo was taken. By mid-afternoon they had developedinto thunderstorms....Problem 1:In Problem 2 you calculated the pressure of earth’s atmosphere as a function of altitude, assuming constant temperature. Ordinarily, however, the temperature of the bottommost 10–15 km of the atmosphere (called the troposphere) decreases with increasing altitude, due to heating from the ground (which is warmed by sunlight). If the temperature gradient |dT/dz| exceeds a certain critical value, convection will occur: Warm, low-density air will rise, while cool, high-density air sinks. The decrease of pressure with altitude causes a rising air mass to expand adiabatically and thus to cool. The condition for convection to occur is that the rising air mass must remain warmer than the surrounding air despite this adiabatic cooling.(a) Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation...(b) Assume that dT/dz is just at the critical value for convection to begin, so that the vertical forces on a convecting air mass are always approximately in balance. Use the result of Problem 2 (b) to find a formula for dT/dz in this case. The result should be a constant, independent of temperature and pressure, which evaluates to approximately - 10°C/km. This fundamental meteorological quantity is known as the dry adiabatic lapse rate.Problem 2:The exponential atmosphere.(a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest , the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air.(b) Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass m of the air molecules. (The information needed to calculate m is given in Problem 3.) Show, then, that the pressure obeys the differential equation...called the barometric equation.(c) Assuming that the temperature of the atmosphere is independent of height (not a great assumption but not terrible either), solve the barometric equation to obtain the pressure as a function of height: P(z) = P(O)e−mgz/kT. Show also that the density obeys a similar equation.(d) Estimate the pressure, in atmospheres, at the following locations: Ogden, Utah (4700 ft or 1430 m above sea level); Leadville, Colorado (10,150 ft , 3090 m) ; Mt. Whitney, California (14,500 ft, 4420 m); Mt. Everest, Nepal/ Tibet (29,000 ft, 8850 m). (Assume that the pressure at sea level is 1 atm.)Problem 3:Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), 02 (21%), and argon (1%).Problem 4:Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour.(a) Use the vapor pressure equation (below Problem 5) and the data in below Figure 2 to plot a graph of the vapor pressure of water from 0°C to 40° C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature.(b) The temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?Problem 5:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔU depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔU to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Figure 2: Phase diagram for H2O (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries). Data from Keenan et al (1978) and Lide (1994).... Get solution
45. In Problem 1 you calculated the atmospheric temperature gradient required for unsaturated air to spontaneously undergo convection. When a rising air mass becomes saturated, however, the condensing water droplets will give up energy, thus slowing the adiabatic cooling process(a) Use the first, law of thermodynamics to show that, as condensation forms during adiabatic expansion, the temperature of an air mass changes by...where nw is the number of moles of water vapour present, L is the latent heat of vaporization per mole, and I’ve set γ = 7/5 for air. You may assume that the H2O makes up only a small fraction of the air mass.(b) Assuming that the air is always saturated during this process, the ratio nw/n is a known function of temperature and pressure, Carefully express dnw/dz in terms of dT/dz, dP/dz, and the vapour pressure Pv(T). Use the Clausius-Clapeyron relation to eliminate dPv/dT.(c) Combine the results of parts (a) and (b) to obtain a formula relating the temperature gradient, dT/dz, to the pressure gradient, dP/dz. Eliminate the latter using the “barometric equation” from Problem 1.16. You should finally obtain....where M is the mass of a mole of air. The prefactor is just the dry adiabatic lapse rate calculated in Problem 1.40, while the rest of the expression gives the correction due to heating from the condensing water vapor. The whole result is called the wet adiabatic lapse rate; it is the critical temperature gradient above which saturated air will spontaneously convect.(d) Calculate the wet adiabatic lapse rate at atmospheric pressure (1 bar) and 25°C, then at atmospheric pressure and 0°C. Explain why the results are different, and discuss their implications. What happens at higher altitudes, where the pressure is lower?Problem 1:In Problem you calculated the pressure of earth’s atmosphere as a function of altitude, assuming constant temperature. Ordinarily, however, the temperature of the bottommost 10–15 km of the atmosphere (called the troposphere) decreases with increasing altitude, due to heating from the ground (which is warmed by sunlight). If the temperature gradient |dT/dz| exceeds a certain critical value, convection will occur: Warm, low-density air will rise, while cool, high-density air sinks. The decrease of pressure with altitude causes a rising air mass to expand adiabatically and thus to cool. The condition for convection to occur is that the rising air mass must remain warmer than the surrounding air despite this adiabatic cooling.(a) Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation...(b) Assume that dT/dz is just at the critical value for convection to begin, so that the vertical forces on a convecting air mass are always approximately in balance. Use the result of Problem 2 (b) to find a formula for dT/dz in this case. The result should be a constant, independent of temperature and pressure, which evaluates to approximately - 10°C/km. This fundamental meteorological quantity is known as the dry adiabatic lapse rate.Problem 2:The exponential atmosphere.(a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest , the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air.(b) Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass m of the air molecules. (The information needed to calculate m is given in Problem 3.) Show, then, that the pressure obeys the differential equation...called the barometric equation.(c) Assuming that the temperature of the atmosphere is independent of height (not a great assumption but not terrible either), solve the barometric equation to obtain the pressure as a function of height: P(z) = P(O)e−mgz/kT. Show also that the density obeys a similar equation.(d) Estimate the pressure, in atmospheres, at the following locations: Ogden, Utah (4700 ft or 1430 m above sea level); Leadville, Colorado (10,150 ft , 3090 m) ; Mt. Whitney, California (14,500 ft, 4420 m); Mt. Everest, Nepal/ Tibet (29,000 ft, 8850 m). (Assume that the pressure at sea level is 1 atm.)Problem 3: Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), 02 (21%), and argon (1%). Get solution
46. Everything in this section so far has ignored the boundary between two phases, as if each molecule were unequivocally part of one phase or the other. In fact, the boundary is a kind of transition zone where molecules are in an environment that differs from both phases. Since the boundary zone is only a few molecules thick, its contribution to the total free energy of a system is very often negligible. One important exception, however, is the first tiny droplets or bubbles or grains that form as a material begins to undergo a phase transformation. The formation of these initial specks of a new phase is called nucleation. In this problem we will consider the nucleation of water droplets in a cloud.The surface forming the boundary between any two given phases generally has a fixed thickness, regardless of its area. The additional Gibbs free energy of this surface is therefore directly proportional to its area; the constant of proportionality is called the surface tension, σ:...If you have a blob of liquid in equilibrium with its vapor and you wish to stretch it into a shape that has the same volume but more surface area, then σ is the minimum work that you must perform, per unit of additional area, at fixed temperature and pressure. For water at 20°C, σ = 0.073 J/m2. a) Consider a spherical droplet of water containing Nl; molecules, surrounded by N – Nl molecules of water vapour. Neglecting surface tension for the moment, write down a formula for the total Gibbs free energy of this system in terms of N, Nl, and the chemical potentials of the liquid and vapour. Rewrite Nl in terms of vt, the volume per molecule in the liquid, and r, the radius of the droplet. b) Now add to your expression for G a term to represent the surface tension, written in terms of r and σ.c) Sketch a qualitative graph of G vs. r for both signs of μg–μl, and discuss the implications. For which sign of μg–μl does there exist a nonzero equilibrium radius? Is this equilibrium stable?d) Let rc represent the critical equilibrium radius that you discussed qualitatively in part (c). Find an expression for rc in terms of μg–μl. Then rewrite the difference of chemical potentials in terms of the relative humidity (see below Problem 1), assuming that the vapuor behaves as an ideal gas. (The relative humidity is defined in terms of equilibrium of a vapour with a flat surface, or with an infinitely large droplet.) Sketch a graph of the critical radius as a function of the relative humidity, including numbers. Discuss the implications. In particular, explain why it is unlikely that the clouds in our atmosphere would form by spontaneous aggregation of water molecules into droplets. (In fact, cloud droplets form around nuclei of dust particles and other foreign material, when the relative humidity is close to 100%.)Problem 1:Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour. (a) Use the vapor pressure equation (below Problem 2) and the data in below Figure to plot a graph of the vapor pressure of water from 0°C to 40° C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature.(b) The temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?Problem 2:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔU depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔU to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Figure: Phase diagram for H2O (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries). Data from Keenan et al (1978) and Lide (1994).... Get solution
47. For a magnetic system held at constant T and H (see Problem 1), the quantity that is minimized is the magnetic analogue of the Gibbs free energy, which obeys the thermodynamic identity...Phase diagrams for two magnetic systems are shown in Figure 5.14; the vertical axis on each of these figures is μ0H.(a) Derive an analogue of the Clausius-Clapeyron relation for the slope of a phase boundary in the H-T plane. Write your equation in terms of the difference in entropy between the two phases.(b) Discuss the application of your equation to the ferromagnet phase diagram in Figure 5.14.(c) In a type-I superconductor, surface currents flow in such a way as to completely cancel the magnetic field (B, not H) inside. Assuming that M is negligible when the material is in its normal (non-superconducting) state, discuss the application of your equation to the superconductor phase diagram in Figure 5.14. Which phase has the greater entropy? What happens to the difference in entropy between the phases at each end of the phase boundary?Problem 1:The enthalpy and Gibbs free energy, as defined in this section, give Special treatment to mechanical (compression-expansion) work, – P dV. Analogous quantities can be defined for other kinds of work, for instance, magnetic work. Consider the situation shown in below Figure, where a long solenoid [N turns, total length L) surrounds a magnetic specimen (perhaps a paramagnetic solid). If the magnetic field inside the specimen is ... and its total magnetic moment is ..., then we define an auxiliary field ... (often called simply the magnetic field) by the relation...where μo is the “permeability of free space,” 4π × 10–7 N/A2. Assuming cylindrical symmetry, all vectors must point either left of right, so we can drop the symbols and agree that rightward is positive, leftward negative. From Ampere’s law, one can also show that when the current in the wire is I, the H field inside the solenoidis NI/L, whether or not the specimen is present.(a) Imagine making an infinitesimal change in the current in the wire, resulting in infinitesimal changes in B, M, and H. Use Faraday’s law to show that the work required (from the power supply) to accomplish this change is Wtotal = VHdB. (Neglect the resistance of the wire.)(b) Rewrite the result of part (a) in terms of H and M, then subtract off the work that would be required even if the specimen wore not present. If we define W, the work done on the system J to be what’s left, show that W = μoHdM.(c) What is the thermodynamic identity for this system? (Include magnetic work but not, mechanical work or particle flow.)(d) How would you define analogues of the enthalpy and Gibbs free energy for a magnetic system? (The Helmholtz free energy is defined in the same way as for a mechanical system.) Derive the thermodynamic identities for each of these quantities, and discuss their interpretations.Figure 1: A long solenoid, surrounding a magnetic specimen, connected to a power supply that can change the current, performing magnetic work.... Get solution
48. As you can see from Figure 5.20, the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives of P with respect to V (at fixed T) are zero. Use this fact to show that... Get solution
49. Use the result of the previous problem and the approximate values of a and b given in the text to estimate Tc, Pc, and Vc/N for N2, H2O, and He. (Tabulated values of a and b are often determined by working backward from the measured critical temperature and pressure.)Problem:As you can see from Figure 5.20, the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives of P with respect to V (at fixed T) are zero. Use this fact to show that... Get solution
51. When plotting graphs and performing numerical calculations, it is convenient to work in terms of reduced variables,...Rewrite the van der Waals equation in terms of these variables, and notice that the constants a and b disappear. Get solution
52. Plot the van der Waals isotherm for T/Tc = 0.95, working in terms of reduced variables. Perform the Maxwell construction (either graphically or numerically) to obtain the vapour pressure. Then plot the Gibbs free energy (in units of NkTc) as a function of pressure for this same temperature and check that this graph predicts the same value for the vapor pressure. Get solution
