Chapter #6 Solutions - An Introduction to Thermal Physics - Daniel V. Schroeder - 1st Edition

 

1. Consider a system of two Einstein solids, where the first “solid” contains just a single oscillator, while the second solid contains 100 oscillators. The total number of energy units in the combined system is fixed at 500. Use a computer to make a table of the multiplicity of the combined system, for each possible value of the energy of the first solid from 0 units to 20. Make a graph of the total multiplicity vs. the energy of the first solid, and discuss, in some detail, whether the shape of the graph is what you would expect. Also plot the logarithm of the total multiplicity, and discuss the shape of this graph. Get solution

2. Prove that the probability of finding an atom in any particular energy level is ...where F = E – TS and the “entropy” of a level is k times the logarithm of the number of degenerate states for that level. Get solution

3. Consider a hypothetical atom that has just two states: a ground state with energy zero and an excited state with energy 2 eV. Draw a graph of the partition function for this system as a function of temperature, and evaluate the partition function numerically at T = 300 K, 3000 K, 30,000 K, and 300,000 K. Get solution

4. Estimate the partition function for the hypothetical system represented in Figure Then estimate the probability of this system being in its ground state.Figure: Bar graph of the relative probabilities of the states of a hypothetical system. The horizontal axis is energy. The smooth curve represents the Boltzmann distribution, equation 6.8, for one particular temperature. At lower temperatures it would fall off more suddenly, while at higher temperatures it would fall off more gradually.... Get solution

5. Imagine a particle that can be in only three states, with energies −0.05 eV, 0, and 0.05 eV. This particle is in equilibrium with a reservoir at 300 K.(a) Calculate the partition function for this particle.(b) Calculate the probability for this particle to be in each of the three states. (c) Because the zero point for measuring energies is arbitrary, we could just as well say that the the energies of the three states are 0, +0.05 eV, and +0.10 eV, respectively. Repeat parts (a) and (b) using these numbers. Explain what changes and what doesn’t. Get solution

6. Estimate the probability that a hydrogen atom at room temperature is in one of its first excited states (relative to the probability of being in the ground state). Don’t forget to take degeneracy into account. Then repeat the calculation for a hydrogen atom in the atmosphere of the star γ UMa, whose surface temperature is approximately 9500 K. Get solution

7. Each of the hydrogen atom states shown in Figure 6.2 is actually twofold degenerate, because the electron can be in two independent spin states, both with essentially the same energy. Repeat the calculation given in the text for the relative probability of being in a first excited state, taking spin degeneracy into account. Show that the results are unaffected. ... Get solution

8. The energy required to ionize a hydrogen atom is 13.6 eV, so you might expect that the number of ionized hydrogen atoms in the sun’s atmosphere would be even less than the number in the first excited state. Yet at the end of Chapter 5 I showed that the fraction of ionized hydrogen is much larger, nearly one atom in 10,000. Explain why this result is not a contradiction, and why it would be incorrect to try to calculate the fraction of ionized hydrogen using the methods of this section. Get solution

9. In the numerical example in the text, I calculated only the ratio of the probabilities of a hydrogen atom being in two different states. At such a low temperature the absolute probability of being in a first excited state is essentially the same as the relative probability compared to the ground state. Proving this rigorously, however, is a bit problematic, because a hydrogen atom has infinitely many states.(a) Estimate the partition function for a hydrogen atom at 5800 K, by adding the Boltzmann factors for all the states shown explicitly in Figure 6.2. (For simplicity you may wish to take the ground state energy to be zero, and shift the other energies accordingly.)(b) Show that if all bound states are included in the sum, then the partition function of a hydrogen atom is infinite, at any nonzero temperature. (See Appendix A for the full energy level structure of a hydrogen atom.) (c) When a hydrogen atom is in energy level n, the approximate radius of the electron wavefunction is a0n2, where a0 is the Bohr radius, about 5 × 10−11m. Going back to equation 6.3, argue that the p dV term is not negligible for the very high-n states, and therefore that the result of part (a), not that of part (b), gives the physically relevant partition function for this problem. Discuss. Get solution

10. A water molecule can vibrate in various ways, but the easiest type of vibration to excite is the “flexing” mode in which the hydrogen atoms move toward and away from each other but the HO bonds do not stretch. The oscillations of this mode are approximately harmonic, with a frequency of 4.8 × 1013 Hz. As for any quantum harmonic oscillator, the energy levels are ... and so on. None of these levels are degenerate.(a) Calculate the probability of a water molecule being in its flexing ground state and in each of the first two excited states, assuming that it is in equilibrium with a reservoir (say the atmosphere) at 300 K. (Hint: Calculate Z by adding up the first few Boltzmann factors, until the rest are negligible.)(b) Repeat the calculation for a water molecule in equilibrium with a reservoir at 700K (perhaps in a steam turbine). Get solution

