PY Introduction to thermodynamics and statistical physics Summer Examination suggested solutions Dr. Asaf Pe er

Transcription

1 PY Introduction to thermodynamics and statistical physics Summer Examination suggested solutions Dr. Asaf Pe er 1) Onemoleofwater(H 2 O)isbeingheatedfrom 20 Cto+80 C,atconstant atmospheric pressure. (i) Calculate the entropy change s during the process. Which stage gives he highest contribution to the entropy? [4 /10] There are 3 stages in the process. (I) heating the ice from 20 C to 0 C. The entropy change is S 1 = C p (Ice)ln(T 2 /T 1 ) = 9 log(273/253) = cal/ K mole. (II) Phase transition: S 2 = L T = = cal/ K mole. (III)Heatingtheliquidwateratconstantpressure: S 3 = C p (water)ln(t 3 /T 2 ) = 18 log(353/273) = cal/ K mole. Thus,thetotalentropychangeis S = S 1 + S 2 + S 3 = 10.6cal/ K mole. Clearly, the phase transition has the largest contribution to the entropy change. (ii) What is the total enthalpy change in the process, H? Using the fact that at constant pressure H = C p T, we find that H 1 = C p (Ice)(T 2 T 1 ) = 180 cal, and H 3 = C p (water)(t 3 T 2 ) = 1440 cal. During the phase transition, we have H = T S+V p = T S, and therefore H 2 = = 1440 cal. Thetotalenthalpychangeisthus H = H 1 + H 2 + H 3 = 3060cal. (iii) What is the change of internal energy, E? Assuming that the change in volume of ice and water is negligible, we have E = H (pv) = H = 3060 cal. (iv) Sketch a phase diagram for water, which need not be to scale but should be qualitatively correct. Explain what it means by triple point and critical point. Plot of the phase diagram appears in Figure 1. At the triple point the three phases - solid, liquid and gas can coexist. At the critical point, the latent heat and volume change in phase transition between liquid and gas becomes zero, and thus at temperatures higher than the critical temperature gas and liquid form a single phase.

2 Figure 1: Phase diagram for water. Note. You may use the following data. C p (Ice) = 9 cal/ K mole; C p (water) = 18 cal/ K mole; Latent heat of water L = 80 cal/gr. Molar mass of Hydrogen is 1 gr/mole and of Oxygen is 16 gr/mole. You may also assume that the volume change of ice and of liquid water during the process can be neglected. 2) The partition function for a perfect gas containing N monatomic particles of mass m at temperature T is Z = 1 N! V N ( 2πmkB T h 2 ) 3N/2. (i) Using the partition function, calculate the Helmholtz free energy. The Helmholtz free energy is F = k B T lnz, or F = k B T ( lnn!+n lnz 1 ) = Nk B T ( ln ( ) ) e ( [ N +lnz1 = Nk B T ln ev N ( 2πmkBT ) 3/2 ]) (1) h 2 (ii) Using the Helmholtz free energy you obtained in the previous section, calculate the pressure of the gas. Compare your result to the thermodynamic equation of state.

3 From the definition of Helmholtz free energy, F = E TS, df = pdv SdT, and thus ( ) F p =. V T Thus, p = V = V = NkBT V { ( [ ev Nk B T ln N { Nk B T lnv Nk B T ( 2πmkBT ) 3/2 ])} h 2 ( [ ln e N ( 2πmkBT h 2 ) 3/2 ])} (iii) Using the partition function given above, write the average number of particles having translational energy E s. (2) The probability of the gas molecule to be in state of translational energy E s is P s = e βes /Z 1, and thus n s = NP s = Ne βes /Z 1 = ( N 2πmkb T V e βes h 2 ) 3/2. (iv) Write the condition of validity of the classical approximation. Express h your answer in terms of de-broglie wavelength, λ db = 3mkBT. The classical approximation is valid as long as n s 1. Using λ db = we can write this as h 3mkBT ( ) 3/2 ( ) 3 3 λdb 1 2π l, where l = (V/N) 1/3 is the mean distance between the particles. Thus, the de-broglie wavelength must be small compared to the mean separation between the molecules. 3) 10 litres of ideal gas at atmospheric pressure at 25 (Celsius) is compressed isothermally to a volume of 1 litre, and then allowed to expand adiabatically to 10 litres. Finally, the pressure of the gas is changed (at constant volume) back to atmospheric pressure. Consider first a monatomic gas (for which C p /C v = 5/3). i) Sketch the processes on a pv diagram. ii) For each of the stages, write the work done on (or by) the gas, and the heat exchange with its surroundings. Let us denote the work done by the gas as positive. During isothermal compression, work is done on the gas, and thus W = PdV = rt1 ln(v 2 /V 1 ), where T 1 is the initial gas temperature, V 1 =