53. Repeat the below problem for T/Tc = 0.8.Problem:Plot the van der Waals isotherm for T/Tc = 0.95, working in terms of reduced variables. Perform the Maxwell construction (either graphically or numerically) to obtain the vapour pressure. Then plot the Gibbs free energy (in units of NkTc) as a function of pressure for this same temperature and check that this graph predicts the same value for the vapor pressure. Get solution
54. Calculate the Helmholtz free energy of a van der Waals fluid, up to an undetermined function of temperature as in equation 5.56. Using reduced variables, carefully plot the Helmholtz free energy (in units of NkTc) as a function of volume for T/Tc = 0.8. Identify the two points on the graph corresponding to the liquid and gas at the vapor pressure. (If you haven’t worked the preceding problem, just read the appropriate values off Figure) Then prove that the Helmholtz free energy of a combination of these two states (part liquid, part gas) can be represented by a straight line connecting these two points on the graph. Explain why the combination is more stable, at a given volume, than the homogeneous state represented by the original curve, and describe how you could have determined the two transition volumes directly from the graph of F.Figure: ... Get solution
55. In this problem you will investigate the behavior of a van der Waals fluid near the critical point. It is easiest to work in terms of reduced variables throughout.(a) Expand the van der Waals equation in a Taylor series in (V − Vc), keeping terms through order (V − Vc)3. Argue that, for T sufficiently close to Tc, the term quadratic in (V − Vc) becomes negligible compared to the others and may be dropped.(b) The resulting expression for P(V) is antisymmetric about the point V = Vc. Use this fact to And an approximate formula for the vapour pressure as a function of temperature. (You may find it helpful to plot the isotherm.) Evaluate the slope of the phase boundary, dP/dT, at the critical point.(c) Still working in the same limit, find an expression for the difference in volume between the gas and liquid phases at the vapor pressure. You should find (Vg − Vl) α (Tc − T)β where β is known as critical exponent, Experiments show that β has a universal value of about 1/3, but the van der Waals model predicts a larger value.(d) Use the previous result to calculate the predicted latent heat of the transformation as a function of temperature, and sketch this function. (e) The shape of the T = TC isotherm defines another critical exponent, called δ: (P − PC) α (V − Vc)δ Calculate δ in the van der Waals model. (Experimental values of δ are typically around 4 or 5.)(f) A third critical exponent describes the temperature dependence of the isothermal compressibility,...This quantity diverges at the critical point, in proportion to a power of (T − Tc) that in principle could differ depending on whether one approaches the critical point from above or below. Therefore the critical exponents γ and γ′ are defined by the relations...Calculate κ on both sides of the critical point in the van der Waals model, and show that γ = γ′ in this model. Get solution
56. Prove that the entropy of mixing of an ideal mixture has an infinite slope, when plotted vs. x, at x = 0 and x = 1. Get solution
57. Consider an ideal mixture of just 100 molecules, varying in composition from pure A to pure B. Use a computer to calculate the mixing entropy as a function of NA and plot this function (in units of k). Suppose you start with all A and then convert one molecule to type B; by how much does the entropy Increase? By how much does the entropy increase when you convert a second molecule, and then a third, from A to B? Discuss. Get solution
58. In this problem you will model the mixing energy of a mixture in a relatively simple way, in order to relate the existence of a solubility gap to molecular behaviour. Consider a mixture of A and B molecules that is ideal in every way but one: The potential energy due to the interaction of neighbouring molecules depends upon whether the molecules are like or unlike. Let n be the average number of nearest neighbours of any given molecule (perhaps 6 or 8 or 10). Let μ0 be the average potential energy associated with the interaction between neighbouring molecules that are the same (A−A or B−B), and let uAB the potential energy associated with the interaction of a neighbouring’ unlike pair (A−B). There are no interactions beyond the range of the nearest neighbors; the values of u0 and uAB are independent, of the amounts of A and B; and the entropy of mixing is the same as for an ideal solution.a) Show that when the system is unmixed, the total potential energy due to all neighbour-neighbour interactions is ½Nnu0. (Hint: Be sure to count each neighbouring pair only once.)b) Find a formula for the total potential energy when the system is mixed, in terms of x, the fraction of B. (Assume that the mixing is totally random.)c) Subtract the results or parts (a) and (b) to obtain the change in energy upon mixing. Simplify the resuit as much as possible; you should obtain an expression proportional to x(1 − x). Sketch this function vs x, for both possible signs of uAB − u0.d) Show that the slope of the mixing energy function is finite at both endpoints, unlike the slope of the mixing entropy function.e) For the case uAB > u0. plot a graph of the Gibbs free energy of this system vs. x at. several temperatures. Discuss the implications. f) Find an expression for the maximum temperature at which this system has a solubility gap.g) Make a very rough estimate of uAB − u0 for a liquid mixture that has a solubility gap below 100oC.h) Plot, the phase diagram (T vs. x) for this system. Get solution
59. Suppose you cool a mixture of 50% nitrogen and 50% oxygen until it liquefies. Describe the cooling sequence in detail, including the temperatures and compositions at which liquefaction begins and ends. Get solution
60. Suppose you start with a liquid mixture of 60% nitrogen and 40% oxygen. Describe what happens as the temperature of this mixture increases. Be sure to give the temperatures and compositions at which boiling begins and ends. Get solution
61. Suppose you need a tank of oxygen that is 95% pure. Describe a process by which you could obtain such a gas, starting with air. Get solution
62. Consider a completely miscible two-component system whose overall composition is x, at a temperature where liquid and gas phases coexist. The composition of the gas phase at this temperature is xa and the composition of the liquid phase is xb. Prove the lever rule, which says that the proportion of liquid to gas is (x - xa) / (xb - x). Interpret this rule graphically on a phase diagram. Get solution
63. Everything in this section assumes that the total pressure of the system is fixed. How would you expect the nitrogen-oxygen phase diagram to change if you increase or decrease the pressure? Justify your answer. Get solution
64. Below Figure shows the phase diagram of plagioclase feldspar, which can be considered a mixture of albite (NaAISi3O8) and anorthite (CaAl2Si2O8).a) Suppose you discover a rock in which each plagioclase crystal varies in composition from center to edge, with the centers of the largest crystals composed of 70% anorthite and the outermost parts of all crystals made of essentially pure albite. Explain in some detail how this variation might arise. What was the composition of the liquid magma from which the rock formed?b) Suppose you discover another rock body in which the crystals near the top are albite-rich while the crystals near the bottom are anorthite-rich. Explain how this variation might arise.Figure: The phase diagram of plagioclase feldspar (at atmospheric pressure). From N. L. Bowen, “The Melting Phenomena of the Plagioclase Feldspars,” American Journal of Science 35, 577-599 (1913).... Get solution
65. In constructing the phase diagram from the free energy graphs in Figure, I assumed that both the liquid and the gas are ideal mixtures. Suppose instead that the liquid has a substantial positive mixing energy, so that its free energy curve, while still concave-up, is much flatter. In this case a portion of the curve may still lie above the gas’s free energy curve at TA. Draw a qualitatively accurate phase diagram for such a system, showing how you obtained the phase diagram from the free energy graphs. Show that there is a particular composition at which this gas mixture will condense with no change in composition. This special composition is called an azeotrope.Figure:... Get solution
66. Repeat the previous problem for the opposite case where the liquid has a substantial negative mixing energy, so that its free energy curve dips below the gas’s free energy curve at a temperature higher than TB. Construct the phase diagram and show that this system also has an azeotrope.Problem:In constructing the phase diagram from the free energy graphs in Figure, I assumed that both the liquid and the gas are ideal mixtures. Suppose instead that the liquid has a substantial positive mixing energy, so that its free energy curve, while still concave-up, is much flatter. In this case a portion of the curve may still lie above the gas’s free energy curve at TA. Draw a qualitatively accurate phase diagram for such a system, showing how you obtained the phase diagram from the free energy graphs. Show that there is a particular composition at which this gas mixture will condense with no change in composition. This special composition is called an azeotrope.Figure... Get solution
67. In this problem you will derive approximate formulas for the shapes of the phase boundary curves in diagrams such as Figures 5.31 and 5.32, assuming that both phases behave as ideal mixtures. For definiteness, suppose that the phases are liquid and gas.a) Show that in an ideal mixture of A and B, the chemical potential of species A can be written...where ...is the chemical potential of pure A (at the same temperature and pressure) and x = NB/(NA + NB). Derive a similar formula for the chemical potential of species B. Note that both formulas can be written for either the liquid phase or the gas phase.b) At any given temperature T, let xl and xg be the compositions of the liquid and gas phases that are in equilibrium with each other. By setting the appropriate chemical potentials equal to each other, show that xl and xg obey the equations...where ΔG° represents the change in G for the pure substance undergoing the phase change at temperature T.c) Over a limited range of temperatures, we can often assume that the main temperature dependence of ΔG° = ΔH° — TΔS° comes from the explicit T; both ΔH° and ΔS° are approximately constant. With this simplification, rewrite the results of part (b) entirely in terms of ...TA, and TB (eliminating ΔG and ΔS). Solve for Xl and Xg as functions of T.d) Plot your results for the nitrogen-oxygcn system. The latent heats of the pure substances are ... = 5570 J/mol and ... = 6820 J/mol. Compare to the experimental diagram, below Figure.e) Show that you can account for the shape of Figure with suitably chosen ΔH° values. What are those values?Figure: Experimental phase diagram for nitrogen and oxygen at atmospheric pressure. Data from International Crilical Tables (volume 3), with endpoints adjusted to values in Lide (1994). ...Figure: The phase diagram of plagioclase feldspar (at atmospheric pressure). From N. L. Bowen, “The Melting Phenomena of the PlagioclaseFeldspars,” American Journal of Science 35, 577-599 (1913). ... Get solution
68. Plumber’s solder is composed of 67% lead and 33% tin by weight. Describe what happens to this mixture as it cools, and explain why this composition might be more suitable than the eutectic composition for joining pipes. Get solution
69. What happens when you spread salt crystals over an icy sidewalk? Why is this procedure rarely used in very cold climates? Get solution
70. What happens when you add salt to the ice bath in an ice cream maker? How is it possible for the temperature to spontaneously drop below 0°C? Explain in as much detail as you can. Get solution
71. Below Figure (left) shows the free energy curves at one particular temperature for a two-component system that has three possible solid phases (crystal structures), one of essentially pure A, one of essentially pure B, and one of intermediate composition. Draw tangent lines to determine which phases are present at which values of x. To determine qualitatively what happens at other temperatures, you can simply shift the liquid free energy curve up or down (since the entropy of the liquid is larger than that of any solid). Do so, and construct a qualitative phase diagram for this system. You should find two eutectic points. Examples of systems with this behaviour include water + ethylene glycol and tin + magnesium.Figure: Free energy diagrams... Get solution
72. Repeat the previous problem for the diagram in below Figure (right), which has an important qualitative difference. In this phase diagram, you should find that β and liquid are in equilibrium only at temperatures below the point where the liquid is in equilibrium with infinitesimal amounts of α and β. This point is called a peritectic point. Examples of systems with this behaviour include water + NaCl and leucite + quartz.Figure: Free energy diagrams... Get solution
73. If below expression is correct, it must be extensive: Increasing both NA and NB by a common factor while holding all intensive variables fixed should increase G by the same factor. Show that expression has tins property. Show that it would not have this property had we not added the term proportional to NB!.Expression:... Get solution
74. Check that below equations 1 and 2, satisfy the identity G = NAμA + NBμB (below equation 1 and 2).Expression 1:...Expression 2:...Equation 1:... Get solution
75. Compare below expression 1 for the Gibbs free energy of a dilute solution to below expression 5.61 for the Gibbs free energy of an ideal mixture. Under what circumstances should these two expressions agree? Show that they do agree under these circumstances, and identify the function f(T,P) in this case.Expression 1:... Get solution
76. Seawater has a salinity of 3.5%, meaning that if you boil away a kilogram of seawater, when you’re finished you’ll have 35 g of solids (mostly NaCl) left in the pot. When dissolved, sodium chloride dissociates into separate Na+ and Cl− ions.a) Calculate the osmotic pressure difference between seawater and fresh water. Assume for simplicity that all the dissolved salts in seawater are NaCl.b) If you apply a pressure difference greater than the osmotic pressure to a solution separated from pure solvent by a semipermeable membrane, you get reverse osmosis: a flow of solvent out of the solution. This process can be used to desalinate seawater. Calculate the minimum work required to desalinate one litre of seawater. Discuss some reasons why the actual work required would be greater than the minimum. Get solution