11. A lithium nucleus has four independent spin orientations, conventionally labeled by the quantum number m = −3/2, −1/2, 1/ 2, 3/2. In a magnetic field B, the energies of these four states are E = − mµB, where the constant µ, is 1.03 × 10 −7 eV / T. In the Purcell-Pound experiment described in Section 3.3, the maximum magnetic field strength was 0.63 T and the temperature was 300 K. Calculate the probability of a lithium nucleus being in each of its four spin states under these conditions. Then show that, if the field is suddenly reversed, the probabilities of the four states obey the Boltzmann distribution for T = −300 K. Get solution

12. Cold interstellar molecular clouds often contain the molecule cyanogen (CN), whose first rotational excited states have an energy of 4.7 × 10−4 eV (above the ground state). There are actually three such excited states, all with the same energy. In 1941, studies of the absorption spectrum of starlight that passes through these molecular clouds showed that for every ten CN molecules that are in the ground state, approximately three others are in the three first excited states (that is, an average of one in each of these states). To account for this data, astronomers suggested that the molecules might be in thermal equilibrium with some “reservoir” with a well-defined temperature. What is that temperature? Get solution

13. At very high temperatures (as in the very early universe), the proton and the neutron can be thought of as two different states of the same particle, called the “nucleon.” (The reactions that convert a proton to a neutron or vice versa require the absorption of an electron or a positron or a neutrino, but all of these particles tend to be very abundant at sufficiently high temperatures.) Since the neutron’s mass is higher than the proton’s by 2.3 × 10−30 kg, its energy is higher by this amount times c2 . Suppose, then , that at some very early time, the nucleons were in thermal equilibrium with the rest of the universe at 10ll K. What fraction of the nucleons at that time were protons, and what fraction were neutrons? Get solution

14. Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere. already derived in Problem 1 and 2. (Hint: Let the system be a single air molecule, let s1 be a state with the molecule at sea level, and let s2 be a state with the molecule atheight z.)Problem 1The 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: Calculate the mass of a mole of dry air, which is a mixture of N2 (78% by volume), 02 (21%), and argon (1%).Problem: 2Consider a monatomic ideal gas that lives at a height z above sea level, so each molecule has potential energy mgz in addition to its kinetic energy.(a) Show that the chemical potential is the same as if the gas were at sea level, plus an additional term mgz:...(You can derive this result from either the definition μ = −T[∂S/∂N)U,V or the formula p = (∂U/∂N)S,V.)(b) Suppose you have two chunks of helium gas, one at sea level and one atheight z, each having the same temperature and volume. Assuming that they are in diffusive equilibrium, show that the number of molecules in the higher chunk isN(z) = N( 0)e−mgz/kT,in agreement with the result of below Problem. Get solution

15. Suppose you have 10 atoms of weberium: 4 with energy 0 eV, 3 with energy 1 eV, 2 with energy 4 eV, and 1 with energy 6 eV.(a) Compute the average energy of all your atoms, by adding up all their energies and dividing by 10. (b) Compute the probability that one of your atoms chosen at random would have energy E, for each of the four values of E that occur. (c) Compute the average energy again, using the formula ... Get solution

16. Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of the energy is...where β = 1/kT. These formulas can be extremely useful when you have an explicit formula for the partition function. Get solution

17. The most common measure of the fluctuations of a set of numbers away from the average is the standard deviation, defined as follows.(a) For each atom in the five-atom toy model of Figure, compute the deviation of the energy from the average energy, that is, ...for i = 1 to 5. Call these deviations ΔEi.(b) Compute the average of the squares of the five deviations, that is, ... Then compute the square root of this quantity, which is the root-meansquare (rms) deviation, or standard deviation. Call this number σE. Does σE give a reasonable measure of how far the individual values tend to stray from the average? (c) Prove in general that...that is, the standard deviation squared is the average of the squares minus the square of the average. This formula usually gives the easier way of computing a standard deviation. (d) Check the preceding formula for the five-atom toy model of Figure.Figure:. Five hypothetical atoms distributed among three different states.... Get solution

18. Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of E2 is...Then use this result and the results of the previous two problems to derive a formula for σ E in terms of the heat capacity, ...You should find... Get solution

19. Apply the result of Problem to obtain a formula for the standard deviation of the energy of a system of N identical harmonic oscillators (such as in an Einstein solid), in the high-temperature limit. Divide by the average energy to obtain a measure of the fractional fluctuation in energy. Evaluate this fraction numerically for N = 1, 104 , and 1020. Discuss the results briefly.Problem:Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of E2 is...Then use this result and the results of the previous two problems to derive a formula for σE in terms of the heat capacity, ...You should find... Get solution