4 10 l, and V 2 = 1 l. Since the internal energy doesn t change, u = 0, we find that the gas absorbs heat Q = rt 0 ln(v 2 /V 1 ). During adiabatic expansion, Q = 0 and W = 1 γ 1 (p 2V 2 P 3 V 3 ), where V 3 = V 1. Since the gas is at higher temperature, (T 3 > T 1 ), the pressure is higher, P 3 > P 1. During pressure decrease at constant volume, W = 0, and Q = c v (T 1 T 3 ) (the gas releases heat). iii) If the gas was diatomic (for which C p /C v = 7/5), would the net work done on (or by) the gas be greater or less than in the monatomic gas case? Explain. During isothermal compression, p 2 = V 1 p 1 /V 2. Thus, the curves of monatomic and diatomic gases are similar. During adiabatic expansion, pv γ = const, or p 3 = (V 2 /V 3 ) γ p 2 = p 2 10 γ. For diatomic gas γ = 7/5, and thus p 3 = p , while for monatomic gas, p 3 = p Thus, the net work done on monatomic gas is greater than on diatomic gas. iv) Does the eficciency of this cycle greater than, equal to or lower than the efficiency of a Carnot cycle? Explain. Hint: there is no need to calculate the actual efficiency, but only to explain the basic physical differences (if there are any), or alternatively explain why this cycle is conceptually similar to a Carnot cycle. The efficiency of this cycle is less than that of a Carnot cycle. The reason is that the isochoric pressure decrease is irreversible. 4) The elasticity of a rubber band can be described in terms of a one-dimensional model of polymer involving N molecules linked together end to end. The anglebetweensuccessivelinksisequallylikely0 or180 (namely, themolecules can be aligned either in the +x or the x directions only). (i) Show that the number of arrangements that give an overall length L = 2md (d is the size of each molecule) is given by g(n,m) = where m is a positive number. 2N! ( N 2 +m)! ( N 2 m)! Assume that there are N + links of 0 angle, and N links of 180 angle, then N + +N = N ; N + N = 2m. Thus, N + = N 2 +m, N = N 2 m.

5 This corresponds to N!/(N +!N!) arrangements. Note that for every arrangement if the angles are reserved, we still get an overall length of 2md. Thus, 2N! g(n,m) = ( N 2 +m)! ( N 2 m)! (ii) For m N, this expression becomes g(n,m) g(n,0)exp( 2m 2 /N). Find the entropy of the system as a function of L for N 1, L Nd. When m N, g(n,m) g(n,0)exp( 2m 2 /N), the entropy of the system becomes S = k B lng(n,m) = k B lng(n,0) k BL 2 2Nd 2. (iii) Find the force required to maintain the length L for L Nd. Hint. Note that in this system, the length L plays a similar role to that of the volume p, and thus the force f plays a similar to that of the pressure, p. You can therefore use the thermodynamic relation de = TdS fdl, where f is the force, as well as F = E TS. Using the thermodynamic relations, du = TdS +fdl (f is the force) and F = U TS, we obtain df = SdT +fdl. Thus, ( ) ( ) f S = = k BL T L Nd 2, L from which f = k BTL Nd 2 +Const. Since f = 0 at L = 0, the constant of integration is 0, and we get T f = k BTL Nd 2. (iv) Find the relationship between the force and the length, without using the condition in the previous section, namely for any possible value of L, with N 1. Consider only one link. When an external force f is exerted, the probability that the angle is 0 or 180 is proportional to e α or e α respectively, where α = fd/k B T. The average length per link is therefore l = d eα e α e α = dtanhα. +e α The overall length of the polymer is then N l = Ndtanh(df/k B T).

6 Asaf Pe er, July 9, 2014