77. Osmotic pressure measurements can be used to determine the molecular weights of largo molecules such as proteins. For a solution of large molecules to qualify as “dilute”, its molar concentration must be very low and hence the osmotic pressure can be too small to measure accurately. For this reason, the usual procedure is to measure the osmotic pressure at a variety of concentrations, then extrapolate the results to the limit of zero concentration. Here are some data* for the protein haemoglobin dissolved in water at 3°C:Concentration(grams/liter)Δh(cm)5.02.016.66.532.512.843.417.654.022.6The quantity Δh is the equilibrium difference in fluid level between the solution and the pure solvent, as shown in below Figure. From these measurements, determine the approximate molecular weight of hemoglobin (in grams per mole) .Figure: An experimental arrangement for measuring osmotic pressure. Solvent flows across the membrane from left to right until the difference in fluid level, Δh, is just enough to supply theosmotic pressure.... Get solution
78. Because osmotic pressures can be quite large, you may wonder whether the approximation made in below equation is valid in practice: Is μ0 really a linear function of P to. the required accuracy? Answer this question by discussing whether the derivative of this function changes significantly, over therelevant pressure range, in realistic examples.Equation:... Get solution
79. Most pasta recipes instruct you to add a teaspoon of salt to a pot of boiling water. Does this have a significant effect on the boiling temperature? Justify your answer with a rough numerical estimate. Get solution
80. Use the Clausius-Clapeyron relation to derive below equation directly from Raoult’s law. Be sure to explain the logic carefully.... Get solution
81. Derive a formula, similar to below equation, for the shift in the freezing temperature of a dilute solution. Assume that the solid phase is pure solvent, no solute. You should find that the shift is negative: The freezing temperature of a solution is less than that of the pure solvent. Explain in general terms why the shift should be negative.Equation:... Get solution
82. Use the result of the below problem to calculate the freezing temperature of seawater.Problem:Derive a formula, similar to below equation, for the shift in the freezing temperature of a dilute solution. Assume that the solid phase is pure solvent, no solute. You should find that the shift is negative: The freezing temperature of a solution is less than that of the pure solvent. Explain in general terms why the shift should be negative.Equation:... Get solution
83. Write down the equilibrium condition for each of the following reactions:a) 2H ↔ H2b) 2CO + O2 ↔ 2CO2c) CH4 + 2O2 ↔ 2H2O + CO2d) H2SO4 ↔ 2H+ + SO2−4e) 2p + 2n ↔ 4 He Get solution
84. A mixture of one part nitrogen and three parts hydrogen is heated, in the presence of a suitable catalyst, to a temperature of 500°C. What fraction of the nitrogen (atom for atom) is converted to ammonia, if the final total pressure is 400 atm? Protend for simplicity that the gases behave ideally despite the very high pressure. The equilibrium constant at 500°C is 6.9 × 10−5. (Hint: You’ll have to solve a quadratic equation.) Get solution
85. Derive the van’t Hoff equation,...which gives the dependence of the equilibrium constant on temperature. Here ΔH° is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if ΔH° is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of ΔH°; solve the equation in this case to obtain... Get solution
86. Use the result of the previous problem to estimate the equilibrium constant of the reaction N2 + 3H2 ↔ 2NH3 at 500°C, using only the room- temperature data at the back of this book. Compare your result to the actual value of K at 500°C quoted in the text. Get solution
87. Sulfuric acid, H2SO4, readily dissociates into H+ and HSO−4 ions:...The hydrogen sulfate ion, in turn, can dissociate again:...The equilibrium constants for these reactions, in aqueous solutions at 298 K, are approximately 102 and 10−19 respectively. (For dissociation of acids it is usually more convenient to look up K than ΔGo, By the way, the negative base-10 logarithm of K for such a reaction is called pK, in analogy to pH. So for the first reaction pK = −2, while for the second reaction pK = 1.9.)a) Argue that the first reaction tends so strongly to the right that we might as well consider it. to have gone to completion, in any solution that could possibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated?b) In industrialized regions whore lots of coal is burned, the concentration of sulfate in rainwater is typically 5 × 10−5 mol/kg. The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of .... What is the pH of this rainwater?c) Explain why you can neglect dissociation of water into H+ and OH− in answering the previous question.d) At what pH would dissolved sulfate be equally distributed between ... and ...? Get solution
88. Express ∂(ΔG°)/∂P in terms of the volumes of solutions of reactants and products, for a chemical reaction of dilute solutes. Plug in some reasonable numbers, to show that a pressure increase of 1 atm has only a negligible effect on the equilibrium constant. Get solution
89. The standard enthalpy change upon dissolving one mole of oxygen at 25°C is −11.7 kJ. Use this number and the van’t Hoff equation (below Problem) to calculate the equilibrium (Henry’s law) constant for oxygen in water at 0°C and at 100°C. Discuss the results briefly.Problem:Derive the van’t Hoff equation,...which gives the dependence of the equilibrium constant on temperature. Here ΔH° is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if ΔH° is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of ΔH°; solve the equation in this case to obtain... Get solution
90. When solid quartz “dissolves” in water, it combines with water molecules in the reaction....a) Use this data in the back of this book to compute the amount of silica dissolved in waler in equilibrium with solid quartz, at 25°C.b) Use the van’t Hoff equation (below Problem) to compute the amount of silica dissolved in water in equilibrium with solid quartz at 100°C.Problem:Derive the van’t Hoff equation,...which gives the dependence of the equilibrium constant on temperature. Here ΔH° is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if ΔH° is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of ΔH°; solve the equation in this case to obtain... Get solution
91. When carbon dioxide “dissolves” in water, essentially all of it reacts to form carbonic acid, H2CO3:CO2(g) + H2O(1) ↔ H2CO3(aq).The carbonic acid can then dissociate into H+ and bicarbonate ions,H2CO3(aq) ↔ H+(aq) + HCO−(aq).(The table at the back of this book gives thermodynamic data for both of these reactions.) Consider a body of otherwise pure water (or perhaps a raindrop) that is in equilibrium with the atmosphere near sea level, where the partial pressure of carbon dioxide is 3.4 × 10−4 bar (or 340 parts per million). Calculate the molality of carbonic acid and of bicarbonate ions in the water, and determine the pH of the solution. Note that even “natural” precipitation is somewhat acidic. Get solution
92. Suppose you have a box of atomic hydrogen, initially at room temperature and atmospheric pressure. You then raise the temperature, keeping the volume fixed.a) Find an expression for the fraction of the hydrogen that is ionized as a function of temperature. (You’ll have to solve a quadratic equation.) Check that your expression has the expected behaviour at very low and very high temperatures.b) At what temperature is exactly half of the hydrogen ionized?c) Would raising the initial pressure cause the temperature you found in part (b) to increase or decrease? Explain.d) Plot the expression you found in part (a) as a function of the dimension-less variable t = kT/I. Choose the range of t values to clearly show the interesting part of the graph. Get solution
2. Consider the production of ammonia from nitrogen and hydrogen,N2 + 3H2 → 2NH3,at 298 K and 1 bar. From the values of ΔH and S tabulated at the back of this book, compute ΔG for this reaction and chock that it is consistent with the value given in the table. Get solution
3. Use the data at the back of this book to verify the values of ΔH and ΔG quoted above for the lead-acid below reaction.Reaction:... Get solution
4. In a hydrogen fuel cell, the steps of the chemical reaction areat – electrode: H2 + 2OH– → 2H2O + 2e–;at + electrode: ½O2 + H2O + 2e– → 2OH–.Calculate the voltage of the cell. What is the minimum voltage required for electrolysis of water? Explain briefly. Get solution
5. Consider a fuel cell that uses methane (“natural gas”) as fuel. The reaction isCH4 + 2O2 → 2H2O + CO2(a) Use the data at the back of this book to determine the values of ΔH and ΔG for this reaction, for one mole of methane. Assume that the reaction takes place at room temperature and atmospheric pressure.(b) Assuming ideal performance, how much electrical work can you get out of the cell, for each mole of methane fuel?(c) How much waste heat is produced, for each mole of methane fuel?(d) The steps of this reaction areat – electrode: CH4 + 2H2O → CO2 + 8H+ + 8e–;at + electrode: 2O2 + 8H+ + 8e– → 4H2O.What is the voltage of the cell? Get solution
6. A muscle can be thought of as a fuel cell, producing work from the metabolism of glucose:C6H12O6 + 6O2 → 6CO2 + 6H2O.(a) Use the data at the back of this book to determine the values of ΔH and ΔG for this reaction, for one mole of glucose. Assume that the reaction takes place at room temperature and atmospheric pressure.(b) What is the maximum amount of work that a muscle can perform, for each mole of glucose consumed, assuming ideal operation?(c) Still assuming ideal operation, how much heat is absorbed or expelled by the chemicals during the metabolism of a mole of glucose? (Be sure to say which direction the heat flows.)(d) Use the concept of entropy to explain why the heat flows in the direction it does.(e) How would your answers to parts (b) and (c) change, if the operation of the muscle is not ideal? Get solution
7. The metabolism of a glucose molecule (see previous problem) occurs in many steps, resulting in the synthesis of 38 molecules of ATP (adenosine triphosphate) out of ADP (adenosine diphosphate) and phosphate ions. When the ATP splits back into ADP and phosphate, it liberates energy that is used in a host of important processes including protein synthesis, active transport of molecules across cell membranes, and muscle contraction. In a muscle, the reaction ATP → ADP + phosphate is catalyzed by an enzyme called myosin that is attached to a muscle filament. As the reaction takes place, the myosin molecule pulls on an adjacent filament, causing the muscle to contract. The force it exerts averages about 4 piconewtons and acts over a distance of about 11 nm. From this data and the results of the previous problem, compute the “efficiency” of a muscle, that is, the ratio of the actual work done to the maximum work that the laws of thermodynamics would allow.Problem:A muscle can be thought of as a fuel cell, producing work from the metabolism of glucose:C6H12O6 + 6O2 → 6CO2 + 6H2O.(a) Use the data at the back of this book to determine the values of ΔH and ΔG for this reaction, for one mole of glucose. Assume that the reaction takes place at room temperature and atmospheric pressure.(b) What is the maximum amount of work that a muscle can perform, for each mole of glucose consumed, assuming ideal operation?(c) Still assuming ideal operation, how much heat is absorbed or expelled by the chemicals during the metabolism of a mole of glucose? (Be sure to say which direction the heat flows.)(d) Use the concept of entropy to explain why the heat flows in the direction it does.(e) How would your answers to parts (b) and (c) change, if the operation of the muscle is not ideal? Get solution
8. Derive the thermodynamic identity for G (below equation), and from it the three partial derivative relations below.EQUATION:dG = – S dT+ V dP+ μ dNPartial Derivative Relations:... Get solution
9. Sketch a qualitatively accurate graph of G vs. T for a pure substance as it changes from solid to liquid to gas at fixed pressure. Think carefully about the slope of the graph. Mark the points of the phase transformations and discuss the features of the graph briefly. Get solution
10. Suppose you have a mole of water at 25°C and atmospheric pressure. Use the data at the back of this book to determine what happens to its Gibbs free energy if you raise the temperature to 30°C. To compensate for this change, you could increase the pressure on the water. How much pressure would be required? Get solution
11. Suppose that a hydrogen fuel cell, as described in the text, is to be operated at 75° C and atmospheric pressure. We wish to estimate the maximum electrical work done by the cell, using only the room-temperature data at the back of this book. It is convenient to first establish a zero-point for each of the three substances, H2, O2, and H2O. Let us take G for both H2 and O2 to be zero at 25° C, so that G for a mole of H2O is −237 kJ at 25°C.(a) Using these conventions, estimate the Gibbs free energy of a mole of H2 at 75°C. Repeat for O2 and H2O.(b) Using the results of part (a), calculate the maximum electrical work done by the cell at 75°C, for one mole of hydrogen fuel. Compare to the ideal performance of the cell at 25°C. Get solution
12. Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,...where each ∂/∂V is taken with S fixed, each ∂/∂S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U) you can evaluate the partial derivatives in parentheses to obtain...a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H, F, and G). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you’ve done four of them the novelty begins to wear off. For applications of these Maxwell relations, see the next four problems. Get solution
13. Use a Maxwell relation from the previous problem 1 and the third law of thermodynamics to prove that the thermal expansion coefficient β (defined in Problem 2) must be zero at T = 0.Problem 1:Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,...where each ∂/∂V is taken with S fixed, each∂/∂S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U) you can evaluate the partial derivatives in parentheses to obtain...a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H, F, and G). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you’ve done four of them the novelty begins to wear off. For applications of these Maxwell relations, see the next four problems.Problem 2:When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:...(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive. Get solution