20. This problem concerns a collection of N identical harmonic oscillators (perhaps an Einstein solid or the internal vibrations of gas molecules) at temperature T. As in Section 2.2, the allowed energies of each oscillator are 0, hf, 2hf, and so on.(a) Prove by long division that...For what values of x does this series have a finite sum?(b) Evaluate the partition function for a single harmonic oscillator. Use the result of part(a) to simplify your answer as much as possible.(c) Use formula to find an expression for the average energy of a single oscillator at temperature T. Simplify your answer as much as possible.(d) What is the total energy of the system of N oscillators at temperature T? Your result should agree with what you found in Problem 1.(e) If you haven’t already done so in Problem 1, compute the heat capacity of this system aud check that it has the expected limits as T → 0 and T → ∞Formula:...Problem 1:In Problem 2 you showed that the multiplicity of an Einstein solid containing N oscillators and q energy units is approximately...(a) Starting with this formula, find an expression for the entropy of an Einstein solid as a function of N and q. Explain why the factors omitted from the formula have no effect on the entropy, when N and q are large. (b) Use the result of part (a) to calculate the temperature of an Einstein solid as a function of its energy. (The energy is U = qϵ, where ϵ is a constant.) Be sure to simplify your result as much as possible.(c) Invert the relation you found in part (b) to find the energy as a function of temperature, then differentiate to find a formula for the heat capacity. (d) Show that, in the limit T → ∞, the heat capacity is C = Nk. (Hint: When x is very small, ex ≈ 1 + x.) Is this the result you would expect? Explain.(e) Make a graph (possibly using a computer) of the result of part (c). To avoid awkward numerical factors, plot C/Nk vs. the dimensionless variable t = kT/ϵ, for t in the range from 0 to about 2. Discuss your prediction for the heat capacity at low temperature, comparing to the data for lead, aluminium, and diamond shown in Figure Estimate the value of ϵ, in electron-volts, for each of those real solids.(f) Derive a more accurate approximation for the heat capacity at high temperatures, by keeping terms through x3 in the expansions of the exponentials and then carefully expanding the denominator and multiplying everything out. Throw away terms that will be smaller than (ϵ/kT)2 in the final answer. When the smoke clears, you should find ...Problem 2:Use Stirling’s approximation to show that the multiplicity of an Einstein solid, for any large values of N and q, is approximately...The square root in the denominator is merely large, and can often be neglected. However, it is needed in below Problem 3 (Hint: First show that ...Do not neglect the ... in Stirling’s approximation.)Problem 3:This problem gives an alternative approach to estimating the width of the peak of the multiplicity function for a system of two large Einstein solids.a) Consider two identical Einstein solids, each with N oscillators, in thermal contact with each other. Suppose that the total number of energy units in the combined system is exactly 2N. How many different macrostates (that is, possible values for the total energy in the first, solid) are there for this combined system?b) Use the result of above Problem 2 to find an approximate expression, for the total number of microstates for the combined system. (Hint: Traet the combined system as a single Einstein solid. Do not throw away factors of “large” numbers, since you will eventually be dividing two “very large” numbers that are nearly equal. c) The most likely macrostate for this system is (of course) the one in which the energy is shared equally between the two solids. Use the result of above Problem 2 to find an approximate expression for the multiplicity of this macrostate.d)You can get a rough idea of the “sharpness” of the multiplicity function by comparing your answers to parts (b) and (c). Part (c) tells you the Height of the peak, while part (b) tells yon the total area under the entire graph. As a very crude approximation, pretend that the peak’s shape is rectangular. In this case, how wide would it be? Out of all the macrostates, what fraction have reasonably large probabilities? Evaluate this fraction numerically for the case N = 1023.Figure:Measured heat capacities at constant pressure (data points) for one mole each of three different elemental solids. The solid curves show the heat capacity at constant volume predicted by the model used in Section 7.5, with the horizontal scale chosen to best fit the data for each substance. At sufficiently high temperatures, Cv for each material approaches the value 3R predicted by the equipartition theorem. The discrepancies between the data and the sohd curves at high T are mostly due to the differences between Cp and Cv. At T = 0 all degrees 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

21. In the real world, most oscillators are not perfectly harmonic. For a quantum oscillator, this means that the spacing between energy levels is not exactly uniform. The vibrational levels of an H2 molecule, for example, are more accurately described by the approximate formula... n = 0, 1, 2, .. . ,where ϵ is the spacing between the two lowest levels. Thus, the level get closer together with increasing energy. (This formula is reasonably accurate only up to about n = 15; for sightly higher n it would say that En decreases with increasing n. In fact, the molecule dissociates and there are no more discrete levels beyond n ≈15.) Use a computer to calculate the partition function, average energy, and heat capacity of a system with this set of energy levels. Include all levels through n = 15, but check to see how the results change when you include fewer levels. Plot the heat capacity as a function of kT/ϵ Compare to the case of a perfectly harmonic oscillator with evenly spaced levels, and also to the vibrational portion of the graph in FigureFigure :Heat capacity at constant volume of one mole of hydrogen (H2) gas. Note that the temperature scale is logarithmic. Below about 100 K only the three translational degrees of freedom are active. Around room temperature the Two rotational degrees of freedom are active as well. Above 1000 K the two vibrational degrees of freedom also become active. At atmospheric pressure, hydrogen liquefies at 20 K and begins to dissociate at about 2000 K. Data from Woolley et al. (1948).... Get solution