14. The partial-derivative relations derived in Problems 1, 4, and 5, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between Cp and CV. (a) With the heat capacity expressions from Problem 4 in mind, first consider S to be a function of T and V. Expand dS in terms of the partial derivatives (∂S/∂T)V and (∂S/∂V)T. Note that one of these derivatives is related to CV.(b) To bring in CP, consider V to be a function of T and P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set dP = 0 and note that you have derived a nontrivial expression for (∂S/∂T)P. This derivative is related to CP, so you now have a formula for the difference CP – CV.(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1. Your final result should be...(d) Check that this formula gives the correct value of CP – CV for an ideal gas.(e) Use this formula to argue that CP cannot be less than CV.(f) Use the data in Problem 1 to evaluate CP – CV for water and for mercury at room temperature. By what percentage do the two heat capacities differ?(g) Figure 1 shows measured values of CP for three elemental solids, compared to predicted values of CV. It turns out that a graph of β vs. T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why CP and CV agree at low temperatures but diverge in the way they do at higher temperatures.Problem 1:Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep V fixed as T increases, as follows. (a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume, dV1, in terms of dT and the thermal expansion coefficient β introduced in Problem 2.(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, dV2, in terms of dP and the isothermal compressibility κT defined as...(This is the reciprocal of the isothermal bulk modulus defined in Problem 3.)(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dP to dT is equal to ... since there is no net change in volume. Express this partial derivative in terms of β and κT. Then express it more abstractly in terms of the partial derivatives used to define β and κT. For the second expression you should obtain...This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.(d) Compute β, κT and ... for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).(e) For water at 25°C, β = 2.57 × 10−4 K−1 and κT = 4.52 × 10−10 Pa−1. Suppose you increase the temperature of some water from 20°C to 30°C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at 25°C) β = 1.81 × 10−4 K−1 and κT = 4.04 × 10−11 Pa−l. Given the choice, would you rather measure the heat capacities of these substances at constant V or at constant P?Problem 2:When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:...(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.Problem 3:By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by...where ρ is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness. More precisely, if we imagine applying an increase in pressure ΔP to a chunk of the material, and this increase results in a (negative) change in volume ΔV, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:...This definition is still ambiguous, however, because I haven’t said whether the compression is to take place isothermally or adiabatically (or in some other way).(a) Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.(b) Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.(c) Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the rms speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.(d) When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not?Problem 4:Use the thermodynamic identity to derive the heat capacity formula...which is occasionally more convenient than the more familial’ expression in terms of U. Then derive a similar formula for CP, by first writing dH in terms of dS and dP.Problem 5:Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,...where each ∂/∂V is taken with S fixed, each ∂/∂S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U) you can evaluate the partial derivatives in parentheses to obtain...a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H, F, and G). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you’ve done four of them the novelty begins to wear off.Figure 1: Measured heat capacities at constant pressure (data points) forone mole each of three different elemental solids. The solid curves show the heatcapacity at constant volume predicted by the model used in Section 7.5, with thehorizontal scale chosen to best fit the data for each substance. At sufficiently hightemperatures, CV for each material approaches the value 3R predicted by theequipartition theorem. The discrepancies between the data and the solid curvesat high T are mostly due to the differences between CP and CV. At T = 0 alldegrees of freedom are frozen out, so both CP and CV go to zero. Data from Y. S.Touloukian, ed., Thermophysical Properties of Matter (Plenum, New York, 1970).... Get solution
15. The formula for CP – CV derived in the previous problem 1 can also be derived starting with the definitions of these quantities in terms of U and H. Do so. Most of the derivation is very similar, but at one point you need to use the relation P = – (∂F/∂V)T. Problem 1: The partial-derivative relations derived in Problems 2, 5, and 6, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between Cp and CV. (a) With the heat capacity expressions from Problem 5 in mind, first consider S to be a function of T and V. Expand dS in terms of the partial derivatives (∂S/∂T)V and (∂S/∂V)T. Note that one of these derivatives is related to CV. (b) To bring in CP, consider V to be a function of T and P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set dP = 0 and note that you have derived a nontrivial expression for (∂S/∂T)P. This derivative is related to CP, so you now have a formula for the difference CP – CV.(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 2. Your final result should be...(d) Check that this formula gives the correct value of CP – CV for an ideal gas.(e) Use this formula to argue that CP cannot be less than CV.(f) Use the data in Problem 2 to evaluate CP – CV for water and for mercury at room temperature. By what percentage do the two heat capacities differ?(g) Figure 1 shows measured values of CP for three elemental solids, compared to predicted values of CV. It turns out that a graph of β vs. T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why CP and CV agree at low temperatures but diverge in the way they do at higher temperatures.Problem 2:Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep V fixed as T increases, as follows. (a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume, dV1, in terms of dT and the thermal expansion coefficient β introduced in Problem 3. (b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, dV2, in terms of dP and the isothermal compressibility κT defined as...(This is the reciprocal of the isothermal bulk modulus defined in Problem 4.)(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dP to dT is equal to ... since there is no net change in volume. Express this partial derivative in terms of β and κT. Then express it more abstractly in terms of the partial derivatives used to define β and κT. For the second expression you should obtain...This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.(d) Compute β, κT and ... for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).(e) For water at 25°C, β = 2.57 × 10−4 K−1 and κT = 4.52 × 10−10 Pa−1. Suppose you increase the temperature of some water from 20°C to 30°C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at 25°C) β = 1.81 × 10−4 K−1 and κT = 4.04 × 10−11 Pa−l. Given the choice, would you rather measure the heat capacities of these substances at constant V or at constant P?Problem 3:When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:...(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.Problem 4:By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by...where ρ is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness. More precisely, if we imagine applying an increase in pressure ΔP to a chunk of the material, and this increase results in a (negative) change in volume ΔV, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:...This definition is still ambiguous, however, because I haven’t said whether the compression is to take place isothermally or adiabatically (or in some other way).(a) Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.(b) Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.(c) Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the rms speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.(d) When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not?Problem 5:Use the thermodynamic identity to derive the heat capacity formula...which is occasionally more convenient than the more familial’ expression in terms of U. Then derive a similar formula for CP, by first writing dH in terms of dS and dP.Problem 6:Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,...where each ∂/∂V is taken with S fixed, each ∂/∂S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U) you can evaluate the partial derivatives in parentheses to obtain...a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H, F, and G). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you’ve done four of them the novelty begins to wear off.Figure 1: Measured heat capacities at constant pressure (data points) forone mole each of three different elemental solids. The solid curves show the heatcapacity at constant volume predicted by the model used in Section 7.5, with thehorizontal scale chosen to best fit the data for each substance. At sufficiently hightemperatures, CV for each material approaches the value 3R predicted by theequipartition theorem. The discrepancies between the data and the solid curvesat high T are mostly due to the differences between CP and CV. At T = 0 alldegrees of freedom are frozen out, so both CP and CV go to zero. Data from Y. S.Touloukian, ed., Thermophysical Properties of Matter (Plenum, New York, 1970).... Get solution
16. A formula analogous to that for CP – CV relates the isothermal and isentropic compressibilities of a material:... (Here κS = – (1/V)(∂V/∂P)S is the reciprocal of the adiabatic bulk modulus considered in Problem.) Derive this formula. Also check that it is true for an ideal gas.Problem:By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by...where ρ is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness. More precisely, if we imagine applying an increase in pressure ΔP to a chunk of the material, and this increase results in a (negative) change in volume ΔV, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:...This definition is still ambiguous, however, because I haven’t said whether the compression is to take place isothermally or adiabatically (or in some other way).(a) Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.(b) Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.(c) Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the rms speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.(d) When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not? Get solution
17. The enthalpy and Gibbs free energy, as defined in this section, give Special treatment to mechanical (compression-expansion) work, – P dV. Analogous quantities can be defined for other kinds of work, for instance, magnetic work. Consider the situation shown in below Figure, where a long solenoid (N turns, total length L) surrounds a magnetic specimen (perhaps a paramagnetic solid). If the magnetic field inside the specimen is ... and its total magnetic moment is ..., then we define an auxiliary field ... (often called simply the magnetic field) by the relation...where μo is the “permeability of free space,” 4π × 10–7 N/A2. Assuming cylindrical symmetry, all vectors must point either left of right, so we can drop the → symbols and agree that rightward is positive, leftward negative. From Ampere’s law, one can also show that when the current in the wire is I, the ... field inside the solenoidis NI/L, whether or not the specimen is present. (a) Imagine making an infinitesimal change in the current in the wire, resulting in infinitesimal changes in B, M, and .... Use Faraday’s law to show that the work required (from the power supply) to accomplish this change is Wtotal = V...dB. (Neglect the resistance of the wire.)(b) Rewrite the result of part (a) in terms of ... and M, then subtract off the work that would be required even if the specimen wore not present. If we define W, the work done on the system to be what’s left, show that W = μo...dM.(c) What is the thermodynamic identity for this system? (Include magnetic work but not, mechanical work or particle flow.)(d) How would you define analogues of the enthalpy and Gibbs free energy for a magnetic system? (The Helmholtz free energy is defined in the same way as for a mechanical system.) Derive the thermodynamic identities for each of these quantities, and discuss their interpretations.Figure: A long solenoid, surrounding a magnetic specimen, connected to a power supply that can change the current, performing magnetic work.... Get solution
18. Imagine that you drop a brick on the ground and it lands with a thud. Apparently the energy of this system tends to spontaneously decrease. Explain why. Get solution
19. In the previous section I derived the formula (∂F/∂V)T = –P. Explain why this formula makes intuitive sense, by discussing graphs of F vs. V with different slopes. Get solution
20. The first excited energy level of a hydrogen atom has an energy of 10.2 eV, if we take the ground-state energy to be zero. However, the first excited level is really four independent states, all with the same energy. We can therefore assign it an entropy of S = k ln 4, since for this given value of the energy, the multiplicity is 4. Question: For what temperatures is the Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what temperatures is it negative? (Comment: When F for the level is negative, the atom will spontaneously go from the ground state into that level, since F = 0 for the ground state and F always tends to decrease. However, for a system this small, the conclusion is only a probabilistic statement; random fluctuations will be very significant.) Get solution
21. Is heat capacity (C) extensive or intensive? What about specific heat (c)? Explain briefly. Get solution
22. Show that below equation is in agreement with the explicit formula for the chemical potential of a monatomic ideal gas derived in Section 3.5. Show how to calculate μ° for a monatomic ideal gas.Equation:μ(T, P) = μ°(T) + kT ln(P/P°). Get solution
24. Go through the arithmetic to verify that diamond becomes more stable than graphite at approximately 15 kbar. Get solution
25. In working high-pressure geochemistry problems it is usually more convenient to express volumes in units of kJ/kbar. Work out the conversion factor between this unit and m3. Get solution
26. How can diamond ever be more stable than graphite, when it has less entropy? Explain how at high pressures the conversion of graphite to diamond can increase the total entropy of the carbon plus its environment. Get solution
27. Graphite is more compressible than diamond.(a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text?(b) The isothermal compressibility of graphite is about 3 × 10–6 bar–1 , while that of diamond is more than ten times less and hence negligible in comparison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem 1.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).Problem 1:Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep V fixed as T increases, as follows. (a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume, dV1, in terms of dT and the thermal expansion coefficient β introduced in Problem 2.(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, dV2, in terms of dP and the isothermal compressibility κT defined as...(This is the reciprocal of the isothermal bulk modulus defined in Problem 3.)(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dP to dT is equal to ... since there is no net change in volume. Express this partial derivative in terms of β and κT. Then express it more abstractly in terms of the partial derivatives used to define β and κT. For the second expression you should obtain...This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.(d) Compute β, κT and ... for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).(e) For water at 25°C, β = 2.57 × 10−4 K−1 and κT = 4.52 × 10−10 Pa−1. Suppose you increase the temperature of some water from 20°C to 30°C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at 25°C) β = 1.81 × 10−4 K−1 and κT = 4.04 × 10−11 Pa−l. Given the choice, would you rather measure the heat capacities of these substances at constant V or at constant P?Problem 2:When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:...(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K−1 = 1.81 × 10−4 K−1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 × 10−4 K−1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of − 0.68 ×10−4 K−1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.Problem 3:By applying Newton’s laws to the oscillations of a continuous medium, one can show that the speed of a sound wave is given by...where ρ is the density of the medium (mass per unit volume) and B is the bulk modulus, a measure of the medium’s stiffness. More precisely, if we imagine applying an increase in pressure ΔP to a chunk of the material, and this increase results in a (negative) change in volume ΔV, then B is defined as the change in pressure divided by the magnitude of the fractional change in volume:...This definition is still ambiguous, however, because I haven’t said whether the compression is to take place isothermally or adiabatically (or in some other way).(a) Compute the bulk modulus of an ideal gas, in terms of its pressure P, for both isothermal and adiabatic compressions.(b) Argue that for purposes of computing the speed of a sound wave, the adiabatic B is the one we should use.(c) Derive an expression for the speed of sound in an ideal gas, in terms of its temperature and average molecular mass. Compare your result to the formula for the rms speed of the molecules in the gas. Evaluate the speed of sound numerically for air at room temperature.(d) When Scotland’s Battlefield Band played in Utah, one musician remarked that the high altitude threw their bagpipes out of tune. Would you expect altitude to affect the speed of sound (and hence the frequencies of the standing waves in the pipes)? If so, in which direction? If not, why not? Get solution
28. Calcium carbonate, CaCO3, has two common crystalline forms, calcite and aragonite. Thermodynamic data for these phases can be found at the back of this book.(a) Which is stable at earth’s surface, calcite or aragonite?(b) Calculate the pressure (still at room temperature) at which the other phase should become stable. Get solution
29. Aluminum silicate. Al2SiO5, has three different crystalline forms: kyanite, andalusite, and sillimanite. Because each is stable under a different, set of temperature-pressure conditions, and all are commonly found in metamorphic rocks, these minerals are important indicators of the geologic history of rock bodies.(a) Referring to the thermodynamic data at the back of this book, argue that at 298 K the stable phase should be kyanite, regardless of pressure.(b) Now consider what happens at fixed pressure as we vary the temperature. Let ΔG be the difference in Gibbs free energies between any two phases, and similarly for ΔS. Show that the T dependence of ΔG is given by...Although the entropy of any given phase will increase significantly as the temperature increases, above room temperature it is often a good approximation to take ΔS. the difference in entropies between two phases, to be independent of T. This is because the temperature dependence of S is a function of the heat capacity (as we saw in Chapter 3), and the heat capacity of a solid at high temperature depends, to a good approximation, only on the number of atoms it contains.(c) Taking ΔS to be independent of T, find the range of temperatures over which kyanite, andalusite, and sillimanite should be stable (at 1 bar).(d) Referring to the room-temperature heat capacities of the three forms of Al2SiO5, discuss the accuracy the approximation ΔS = constant. Get solution
30. Sketch qualitatively accurate graphs of G vs. T for the three phases of H2O (ice, water, and steam) at atmospheric pressure. Put all three graphs on the same set of axes, and label the temperatures 0°C and 100°C. How would the graphs differ at a pressure of 0.001 bar? Get solution
31. Sketch qualitatively accurate graphs of G vs. P for the three phases of H2O (ice, water, and steam) at 0°C. Put all three graphs on the same set of axes, and label the point corresponding to atmospheric pressure. How would the graphs differ at slightly higher temperatures? Get solution
32. The density of ice is 917 kg/m3.(a) Use the Clausius-Clapeyron relation to explain why the slope of the phase boundary between water and ice is negative.(b) How much pressure would you have to put on an ice cube to make it melt at –1°C?(c) Approximately how deep under a glacier would you have to be before the weight of the ice above gives the pressure you found in part (b)? (Note that the pressure can be greater at some locations, as where the glacier flows over a protruding rock.)(d) Make a rough estimate of the pressure under the blade of an ice skate, and calculate the melting temperature of ice at this pressure. Some authors have claimed that skaters glide with very little friction because the increased pressure under the blade melts the ice to create a thin layer of water. What do you think of this explanation? Get solution
33. An inventor proposes to make a heat engine using water/ice as the working substance, taking advantage of the fact that water expands as it freezes. A weight to be lifted is placed on top of a piston over a cylinder of water at 1°C. The system is then placed in thermal contact with a low-temperature reservoir at –1°C until the water freezes into ice, lifting the weight. The weight is then removed and the ice is melted by putting it in contact with a high-temperature reservoir at 1°C. The inventor is pleased with this device because it can seemingly perform an unlimited amount of work while absorbing only a finite amount of heat. Explain the flaw in the inventor’s reasoning, and use the Clausius-Clapeyron relation to prove that the maximum efficiency of this engine is still given by the Carnot formula, 1 – Tc/Th. Get solution
34. Below 0.3 K the slope of the 3He solid-liquid phase boundary is negative (see Figure 5.13).(a) Which phase, solid or liquid, is more dense? Which phase has more entropy (per mole)? Explain your reasoning carefully.(b) Use the third law of thermodynamics to argue that the slope of the phase boundary must go to zero at T = 0. (Note that the 4He solid-liquid phase boundary is essentially horizontal below 1 K.)(c) Suppose that you compress liquid 3He adiabatically until it becomes a solid. If the temperature just before the phase change is 0.1 K, will the temperature after the phase change be higher or lower? Explain your reasoning carefully. Get solution
35. The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔV depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔV to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:... Get solution
36. Effect of altitude on boiling water.(a) Use the result of the previous problem 1 and the data in below Figure to plot a graph of the vapor pressure of water between 50°C and 100°C. How well can you match the data at the two endpoints?(b) Reading the graph backwards, estimate the boiling temperature of water at each of the locations for which you determined the pressure in Problem 2. Explain why it takes longer to cook noodles when you’re camping in the mountains.(c) Show that the dependence of boiling temperature on altitude is very nearly (though not exactly) a linear function, and calculate the slope in degrees Celsius per thousand feet (or in degrees Celsius per kilometer).Figure: Phase diagram for H2O (not to scale). The table gives the vaporpressure and molar latent heat for the solid-gas transformation (first three entries)and the liquid-gas transformation (remaining entries). Data from Keenan et al.(1978) and Lide (1994)....Problem 1:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔV depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔV to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Problem 2:The exponential atmosphere. (a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest, the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air.(b) Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass m of the air molecules. (The information needed to calculate m is given in Problem.) Show, then, that the pressure obeys the differential equation...called the barometric equation.(c) Assuming that the temperature of the atmosphere is independent of height (not a great assumption but not terrible either), solve the barometric equation to obtain the pressure as a function of height: P(z) = P(0)e−mgz/kT. Show also that the density obeys a similar equation.(d) Estimate the pressure, in atmospheres, at the following locations: Ogden, Utah (4700 ft or 1430 m above sea level); Leadville, Colorado (10,150 ft, 3090 m) ; Mt. Whitney, California (14,500 ft, 4420 m); Mt. Everest, Nepal/Tibet (29,000 ft, 8850 m). (Assume that the pressure at sea level is 1 atm.)Problem 3:Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), O2 (21%), and argon (1%). Get solution
37. Use the data at the back of this book to calculate the slope of the calcite-aragonite phase boundary (at 298 K). You located one point on this phase boundary in below Problem; use this information to sketch the phase diagram of calcium carbonate.Problem:Calcium carbonate, CaCO3, has two common crystalline forms, calcite and aragonite. Thermodynamic data for these phases can be found at the back of this book.(a) Which is stable at earth’s surface, calcite or aragonite?(b) Calculate the pressure (still at room temperature) at which the other phase should become stable. Get solution
38. In Problems 1 and 2, you calculated the entropies of diamond and graphite at 500 K. Use these values to predict the slope of the graphite-diamond phase boundary at 500 K, and compare to below Figure 1. Why is the slope almost constant at still higher temperatures? Why is the slope zero at T = 0?Figure 1: The experimentalphase diagram of carbon.The stability region of the gasphase is not visible on this scale;the graphite-liquid-gas triplepoint is at the bottom of thegraphite-liquid phase boundary,at 110 bars pressure. FromDavid A. Young, Phase Diagramsof the Elements (Universityof California Press, Berkeley,1991)....Problem 1:As shown in below Figure 2, the heat capacity of diamond near room temperature is approximately linear in T. Extrapolate this function up to 500 K, and estimate the change in entropy of a mole of diamond as its temperature is raised from 298 K to 500 K. Add on the tabulated value at 298 K (from the backof this book) to obtain S(500 K).Figure 2: Measured heat capacities at constant pressure (data points) forone mole each of three different elemental solids. The solid curves show the heatcapacity at constant volume predicted by the model used in Section 7.5, with thehorizontal scale chosen to best fit the data for each substance. At sufficiently hightemperatures, CV for each material approaches the value 3R predicted by theequipartition theorem. The discrepancies between the data and the solid curvesat high T are mostly due to the differences between CP and CV. At T = 0 alldegrees of freedom are frozen out, so both CP and CV go to zero. Data from Y. S.Touloukian, ed., Thermophysical Properties of Matter (Plenum, New York, 1970)....Problem 2:Experimental measurements of heat capacities are often represented in reference works as empirical formulas. For graphite, a formula that works well over a fairly wide range of temperatures is (for one mole)...where a = 16.86 J/K, b = 4.77 × 10–3 J/K2, and c = 8.54 × 105 J∙K. Suppose, then, that a mole of graphite is heated at constant pressure from 298 K to 500 K. Calculate the increase in its entropy during this process. Add on the tabulated value of S(298 K) (from the back of this book) to obtain S(500 K). Get solution