22. In most paramagnetic materials, the individual magnetic particles have more than two independent states (orientations). The number of independent states depends on the particle’s angular momentum “quantum number” j, which must be a multiple of 1/ 2. For j = 1/ 2 there are just two independent states, as discussed in the text above and in Section 3.3. More generally, the allowed values of the z component of a particle’s magnetic moment are...where δμ is a constant, equal to the difference in µz between one state and the next. (When the particle’s angular momentum comes entirely from electron spins, δμ equals twice the Bohr magneton. When orbital angular momentum also contributes, δμ is somewhat different but comparable in magnitude. For an atomic nucleus, δμ is roughly a thousand times smaller.) Thus the number of states is 2j + 1. In the presence of a magnetic field B pointing in the z direction, the particle’s magnetic energy (neglecting interactions between dipoles) is − µzB.(a) Prove the following identity for the sum of a finite geometric series:...(Hint: Either prove this formula by induction on n, or write the series as a difference between two infinite series and use the result of Problem 1). (b) Show that the partition function of a single magnetic particle is...where b = βδμB (c) Show that the total magnetization of a system of N such particles is...Where coth x is the hyperbolic cotangent, equal to cosh x /sin h x. Plot the quantity ...vs. b, for a few different values of j . (d) Show that the magnetization has the expected behaviour as T → 0. (e) Show that the magnetization is proportional to 1/T (Curie’s law) in the limit T → ∞ (Hint: First show that coth...when x ≪ 1.) (f) Show hat for j = 1/ 2, the result of part (c) reduces to the formula derived in the text for a two-state paramagnet.Problem 1:This problem concerns a collection of N identical harmonic oscillators (perhaps an Einstein solid or the internal vibrations of gas molecules) at temperature T. As in Section 2.2, the allowed energies of each oscillator arc 0, hf, 2hf, and so on.(a) Prove by long division that...For what values of x does this series have a finite sum?(b) Evaluate the partition function for a single harmonic oscillator. Use the result of part(a) to simplify your answer as much as possible.(c) Use formula to find an expression for the average energy of a single oscillator at temperature T. Simplify your answer as much as possible.(d) What is the total energy of the system of N oscillators at temperature T? Your result should agree with what you found in Problem 1.(e) If you haven’t already done so in Problem 1, compute the heat capacity of this system aud check that it has the expected limits as T → 0 and T → ∞Formula:...Problem 2:In Problem 3 you showed that the multiplicity of an Einstein solid containing N oscillators and q energy units is approximately....(a) Starting with this formula, find an expression for the entropy of an Einstein solid as a function of N and q. Explain why the factors omitted from the formula have no effect on the entropy, when N and q are large. (b) Use the result of part (a) to calculate the temperature of an Einstein solid as a function of its energy. (The energy is U = qϵ, where ϵ is a constant.) Be sure to simplify your result as much as possible.(c) Invert the relation you found in part (b) to find the energy as a function of temperature, then differentiate to find a formula for the heat capacity. (d) Show that, in the limit T → ∞, the heat capacity is C = Nk. (Hint: When x is very small, ex ≈ 1 + x.) Is this the result you would expect? Explain.(e) Make a graph (possibly using a computer) of the result of part (c). To avoid awkward numerical factors, plot C/Nk vs. the dimensionless variable t = kT/ϵ, for t in the range from 0 to about 2. Discuss your prediction for the heat capacity at low temperature, comparing to the data for lead, aluminium, and diamond shown in Figure Estimate the value of ϵ, in electron-volts, for each of those real solids.(f) Derive a more accurate approximation for the heat capacity at high temperatures, by keeping terms through x3 in the expansions of the exponentials and then carefully expanding the denominator and multiplying everything out. Throw away terms that will be smaller than (ϵ/kT)2 in the final answer. When the smoke clears, you should find ...Problem 3:Use Stirling’s approximation to show that the multiplicity of an Einstein solid, for any large values of N and q, is approximately...The square root in the denominator is merely large, and can often be neglected. However, it is needed in below Problem 4 (Hint: First show that ...Do not neglect the ... in Stirling’s approximation.)Problem 4:This problem gives an alternative approach to estimating the width of the peak of the multiplicity function for a system of two large Einstein solids.a) Consider two identical Einstein solids, each with N oscillators, in thermal contact with each other. Suppose that the total number of energy units in the combined system is exactly 2N. How many different macrostates (that is, possible values for the total energy in the first, solid) are there for this combined system?b) Use the result of above Problem 2 to find an approximate expression, for the total number of microstates for the combined system. (Hint: Traet the combined system as a single Einstein solid. Do not throw away factors of “large” numbers, since you will eventually be dividing two “very large” numbers that are nearly equal. c) The most likely macrostate for this system is (of course) the one in which the energy is shared equally between the two solids. Use the result of above Problem 2 to find an approximate expression for the multiplicity of this macrostate.d)You can get a rough idea of the “sharpness” of the multiplicity function by comparing your answers to parts (b) and (c). Part (c) tells you the Height of the peak, while part (b) tells yon the total area under the entire graph. As a very crude approximation, pretend that the peak’s shape is rectangular. In this case, how wide would it be? Out of all the macrostates, what fraction have reasonably large probabilities? Evaluate this fraction numerically for the case N = 1023.Figure:Measured heat capacities at constant pressure (data points) for one mole each of three different elemental solids. The solid curves show the heat capacity at constant volume predicted by the model used in Section 7.5, with the horizontal scale chosen to best fit the data for each substance. At sufficiently high temperatures, Cv for each material approaches the value 3R predicted by the equipartition theorem. The discrepancies between the data and the sohd curves at high T are mostly due to the differences between Cp and Cv. At T = 0 all degrees 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

23. For a CO molecule, the constant ϵ is approximately 0.00024 eV. (This number is measured using microwave spectroscopy, that is, by measuring the microwave frequencies needed to excite the molecules into higher rotational states.) Calculate the rotational partition function for a CO molecule at room temparature (300K), first using the exact formula and then using the approximate formulaformula 1:...Formula 2:... Get solution

24. For an O2 molecule, the constant ϵ is approximately 0.00018 eV Estimate the rotational partition function for an O2 molecule at room temperature. Get solution

25. The analysis of this section applies also to linear polyatomic molecules, for which no rotation about the axis of symmetry is possible. An example is CO2, with ϵ = 0.000049 eV. Estimate the rotational partition function for a CO2 molecule at room temperature. (Note that the arrangement of the atoms is OCO, and the two oxygen atoms are identical.) Get solution

27. Use a computer to sum the exact rotational partition function (equation) numerically, and plot the result as a function of KT /ϵ. Keep enough terms in the sum to be confident that the series has converged. Show that the approximation in equation is a bit low, and estimate by how much. Explain the discrepancy by referring to FigureEquation 1:...Equation 2:...Figure. Bar-graph representations of the partition sum, for two different temperatures. At high temperatures the sum can be approximated as the area under a smooth curve.... Get solution

28. Use a computer to sum the rotational partition function (equation) algebraically, keeping terms through j = 6. Then calculate the average energy and the heat capacity. Plot the heat capacity for values of kT /ϵ. ranging from 0 to 3. Have you kept enough terms in Z to give accurate results within this temperature range?Equation:... Get solution

29. Although an ordinary H2 molecule consists of two identical atoms, this is not the case for the molecule HD, with one atom of deuterium (i.e., heavy hydrogen, 2H). Because of its small moment of inertia, the HD molecule has a relatively large value of ϵ: 0.0057 eV. At approximately what temperature would you expect the rotational heat capacity of a gas of HD molecules to “freeze out,” that is, to fall significantly below the constant value predicted by the equipartition theorem? Get solution

30. In this problem you will investigate the behaviour of ordinary hydrogen, H2, at low temperatures. The constant ϵ is 0.0076 eV. As noted in the text, only half of the terms in the rotational partition function, equation, contribute for any given molecule. More precisely, the set of allowed j values is determined by the spin configuration of the two atomic nuclei. There are four independent spin configurations, classified as a single “singlet” state and three “triplet” states. The time required for a molecule to convert between the singlet and triplet configurations is ordinarily quite long, so the properties of the two types of molecules can be studied independently. The singlet molecules are known as parahydrogen while the triplet molecules are known as orthohydrogen.(a)For parahydrogen, only the rotational states with even values of j are allowed.” Use a computer (as in Problem) to calculate the rotational partition function, average energy, and heat capacity of a parahydrogen molecule. Plot the heat capacity as a function of kT/ϵ.(b) For orthohydrogen, only the rotational states with odd values of j are allowed. Repeat part (a) for orthohydrogen. (c) At high temperature, where the number of accessible even-j states is essentially the same as the number of accessible odd-j states, a sample of hydrogen gas will ordinarily consist of a mixture of 1/4 parahydrogen and 3/4 orthohydrogen. A mixture with these proportions is called normal hydrogen. Suppose that normal hydrogen is cooled to low temperature without allowing the spin configurations of the molecules to change. Plot the rotational heat capacity of this mixture as a function of temperature. At what temperature does the rotational heat capacity fall to half its high temperature value (i.e., to k/2 per molecule)? (d) Suppose now that some hydrogen is cooled in the presence of a catalyst that allows the nuclear spins to frequently change alignment. In this case all terms in the original partition function are allowed, but the odd-j terms should be counted three times each because of the nuclear spin degeneracy. Calculate the rotational partition function, average energy, and heat capacity of this system, and plot the heat capacity as a function of kT/ϵ.(e) A deuterium molecule, D2, has nine independent nuclear spin configurations, of which six are “symmetric” and three are “anti symmetric.” The rule for nomenclature is that the variety with more independent states gels called “ortho-,” while the other gets called “para-.” For orthodeuterium only even-j rotational states are allowed, while for paradeuterium only odd. j states are allowed. Suppose, then , that a sample of D2 gas, consisting of a normal equilibrium mixture of 2/3 ortho and 1/3 para, is cooled without allowing the nuclear spin configurations to change. Calculate and plot the, rotational heat capacity of this system as a function of temperature....Problem:. Use a computer to sum the rotational partition function (equation) algebraically, keeping terms through j = 6. Then calculate the average energy and the heat capacity. Plot the heat capacity for values of kT /ϵ. ranging from 0 to 3. Have you kept enough terms in Z to give accurate results within this temperature range? Get solution