39. Consider again the aluminosilicate system treated in below Problem. Calculate the slopes of all three phase boundaries for this system: kyanite-andalusitc, kyanite-sillimanite, and andalusite-sillimanite. Sketch the phase diagram, and calculate the temperature and pressure of the triple point.Problem:Aluminum silicate. Al2SiO5, has three different crystalline forms: kyanite, andalusite, and sillimanite. Because each is stable under a different, set of temperature-pressure conditions, and all are commonly found in metamorphic rocks, these minerals are important indicators of the geologic history of rock bodies.(a) Referring to the thermodynamic data at the back of this book, argue that at 298 K the stable phase should be kyanite, regardless of pressure.(b) Now consider what happens at fixed pressure as we vary the temperature. Let ΔG be the difference in Gibbs free energies between any two phases, and similarly for ΔS. Show that the T dependence of ΔG is given by...Although the entropy of any given phase will increase significantly as the temperature increases, above room temperature it is often a good approximation to take ΔS. the difference in entropies between two phases, to be independent of T. This is because the temperature dependence of S is a function of the heat capacity (as we saw in Chapter 3), and the heat capacity of a solid at high temperature depends, to a good approximation, only on the number of atoms it contains.(c) Taking ΔS to be independent of T, find the range of temperatures over which kyanite, andalusite, and sillimanite should be stable (at 1 bar).(d) Referring to the room-temperature heat capacities of the three forms of Al2SiO5, discuss the accuracy the approximation ΔS = constant. Get solution
40. The methods of this section can also be applied to reactions in which one set of solids converts to another. A geologically important example is the transformation of albite into jadeite + quartz:...Use the data at the back of this book to determine the temperatures and pressures under which a combination of jadeite and quartz is more stable than albite. Sketch the phase diagram of this system. For simplicity, neglect the temperature and pressure dependence of both ΔS and ΔV. Get solution
41. Suppose you have a liquid (say, water) in equilibrium with its gas phase, inside some closed container. You then pump in an inert gas (say, air), thus raising the pressure exerted on the liquid. What happens?(a) For the liquid to remain in diffusive equilibrium with its gas phase, the chemical potentials of each must change by the same amount: dμl = dμg. Use this fact and below equation to derive a differential equation for the equilibrium vapor pressure, Pυ, as a function of the total pressure P. (Treat the gases as ideal, and assume that none of the inert gas dissolves in the liquid.)(b) Solve the differential equation to obtain...where the ratio V/N in the exponent is that of the liquid. (The quantity ...) is just the vapour pressure in the absence of the inert gas.) Thus, the presence of the inert gas leads to a slight increase in the vapour pressure: It causes more of the liquid to evaporate.(c) Calculate the percent increase in vapour pressure when air at atmospheric pressure is added to a system of water and water vapour in equilibrium at 25°C. Argue more generally that the increase in vapour pressure due to the presence of an inert gas will be negligible except under extreme conditions.Equation... Get solution
42. Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour.(a) Use the vapor pressure equation (below Problem) and the data in below Figure to plot a graph of the vapor pressure of water from 0°C to 40° C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature.(b) The temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?Problem:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔU depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔU to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Figure: Phase diagram for H2O (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries). Data from Keenan et al (1978) and Lide (1994).... Get solution
43. Assume that the air you exhale is at 35°C, with a relative humidity of 90%. This air immediately mixes with environmental air at 10°C and unknown relative humidity; during the mixing, a variety of intermediate temperatures and water vapour percentages temporarily occur, If you are able to “see your breath” due to the formation of cloud droplets during this mixing, what can you conclude about the relative humidity of your environment? (Refer to the vapour pressure graph drawn below Problem 1.)Problem 1:Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour. (a) Use the vapor pressure equation (below Problem 2) and the data in below Figure to plot a graph of the vapor pressure of water from 0°C to 40° C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature. (b) The temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?Problem 2:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔU depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔU to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Figure: Phase diagram for H2O (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries). Data from Keenan et al (1978) and Lide (1994).... Get solution
44. Suppose that an unsaturated air mass is rising and cooling at the dry adiabatic lapse rate found in Problem 1. If the temperature at ground level is 25° C and the relative humidity there is 50%, at what altitude will this air mass become saturated so that condensation begins and a cloud forms (see below Figure 1)? (Refer to the vapour pressure graph drawn shown below Problem 5.)Figure 1: Cumulus clouds form when rising air expands adiabaticallyand cools to the dew point the onset of condensation slowsthe cooling, increasing the tendency of the air to rise further.These clouds began to form in late morning, in a sky that was clear onlyan hour before the photo was taken. By mid-afternoon they had developedinto thunderstorms....Problem 1:In Problem 2 you calculated the pressure of earth’s atmosphere as a function of altitude, assuming constant temperature. Ordinarily, however, the temperature of the bottommost 10–15 km of the atmosphere (called the troposphere) decreases with increasing altitude, due to heating from the ground (which is warmed by sunlight). If the temperature gradient |dT/dz| exceeds a certain critical value, convection will occur: Warm, low-density air will rise, while cool, high-density air sinks. The decrease of pressure with altitude causes a rising air mass to expand adiabatically and thus to cool. The condition for convection to occur is that the rising air mass must remain warmer than the surrounding air despite this adiabatic cooling.(a) Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation...(b) Assume that dT/dz is just at the critical value for convection to begin, so that the vertical forces on a convecting air mass are always approximately in balance. Use the result of Problem 2 (b) to find a formula for dT/dz in this case. The result should be a constant, independent of temperature and pressure, which evaluates to approximately - 10°C/km. This fundamental meteorological quantity is known as the dry adiabatic lapse rate.Problem 2:The exponential atmosphere.(a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest , the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air.(b) Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass m of the air molecules. (The information needed to calculate m is given in Problem 3.) Show, then, that the pressure obeys the differential equation...called the barometric equation.(c) Assuming that the temperature of the atmosphere is independent of height (not a great assumption but not terrible either), solve the barometric equation to obtain the pressure as a function of height: P(z) = P(O)e−mgz/kT. Show also that the density obeys a similar equation.(d) Estimate the pressure, in atmospheres, at the following locations: Ogden, Utah (4700 ft or 1430 m above sea level); Leadville, Colorado (10,150 ft , 3090 m) ; Mt. Whitney, California (14,500 ft, 4420 m); Mt. Everest, Nepal/ Tibet (29,000 ft, 8850 m). (Assume that the pressure at sea level is 1 atm.)Problem 3:Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), 02 (21%), and argon (1%).Problem 4:Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour.(a) Use the vapor pressure equation (below Problem 5) and the data in below Figure 2 to plot a graph of the vapor pressure of water from 0°C to 40° C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature.(b) The temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?Problem 5:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔU depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔU to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Figure 2: Phase diagram for H2O (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries). Data from Keenan et al (1978) and Lide (1994).... Get solution
45. In Problem 1 you calculated the atmospheric temperature gradient required for unsaturated air to spontaneously undergo convection. When a rising air mass becomes saturated, however, the condensing water droplets will give up energy, thus slowing the adiabatic cooling process(a) Use the first, law of thermodynamics to show that, as condensation forms during adiabatic expansion, the temperature of an air mass changes by...where nw is the number of moles of water vapour present, L is the latent heat of vaporization per mole, and I’ve set γ = 7/5 for air. You may assume that the H2O makes up only a small fraction of the air mass.(b) Assuming that the air is always saturated during this process, the ratio nw/n is a known function of temperature and pressure, Carefully express dnw/dz in terms of dT/dz, dP/dz, and the vapour pressure Pv(T). Use the Clausius-Clapeyron relation to eliminate dPv/dT.(c) Combine the results of parts (a) and (b) to obtain a formula relating the temperature gradient, dT/dz, to the pressure gradient, dP/dz. Eliminate the latter using the “barometric equation” from Problem 1.16. You should finally obtain....where M is the mass of a mole of air. The prefactor is just the dry adiabatic lapse rate calculated in Problem 1.40, while the rest of the expression gives the correction due to heating from the condensing water vapor. The whole result is called the wet adiabatic lapse rate; it is the critical temperature gradient above which saturated air will spontaneously convect.(d) Calculate the wet adiabatic lapse rate at atmospheric pressure (1 bar) and 25°C, then at atmospheric pressure and 0°C. Explain why the results are different, and discuss their implications. What happens at higher altitudes, where the pressure is lower?Problem 1:In Problem you calculated the pressure of earth’s atmosphere as a function of altitude, assuming constant temperature. Ordinarily, however, the temperature of the bottommost 10–15 km of the atmosphere (called the troposphere) decreases with increasing altitude, due to heating from the ground (which is warmed by sunlight). If the temperature gradient |dT/dz| exceeds a certain critical value, convection will occur: Warm, low-density air will rise, while cool, high-density air sinks. The decrease of pressure with altitude causes a rising air mass to expand adiabatically and thus to cool. The condition for convection to occur is that the rising air mass must remain warmer than the surrounding air despite this adiabatic cooling.(a) Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation...(b) Assume that dT/dz is just at the critical value for convection to begin, so that the vertical forces on a convecting air mass are always approximately in balance. Use the result of Problem 2 (b) to find a formula for dT/dz in this case. The result should be a constant, independent of temperature and pressure, which evaluates to approximately - 10°C/km. This fundamental meteorological quantity is known as the dry adiabatic lapse rate.Problem 2:The exponential atmosphere.(a) Consider a horizontal slab of air whose thickness (height) is dz. If this slab is at rest , the pressure holding it up from below must balance both the pressure from above and the weight of the slab. Use this fact to find an expression for dP/dz, the variation of pressure with altitude, in terms of the density of air.(b) Use the ideal gas law to write the density of air in terms of pressure, temperature, and the average mass m of the air molecules. (The information needed to calculate m is given in Problem 3.) Show, then, that the pressure obeys the differential equation...called the barometric equation.(c) Assuming that the temperature of the atmosphere is independent of height (not a great assumption but not terrible either), solve the barometric equation to obtain the pressure as a function of height: P(z) = P(O)e−mgz/kT. Show also that the density obeys a similar equation.(d) Estimate the pressure, in atmospheres, at the following locations: Ogden, Utah (4700 ft or 1430 m above sea level); Leadville, Colorado (10,150 ft , 3090 m) ; Mt. Whitney, California (14,500 ft, 4420 m); Mt. Everest, Nepal/ Tibet (29,000 ft, 8850 m). (Assume that the pressure at sea level is 1 atm.)Problem 3: Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), 02 (21%), and argon (1%). Get solution
46. Everything in this section so far has ignored the boundary between two phases, as if each molecule were unequivocally part of one phase or the other. In fact, the boundary is a kind of transition zone where molecules are in an environment that differs from both phases. Since the boundary zone is only a few molecules thick, its contribution to the total free energy of a system is very often negligible. One important exception, however, is the first tiny droplets or bubbles or grains that form as a material begins to undergo a phase transformation. The formation of these initial specks of a new phase is called nucleation. In this problem we will consider the nucleation of water droplets in a cloud.The surface forming the boundary between any two given phases generally has a fixed thickness, regardless of its area. The additional Gibbs free energy of this surface is therefore directly proportional to its area; the constant of proportionality is called the surface tension, σ:...If you have a blob of liquid in equilibrium with its vapor and you wish to stretch it into a shape that has the same volume but more surface area, then σ is the minimum work that you must perform, per unit of additional area, at fixed temperature and pressure. For water at 20°C, σ = 0.073 J/m2. a) Consider a spherical droplet of water containing Nl; molecules, surrounded by N – Nl molecules of water vapour. Neglecting surface tension for the moment, write down a formula for the total Gibbs free energy of this system in terms of N, Nl, and the chemical potentials of the liquid and vapour. Rewrite Nl in terms of vt, the volume per molecule in the liquid, and r, the radius of the droplet. b) Now add to your expression for G a term to represent the surface tension, written in terms of r and σ.c) Sketch a qualitative graph of G vs. r for both signs of μg–μl, and discuss the implications. For which sign of μg–μl does there exist a nonzero equilibrium radius? Is this equilibrium stable?d) Let rc represent the critical equilibrium radius that you discussed qualitatively in part (c). Find an expression for rc in terms of μg–μl. Then rewrite the difference of chemical potentials in terms of the relative humidity (see below Problem 1), assuming that the vapuor behaves as an ideal gas. (The relative humidity is defined in terms of equilibrium of a vapour with a flat surface, or with an infinitely large droplet.) Sketch a graph of the critical radius as a function of the relative humidity, including numbers. Discuss the implications. In particular, explain why it is unlikely that the clouds in our atmosphere would form by spontaneous aggregation of water molecules into droplets. (In fact, cloud droplets form around nuclei of dust particles and other foreign material, when the relative humidity is close to 100%.)Problem 1:Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour. (a) Use the vapor pressure equation (below Problem 2) and the data in below Figure to plot a graph of the vapor pressure of water from 0°C to 40° C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature.(b) The temperature on a certain summer day is 30°C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?Problem 2:The Clausius-Clapeyron below relation is a differential equation that can, in principle, be solved to find the shape of the entire phase-boundary curve. To solve it, however, you have to know how both L and ΔU depend on temperature and pressure. Often, over a reasonably small section of the curve, you can take L to be constant. Moreover, if one of the phases is a gas, you can usually neglect the volume of the condensed phase and just take ΔU to be the volume of the gas, expressed in terms of temperature and pressure using the ideal gas law. Making all these assumptions, solve the differential equation explicitly to obtain the following formula for the phase boundary curve:P = (constant) × e–L/RT,This result is called the vapour pressure equation. Caution: Be sure to use this formula only when all the assumptions just listed are valid.Relation:...Figure: Phase diagram for H2O (not to scale). The table gives the vapor pressure and molar latent heat for the solid-gas transformation (first three entries) and the liquid-gas transformation (remaining entries). Data from Keenan et al (1978) and Lide (1994).... Get solution
47. For a magnetic system held at constant T and H (see Problem 1), the quantity that is minimized is the magnetic analogue of the Gibbs free energy, which obeys the thermodynamic identity...Phase diagrams for two magnetic systems are shown in Figure 5.14; the vertical axis on each of these figures is μ0H.(a) Derive an analogue of the Clausius-Clapeyron relation for the slope of a phase boundary in the H-T plane. Write your equation in terms of the difference in entropy between the two phases.(b) Discuss the application of your equation to the ferromagnet phase diagram in Figure 5.14.(c) In a type-I superconductor, surface currents flow in such a way as to completely cancel the magnetic field (B, not H) inside. Assuming that M is negligible when the material is in its normal (non-superconducting) state, discuss the application of your equation to the superconductor phase diagram in Figure 5.14. Which phase has the greater entropy? What happens to the difference in entropy between the phases at each end of the phase boundary?Problem 1:The enthalpy and Gibbs free energy, as defined in this section, give Special treatment to mechanical (compression-expansion) work, – P dV. Analogous quantities can be defined for other kinds of work, for instance, magnetic work. Consider the situation shown in below Figure, where a long solenoid [N turns, total length L) surrounds a magnetic specimen (perhaps a paramagnetic solid). If the magnetic field inside the specimen is ... and its total magnetic moment is ..., then we define an auxiliary field ... (often called simply the magnetic field) by the relation...where μo is the “permeability of free space,” 4π × 10–7 N/A2. Assuming cylindrical symmetry, all vectors must point either left of right, so we can drop the symbols and agree that rightward is positive, leftward negative. From Ampere’s law, one can also show that when the current in the wire is I, the H field inside the solenoidis NI/L, whether or not the specimen is present.(a) Imagine making an infinitesimal change in the current in the wire, resulting in infinitesimal changes in B, M, and H. Use Faraday’s law to show that the work required (from the power supply) to accomplish this change is Wtotal = VHdB. (Neglect the resistance of the wire.)(b) Rewrite the result of part (a) in terms of H and M, then subtract off the work that would be required even if the specimen wore not present. If we define W, the work done on the system J to be what’s left, show that W = μoHdM.(c) What is the thermodynamic identity for this system? (Include magnetic work but not, mechanical work or particle flow.)(d) How would you define analogues of the enthalpy and Gibbs free energy for a magnetic system? (The Helmholtz free energy is defined in the same way as for a mechanical system.) Derive the thermodynamic identities for each of these quantities, and discuss their interpretations.Figure 1: A long solenoid, surrounding a magnetic specimen, connected to a power supply that can change the current, performing magnetic work.... Get solution
48. As you can see from Figure 5.20, the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives of P with respect to V (at fixed T) are zero. Use this fact to show that... Get solution
49. Use the result of the previous problem and the approximate values of a and b given in the text to estimate Tc, Pc, and Vc/N for N2, H2O, and He. (Tabulated values of a and b are often determined by working backward from the measured critical temperature and pressure.)Problem:As you can see from Figure 5.20, the critical point is the unique point on the original van der Walls isotherms (before the Maxwell construction) where both the first and second derivatives of P with respect to V (at fixed T) are zero. Use this fact to show that... Get solution
51. When plotting graphs and performing numerical calculations, it is convenient to work in terms of reduced variables,...Rewrite the van der Waals equation in terms of these variables, and notice that the constants a and b disappear. Get solution
52. Plot the van der Waals isotherm for T/Tc = 0.95, working in terms of reduced variables. Perform the Maxwell construction (either graphically or numerically) to obtain the vapour pressure. Then plot the Gibbs free energy (in units of NkTc) as a function of pressure for this same temperature and check that this graph predicts the same value for the vapor pressure. Get solution
53. Repeat the below problem for T/Tc = 0.8.Problem:Plot the van der Waals isotherm for T/Tc = 0.95, working in terms of reduced variables. Perform the Maxwell construction (either graphically or numerically) to obtain the vapour pressure. Then plot the Gibbs free energy (in units of NkTc) as a function of pressure for this same temperature and check that this graph predicts the same value for the vapor pressure. Get solution
54. Calculate the Helmholtz free energy of a van der Waals fluid, up to an undetermined function of temperature as in equation 5.56. Using reduced variables, carefully plot the Helmholtz free energy (in units of NkTc) as a function of volume for T/Tc = 0.8. Identify the two points on the graph corresponding to the liquid and gas at the vapor pressure. (If you haven’t worked the preceding problem, just read the appropriate values off Figure) Then prove that the Helmholtz free energy of a combination of these two states (part liquid, part gas) can be represented by a straight line connecting these two points on the graph. Explain why the combination is more stable, at a given volume, than the homogeneous state represented by the original curve, and describe how you could have determined the two transition volumes directly from the graph of F.Figure: ... Get solution
55. In this problem you will investigate the behavior of a van der Waals fluid near the critical point. It is easiest to work in terms of reduced variables throughout.(a) Expand the van der Waals equation in a Taylor series in (V − Vc), keeping terms through order (V − Vc)3. Argue that, for T sufficiently close to Tc, the term quadratic in (V − Vc) becomes negligible compared to the others and may be dropped.(b) The resulting expression for P(V) is antisymmetric about the point V = Vc. Use this fact to And an approximate formula for the vapour pressure as a function of temperature. (You may find it helpful to plot the isotherm.) Evaluate the slope of the phase boundary, dP/dT, at the critical point.(c) Still working in the same limit, find an expression for the difference in volume between the gas and liquid phases at the vapor pressure. You should find (Vg − Vl) α (Tc − T)β where β is known as critical exponent, Experiments show that β has a universal value of about 1/3, but the van der Waals model predicts a larger value.(d) Use the previous result to calculate the predicted latent heat of the transformation as a function of temperature, and sketch this function. (e) The shape of the T = TC isotherm defines another critical exponent, called δ: (P − PC) α (V − Vc)δ Calculate δ in the van der Waals model. (Experimental values of δ are typically around 4 or 5.)(f) A third critical exponent describes the temperature dependence of the isothermal compressibility,...This quantity diverges at the critical point, in proportion to a power of (T − Tc) that in principle could differ depending on whether one approaches the critical point from above or below. Therefore the critical exponents γ and γ′ are defined by the relations...Calculate κ on both sides of the critical point in the van der Waals model, and show that γ = γ′ in this model. Get solution
56. Prove that the entropy of mixing of an ideal mixture has an infinite slope, when plotted vs. x, at x = 0 and x = 1. Get solution
57. Consider an ideal mixture of just 100 molecules, varying in composition from pure A to pure B. Use a computer to calculate the mixing entropy as a function of NA and plot this function (in units of k). Suppose you start with all A and then convert one molecule to type B; by how much does the entropy Increase? By how much does the entropy increase when you convert a second molecule, and then a third, from A to B? Discuss. Get solution
58. In this problem you will model the mixing energy of a mixture in a relatively simple way, in order to relate the existence of a solubility gap to molecular behaviour. Consider a mixture of A and B molecules that is ideal in every way but one: The potential energy due to the interaction of neighbouring molecules depends upon whether the molecules are like or unlike. Let n be the average number of nearest neighbours of any given molecule (perhaps 6 or 8 or 10). Let μ0 be the average potential energy associated with the interaction between neighbouring molecules that are the same (A−A or B−B), and let uAB the potential energy associated with the interaction of a neighbouring’ unlike pair (A−B). There are no interactions beyond the range of the nearest neighbors; the values of u0 and uAB are independent, of the amounts of A and B; and the entropy of mixing is the same as for an ideal solution.a) Show that when the system is unmixed, the total potential energy due to all neighbour-neighbour interactions is ½Nnu0. (Hint: Be sure to count each neighbouring pair only once.)b) Find a formula for the total potential energy when the system is mixed, in terms of x, the fraction of B. (Assume that the mixing is totally random.)c) Subtract the results or parts (a) and (b) to obtain the change in energy upon mixing. Simplify the resuit as much as possible; you should obtain an expression proportional to x(1 − x). Sketch this function vs x, for both possible signs of uAB − u0.d) Show that the slope of the mixing energy function is finite at both endpoints, unlike the slope of the mixing entropy function.e) For the case uAB > u0. plot a graph of the Gibbs free energy of this system vs. x at. several temperatures. Discuss the implications. f) Find an expression for the maximum temperature at which this system has a solubility gap.g) Make a very rough estimate of uAB − u0 for a liquid mixture that has a solubility gap below 100oC.h) Plot, the phase diagram (T vs. x) for this system. Get solution
59. Suppose you cool a mixture of 50% nitrogen and 50% oxygen until it liquefies. Describe the cooling sequence in detail, including the temperatures and compositions at which liquefaction begins and ends. Get solution
60. Suppose you start with a liquid mixture of 60% nitrogen and 40% oxygen. Describe what happens as the temperature of this mixture increases. Be sure to give the temperatures and compositions at which boiling begins and ends. Get solution
61. Suppose you need a tank of oxygen that is 95% pure. Describe a process by which you could obtain such a gas, starting with air. Get solution
62. Consider a completely miscible two-component system whose overall composition is x, at a temperature where liquid and gas phases coexist. The composition of the gas phase at this temperature is xa and the composition of the liquid phase is xb. Prove the lever rule, which says that the proportion of liquid to gas is (x - xa) / (xb - x). Interpret this rule graphically on a phase diagram. Get solution
63. Everything in this section assumes that the total pressure of the system is fixed. How would you expect the nitrogen-oxygen phase diagram to change if you increase or decrease the pressure? Justify your answer. Get solution
64. Below Figure shows the phase diagram of plagioclase feldspar, which can be considered a mixture of albite (NaAISi3O8) and anorthite (CaAl2Si2O8).a) Suppose you discover a rock in which each plagioclase crystal varies in composition from center to edge, with the centers of the largest crystals composed of 70% anorthite and the outermost parts of all crystals made of essentially pure albite. Explain in some detail how this variation might arise. What was the composition of the liquid magma from which the rock formed?b) Suppose you discover another rock body in which the crystals near the top are albite-rich while the crystals near the bottom are anorthite-rich. Explain how this variation might arise.Figure: The phase diagram of plagioclase feldspar (at atmospheric pressure). From N. L. Bowen, “The Melting Phenomena of the Plagioclase Feldspars,” American Journal of Science 35, 577-599 (1913).... Get solution
65. In constructing the phase diagram from the free energy graphs in Figure, I assumed that both the liquid and the gas are ideal mixtures. Suppose instead that the liquid has a substantial positive mixing energy, so that its free energy curve, while still concave-up, is much flatter. In this case a portion of the curve may still lie above the gas’s free energy curve at TA. Draw a qualitatively accurate phase diagram for such a system, showing how you obtained the phase diagram from the free energy graphs. Show that there is a particular composition at which this gas mixture will condense with no change in composition. This special composition is called an azeotrope.Figure:... Get solution