31. Consider a classical “degree of freedom” that is linear rather than quadratic: E = c|q| for some constant c. (An example would be the kinetic energy of a highly relativistic particle in one dimension, written in terms of its momentum.) Repeat the derivation of the equipartition theorem for this system, and show that the average energy is ... Get solution

32. Consider a classical particle moving in a one-dimensional potential well u (x), as shown in Figure The particle is in thermal equilibrium with a reservoir at temperature T, so the probabilities of its various states are determined by Boltzmann statistics.(a) Show that the average position of the particle is given by...where each integral is over the entire x axis.Figure: A one-dimensional potential well. The higher the temperature, the farther the particle will stray from the equilibrium point....(b) If the temperature is reasonably low (but still high enough for classical mechanics to apply), the particle will spend most of its time near the bottom of the potential well. In that case we can expand u(x) in a Taylor series about the equilibrium point x0...Show that the linear term must be zero, and that truncating the series after the quadratic term results in the trivial prediction ...(c) If we keep the cubic term in the Taylor series as well, the integrals in the formula for ... become difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent). Keeping only the smallest temperature-dependent term, show that in this limit ... differs from x0 by a term proportional to kT. Express the coefficient of this term in terms of the coeffcient of the Taylor series for u(x). (d) The interaction of noble gas atoms can be modeled using the Lennard Jones potential,...Sketch this function, and show that the minimum of the potential well is at x = xo, with depth uo. For argon, xo = 3.9 Å and uo = 0.010 eV. Expand the Lennard-Jones potential in a Taylor series about the equilibrium point, and use the result of part (c) to predict the linear thermal expansion coefficient of a noble gas crystal in terms of u0. Evaluate the result numerically for argon, and compare to the measured value α = 0.0007 K−1(at, 80 K).For a solid, we also define the linear thermal expansion coefficient, α, as the fractional increase in length per degree:...(a) For steel, α is 1.1 × 10−5 K−1. Estimate the total variation in length of a 1-km steel bridge between a cold winter night and a hot summer day.(b) The dial thermometer in Figure uses a coiled metal strip made of two different metals laminated together. Explain how this works.(c) Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three directions: β = αx + αy + αz. (So for an isotropic solid, which expands the same in all directions, β = 3α.)Figure: A selection of thermometers. In the center are two liquid-in-glass thermometers, which measure the expansion of mercury (for higher temperatures) and alcohol (for lower temperatures). The dial thermometer to the right measures the turning of a coil of metal, while the bulb apparatus behind it measures the pressure of a fixed volume of gas. The digital thermometer at left-rear uses a thermocouple—a junction of two metals—which generates a small temperature dependent voltage. At left-front is a set of three potter’s cones, which melt and droop at specified clay-firing temperatures.... Get solution

33. Calculate the most probable speed, average speed, and rms speed for oxygen (O2) molecules at room temperature. Get solution

34. Carefully plot the Maxwell speed distribution for nitrogen molecules at T = 300 K and at T = 600 K. Plot both graphs on the same axes, and label the axes with numbers. Get solution

35. Verify from the Maxwell speed distribution that the most likely speed of a molecule is ... Get solution

36. Fill in the steps between equations to determine the average speed of the molecules in an ideal gas. Get solution

37. Use the Maxwell distribution to calculate the average value of v2 for the molecules in an ideal gas. Check that your answer agrees with equation. Get solution

38. At room temperature, what fraction of the nitrogen molecules in the air are moving at less than 300 m/s? Get solution

39. A particle near earth’s surface traveling faster than about 11 km/s has enough kinetic energy to completely escape from the earth, despite earth’s gravitational pull. Molecules in the upper atmosphere that are moving faster than this will therefore escape if they do not suffer any collisions on the way out.(a) The temperature of earth’supper atmosphere is actually quite high, around 1000 K. Calculate the probability of a nitrogen molecule at this temperature moving faster than 11 km/s, and comment on the result. (b) Repeat the calculation for a hydrogen molecule (H2) and for a helium atom, and discuss the implications. (c) Escape speed from the moon’s surface is only about 2.4 km/s. Explain why the moon has no atmosphere Get solution

40. You might wonder why all the molecules in a gas in thermal equilibrium don’t have exactly the same speed. After all, when two molecules collide, doesn’t the faster one always lose energy and the slower one gain energy? And if so, wouldn’t repeated collisions eventually bring all the molecules to some common speed? Describe an example of a billiard-ball collision in which this is not the case: The faster ball gains energy and the slower ball loses energy. Include numbers, and be sure that your collision conserves both energy and momentum. Get solution

41. Imagine a world in which space is two-dimensional, but the laws of physics are otherwise the same. Derive the speed distribution formula for an ideal gas of nonrelativistic particles in this fictitious world, and sketch this distribution. Carefully explain the similarities and differences between the two-dimensional and three-dimensional cases. What is the most likely velocity vector? What is the most likely speed? Get solution

42. In Problem 1 you computed the partition function for a quantum harmonic oscillator: ... where ϵ = h f is the spacing between energy levels.(a) Find an expression for the Helmholtz free energy of a system of N harmonic oscillators.(b) Find an expression for the entropy of this system as a function of temperature. (Don’t worry, the result is fairly complicated .)Problem 1:This problem concerns a collection of N identical harmonic oscillators (perhaps an Einstein solid or the internal vibrations of gas molecules) at temperature T. As in Section 2.2, the allowed energies of each oscillator are 0, hf, 2hf, and so on.(a) Prove by long division that...For what values of x does this series have a finite sum?(b) Evaluate the partition function for a single harmonic oscillator. Use the result of part(a) to simplify your answer as much as possible.(c) Use formula to find an expression for the average energy of a single oscillator at temperature T. Simplify your answer as much as possible.(d) What is the total energy of the system of N oscillators at temperature T? Your result should agree with what you found in Problem 1.(e) If you haven’t already done so in Problem 1, compute the heat capacity of this system aud check that it has the expected limits as T → 0 and T → ∞Formula:...Problem 2:In Problem 3 you showed that the multiplicity of an Einstein solid containing N oscillators and q energy units is approximately...(a) Starting with this formula, find an expression for the entropy of an Einstein solid as a function of N and q. Explain why the factors omitted from the formula have no effect on the entropy, when N and q are large. (b) Use the result of part (a) to calculate the temperature of an Einstein solid as a function of its energy. (The energy is U = qϵ, where ϵ is a constant.) Be sure to simplify your result as much as possible.(c) Invert the relation you found in part (b) to find the energy as a function of temperature, then differentiate to find a formula for the heat capacity. (d) Show that, in the limit T → ∞, the heat capacity is C = Nk. (Hint: When x is very small, ex ≈ 1 + x.) Is this the result you would expect? Explain.(e) Make a graph (possibly using a computer) of the result of part (c). To avoid awkward numerical factors, plot C/Nk vs. the dimensionless variable t = kT/ϵ, for t in the range from 0 to about 2. Discuss your prediction for the heat capacity at low temperature, comparing to the data for lead, aluminium, and diamond shown in Figure Estimate the value of ϵ, in electron-volts, for each of those real solids.(f) Derive a more accurate approximation for the heat capacity at high temperatures, by keeping terms through x3 in the expansions of the exponentials and then carefully expanding the denominator and multiplying everything out. Throw away terms that will be smaller than (ϵ/kT)2 in the final answer. When the smoke clears, you should find ...Problem 3:Use Stirling’s approximation to show that the multiplicity of an Einstein solid, for any large values of N and q, is approximately...The square root in the denominator is merely large, and can often be neglected. However, it is needed in below Problem 4 (Hint: First show that ...Do not neglect the ... in Stirling’s approximation.)Problem 4:This problem gives an alternative approach to estimating the width of the peak of the multiplicity function for a system of two large Einstein solids.a) Consider two identical Einstein solids, each with N oscillators, in thermal contact with each other. Suppose that the total number of energy units in the combined system is exactly 2N. How many different macrostates (that is, possible values for the total energy in the first, solid) are there for this combined system?b) Use the result of above Problem 2 to find an approximate expression, for the total number of microstates for the combined system. (Hint: Traet the combined system as a single Einstein solid. Do not throw away factors of “large” numbers, since you will eventually be dividing two “very large” numbers that are nearly equal. c) The most likely macrostate for this system is (of course) the one in which the energy is shared equally between the two solids. Use the result of above Problem 2 to find an approximate expression for the multiplicity of this macrostate.d)You can get a rough idea of the “sharpness” of the multiplicity function by comparing your answers to parts (b) and (c). Part (c) tells you the Height of the peak, while part (b) tells yon the total area under the entire graph. As a very crude approximation, pretend that the peak’s shape is rectangular. In this case, how wide would it be? Out of all the macrostates, what fraction have reasonably large probabilities? Evaluate this fraction numerically for the case N = 1023.Figure:Measured heat capacities at constant pressure (data points) for one mole each of three different elemental solids. The solid curves show the heat capacity at constant volume predicted by the model used in Section 7.5, with the horizontal scale chosen to best fit the data for each substance. At sufficiently high temperatures, Cv for each material approaches the value 3R predicted by the equipartition theorem. The discrepancies between the data and the sohd curves at high T are mostly due to the differences between Cp and Cv. At T = 0 all degrees 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