66. Repeat the previous problem for the opposite case where the liquid has a substantial negative mixing energy, so that its free energy curve dips below the gas’s free energy curve at a temperature higher than TB. Construct the phase diagram and show that this system also has an azeotrope.Problem:In constructing the phase diagram from the free energy graphs in Figure, I assumed that both the liquid and the gas are ideal mixtures. Suppose instead that the liquid has a substantial positive mixing energy, so that its free energy curve, while still concave-up, is much flatter. In this case a portion of the curve may still lie above the gas’s free energy curve at TA. Draw a qualitatively accurate phase diagram for such a system, showing how you obtained the phase diagram from the free energy graphs. Show that there is a particular composition at which this gas mixture will condense with no change in composition. This special composition is called an azeotrope.Figure... Get solution
67. In this problem you will derive approximate formulas for the shapes of the phase boundary curves in diagrams such as Figures 5.31 and 5.32, assuming that both phases behave as ideal mixtures. For definiteness, suppose that the phases are liquid and gas.a) Show that in an ideal mixture of A and B, the chemical potential of species A can be written...where ...is the chemical potential of pure A (at the same temperature and pressure) and x = NB/(NA + NB). Derive a similar formula for the chemical potential of species B. Note that both formulas can be written for either the liquid phase or the gas phase.b) At any given temperature T, let xl and xg be the compositions of the liquid and gas phases that are in equilibrium with each other. By setting the appropriate chemical potentials equal to each other, show that xl and xg obey the equations...where ΔG° represents the change in G for the pure substance undergoing the phase change at temperature T.c) Over a limited range of temperatures, we can often assume that the main temperature dependence of ΔG° = ΔH° — TΔS° comes from the explicit T; both ΔH° and ΔS° are approximately constant. With this simplification, rewrite the results of part (b) entirely in terms of ...TA, and TB (eliminating ΔG and ΔS). Solve for Xl and Xg as functions of T.d) Plot your results for the nitrogen-oxygcn system. The latent heats of the pure substances are ... = 5570 J/mol and ... = 6820 J/mol. Compare to the experimental diagram, below Figure.e) Show that you can account for the shape of Figure with suitably chosen ΔH° values. What are those values?Figure: Experimental phase diagram for nitrogen and oxygen at atmospheric pressure. Data from International Crilical Tables (volume 3), with endpoints adjusted to values in Lide (1994). ...Figure: The phase diagram of plagioclase feldspar (at atmospheric pressure). From N. L. Bowen, “The Melting Phenomena of the PlagioclaseFeldspars,” American Journal of Science 35, 577-599 (1913). ... Get solution
68. Plumber’s solder is composed of 67% lead and 33% tin by weight. Describe what happens to this mixture as it cools, and explain why this composition might be more suitable than the eutectic composition for joining pipes. Get solution
69. What happens when you spread salt crystals over an icy sidewalk? Why is this procedure rarely used in very cold climates? Get solution
70. What happens when you add salt to the ice bath in an ice cream maker? How is it possible for the temperature to spontaneously drop below 0°C? Explain in as much detail as you can. Get solution
71. Below Figure (left) shows the free energy curves at one particular temperature for a two-component system that has three possible solid phases (crystal structures), one of essentially pure A, one of essentially pure B, and one of intermediate composition. Draw tangent lines to determine which phases are present at which values of x. To determine qualitatively what happens at other temperatures, you can simply shift the liquid free energy curve up or down (since the entropy of the liquid is larger than that of any solid). Do so, and construct a qualitative phase diagram for this system. You should find two eutectic points. Examples of systems with this behaviour include water + ethylene glycol and tin + magnesium.Figure: Free energy diagrams... Get solution
72. Repeat the previous problem for the diagram in below Figure (right), which has an important qualitative difference. In this phase diagram, you should find that β and liquid are in equilibrium only at temperatures below the point where the liquid is in equilibrium with infinitesimal amounts of α and β. This point is called a peritectic point. Examples of systems with this behaviour include water + NaCl and leucite + quartz.Figure: Free energy diagrams... Get solution
73. If below expression is correct, it must be extensive: Increasing both NA and NB by a common factor while holding all intensive variables fixed should increase G by the same factor. Show that expression has tins property. Show that it would not have this property had we not added the term proportional to NB!.Expression:... Get solution
74. Check that below equations 1 and 2, satisfy the identity G = NAμA + NBμB (below equation 1 and 2).Expression 1:...Expression 2:...Equation 1:... Get solution
75. Compare below expression 1 for the Gibbs free energy of a dilute solution to below expression 5.61 for the Gibbs free energy of an ideal mixture. Under what circumstances should these two expressions agree? Show that they do agree under these circumstances, and identify the function f(T,P) in this case.Expression 1:... Get solution
76. Seawater has a salinity of 3.5%, meaning that if you boil away a kilogram of seawater, when you’re finished you’ll have 35 g of solids (mostly NaCl) left in the pot. When dissolved, sodium chloride dissociates into separate Na+ and Cl− ions.a) Calculate the osmotic pressure difference between seawater and fresh water. Assume for simplicity that all the dissolved salts in seawater are NaCl.b) If you apply a pressure difference greater than the osmotic pressure to a solution separated from pure solvent by a semipermeable membrane, you get reverse osmosis: a flow of solvent out of the solution. This process can be used to desalinate seawater. Calculate the minimum work required to desalinate one litre of seawater. Discuss some reasons why the actual work required would be greater than the minimum. Get solution
77. Osmotic pressure measurements can be used to determine the molecular weights of largo molecules such as proteins. For a solution of large molecules to qualify as “dilute”, its molar concentration must be very low and hence the osmotic pressure can be too small to measure accurately. For this reason, the usual procedure is to measure the osmotic pressure at a variety of concentrations, then extrapolate the results to the limit of zero concentration. Here are some data* for the protein haemoglobin dissolved in water at 3°C:Concentration(grams/liter)Δh(cm)5.02.016.66.532.512.843.417.654.022.6The quantity Δh is the equilibrium difference in fluid level between the solution and the pure solvent, as shown in below Figure. From these measurements, determine the approximate molecular weight of hemoglobin (in grams per mole) .Figure: An experimental arrangement for measuring osmotic pressure. Solvent flows across the membrane from left to right until the difference in fluid level, Δh, is just enough to supply theosmotic pressure.... Get solution
78. Because osmotic pressures can be quite large, you may wonder whether the approximation made in below equation is valid in practice: Is μ0 really a linear function of P to. the required accuracy? Answer this question by discussing whether the derivative of this function changes significantly, over therelevant pressure range, in realistic examples.Equation:... Get solution
79. Most pasta recipes instruct you to add a teaspoon of salt to a pot of boiling water. Does this have a significant effect on the boiling temperature? Justify your answer with a rough numerical estimate. Get solution
80. Use the Clausius-Clapeyron relation to derive below equation directly from Raoult’s law. Be sure to explain the logic carefully.... Get solution
81. Derive a formula, similar to below equation, for the shift in the freezing temperature of a dilute solution. Assume that the solid phase is pure solvent, no solute. You should find that the shift is negative: The freezing temperature of a solution is less than that of the pure solvent. Explain in general terms why the shift should be negative.Equation:... Get solution
82. Use the result of the below problem to calculate the freezing temperature of seawater.Problem:Derive a formula, similar to below equation, for the shift in the freezing temperature of a dilute solution. Assume that the solid phase is pure solvent, no solute. You should find that the shift is negative: The freezing temperature of a solution is less than that of the pure solvent. Explain in general terms why the shift should be negative.Equation:... Get solution
83. Write down the equilibrium condition for each of the following reactions:a) 2H ↔ H2b) 2CO + O2 ↔ 2CO2c) CH4 + 2O2 ↔ 2H2O + CO2d) H2SO4 ↔ 2H+ + SO2−4e) 2p + 2n ↔ 4 He Get solution
84. A mixture of one part nitrogen and three parts hydrogen is heated, in the presence of a suitable catalyst, to a temperature of 500°C. What fraction of the nitrogen (atom for atom) is converted to ammonia, if the final total pressure is 400 atm? Protend for simplicity that the gases behave ideally despite the very high pressure. The equilibrium constant at 500°C is 6.9 × 10−5. (Hint: You’ll have to solve a quadratic equation.) Get solution
85. Derive the van’t Hoff equation,...which gives the dependence of the equilibrium constant on temperature. Here ΔH° is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if ΔH° is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of ΔH°; solve the equation in this case to obtain... Get solution
86. Use the result of the previous problem to estimate the equilibrium constant of the reaction N2 + 3H2 ↔ 2NH3 at 500°C, using only the room- temperature data at the back of this book. Compare your result to the actual value of K at 500°C quoted in the text. Get solution
87. Sulfuric acid, H2SO4, readily dissociates into H+ and HSO−4 ions:...The hydrogen sulfate ion, in turn, can dissociate again:...The equilibrium constants for these reactions, in aqueous solutions at 298 K, are approximately 102 and 10−19 respectively. (For dissociation of acids it is usually more convenient to look up K than ΔGo, By the way, the negative base-10 logarithm of K for such a reaction is called pK, in analogy to pH. So for the first reaction pK = −2, while for the second reaction pK = 1.9.)a) Argue that the first reaction tends so strongly to the right that we might as well consider it. to have gone to completion, in any solution that could possibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated?b) In industrialized regions whore lots of coal is burned, the concentration of sulfate in rainwater is typically 5 × 10−5 mol/kg. The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of .... What is the pH of this rainwater?c) Explain why you can neglect dissociation of water into H+ and OH− in answering the previous question.d) At what pH would dissolved sulfate be equally distributed between ... and ...? Get solution
88. Express ∂(ΔG°)/∂P in terms of the volumes of solutions of reactants and products, for a chemical reaction of dilute solutes. Plug in some reasonable numbers, to show that a pressure increase of 1 atm has only a negligible effect on the equilibrium constant. Get solution
89. The standard enthalpy change upon dissolving one mole of oxygen at 25°C is −11.7 kJ. Use this number and the van’t Hoff equation (below Problem) to calculate the equilibrium (Henry’s law) constant for oxygen in water at 0°C and at 100°C. Discuss the results briefly.Problem:Derive the van’t Hoff equation,...which gives the dependence of the equilibrium constant on temperature. Here ΔH° is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if ΔH° is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of ΔH°; solve the equation in this case to obtain... Get solution
90. When solid quartz “dissolves” in water, it combines with water molecules in the reaction....a) Use this data in the back of this book to compute the amount of silica dissolved in waler in equilibrium with solid quartz, at 25°C.b) Use the van’t Hoff equation (below Problem) to compute the amount of silica dissolved in water in equilibrium with solid quartz at 100°C.Problem:Derive the van’t Hoff equation,...which gives the dependence of the equilibrium constant on temperature. Here ΔH° is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if ΔH° is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of ΔH°; solve the equation in this case to obtain... Get solution
91. When carbon dioxide “dissolves” in water, essentially all of it reacts to form carbonic acid, H2CO3:CO2(g) + H2O(1) ↔ H2CO3(aq).The carbonic acid can then dissociate into H+ and bicarbonate ions,H2CO3(aq) ↔ H+(aq) + HCO−(aq).(The table at the back of this book gives thermodynamic data for both of these reactions.) Consider a body of otherwise pure water (or perhaps a raindrop) that is in equilibrium with the atmosphere near sea level, where the partial pressure of carbon dioxide is 3.4 × 10−4 bar (or 340 parts per million). Calculate the molality of carbonic acid and of bicarbonate ions in the water, and determine the pH of the solution. Note that even “natural” precipitation is somewhat acidic. Get solution
92. Suppose you have a box of atomic hydrogen, initially at room temperature and atmospheric pressure. You then raise the temperature, keeping the volume fixed.a) Find an expression for the fraction of the hydrogen that is ionized as a function of temperature. (You’ll have to solve a quadratic equation.) Check that your expression has the expected behaviour at very low and very high temperatures.b) At what temperature is exactly half of the hydrogen ionized?c) Would raising the initial pressure cause the temperature you found in part (b) to increase or decrease? Explain.d) Plot the expression you found in part (a) as a function of the dimension-less variable t = kT/I. Choose the range of t values to clearly show the interesting part of the graph. Get solution
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