43. Some advanced textbooks define entropy by the formula...where the sum runs over all. microstates accessible to the system and P(s) is the probability of the system being in microstate s . (a) For an isolated system, P(s) = l/ Ω for all accessible states s. Show that in this case the preceding formula reduces to our familiar definition of entropy. (b) For a system in thermal equilibrium with a reservoir at temperature T, P(s) =.... Show that in this case as well, the preceding formula agrees with what we already know about entropy. Get solution

44. Consider a large system of N indistinguishable, noninteracting molecules. (perhaps in an Ideal gas or a dilute solution) . find an expression for the Helmholtz free energy of this system, in terms of Z1 , the partition function for a single molecule. (Use Stirling’ s approximation to eliminate the N!) Then use your result to find the chemical potential, again in terms of Zl. Get solution

45. Derive equations and for entropy and chemical potential of an ideal gas.Equation 1:...Equation 2:... Get solution

46. Equations and for the entropy and chemical potential involve the logarithm of the quantity .... Is this logarithm normally positive or negative? Plug in some numbers for an ordinary gas and discuss.Equation 1:...Equation 2:... Get solution

47. Estimate the temperature at which the translational motion of a nitrogen molecule would freeze out, in a box of width 1 cm. Get solution

48. For a diatomic gas near room temperature, the internal parti­tion function is simply the rotational partition function computed in Section 6.2, multiplied by the degeneracy Ze of tlie electronic ground state.(a) Show that the entropy in this case is...Calculate the entropy of a mole of oxygen (Ze = 3) at room temperature and atmospheric pressure, and compare to the measured value in the table at the back of this book. (b)Calculate the chemical potential of oxygen in earth’s atmosphere near sea level, at room temperature. Express the answer in electron-volts. Get solution

49. For a mole of nitrogen (N2) gas at room temperature and atmospheric pressure, compute the following: U, H, F, G, S, and µ. (The rotational constant ϵ for N2 is 9.00025 eV. The electronic ground state is not degenerate.) Get solution

50. Show explicitly from the results of this section that G = N/µ for an ideal gas. Get solution

51. In this section we computed the single-particle translational partition function, Ztr by summing over all definite-energy wavefunctions. An alternative approach, however, is to sum over all possible position and momentum vectors, as we did in Section 2.5. Because position and momentum are continuous variables, the sums are really integrals, and we need to slip in a factor of 1/h3 to get a unitless number that actually counts the independent wave functions. Thus, we might guess the formula...where the single integral sign actually represents six integrals, three over the position components (denoted d3r) and three over the momentum components (denoted d3p). The region of integration includes all momentum vectors, but only those position vectors that lie within a box of volume V. By evaluating the integrals explicitly, show that this expression yields the same result for the translational partition function as that obtained in the text. (The only time this formula would not be valid would be when the box is so small that we could not justify converting the sum in equation to an integral.) Get solution

52. Consider an ideal gas of highly relativistic particles (such as photons or fast-moving electrons), whose energy-momentum relation is E = pc instead of E = p2/2m. Assume that these particles live in a one-dimensional universe. By following the same logic as above, derive a formula for the single-particle partition function, Z1, for one particle in this gas. Get solution

53. The dissociation of molecular hydrogen into atomic hydrogen, ... can be treated as an ideal gas reaction using the techniques of Section 5.6. The equilibrium constant K for this reaction is defined as...where P° is a reference pressure conventionally taken to be 1 bar, and the other P’s are the partial pressures of the two species at equilibrium. Now, using the methods of Boltzmann statistics developed in this chapter, you are ready to calculate K from first principles. Do so. That is, derive a formula for K in terms of more basic quantities such as the energy needed to dissociate one molecule (see Problem) and the internal partition function for molecular hydrogen. This internal partition function is a product of rotational and vibrational contributions, which you can estimate using the methods and data in Section 6.2. (An H2 molecule doesn’t have any electronic spin degeneracy, but an H atom does—the electron can be in two different spin states. Neglect electronic excited states, which are important only at very high temperatures. The degeneracy due to nuclear spin alignments cancels, but include it if you wish.) Calculate K numerically at T = 300 K, 1000 K, 3000 K, and 6000 K. Discuss the implications, working out a couple of numerical examples to show when hydrogen is mostly dissociated and when it is not.Problem:Look up the enthalpy of formation of atomic hydrogen in the back of this book. This is the enthalpy change when a mole of atomic hydrogen is formed by dissociating 1/2 mole of molecular hydrogen (the more stable State of the element). From this number, determine the energy needed to dissociate a single H2 molecule, in electron-volts. Get solution


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