Thermodynamics and Statistical Physics Exam

Transcription

1 Thermodynamics and Statistical Physics Exam You may use your textbook (Thermal Physics by Schroeder) and a calculator. 1. Short questions. No calculation needed. (a) Two rooms A and B in a building are connected by an open door. One room (say, A) is on the sunny side of the building and is warmer than the other room. Which room has more oxygen per unit volume? Explain. SOLUTION: The system must at least be in mechanical equilibrium, even if not in thermal equilibrium. Thus, the pressures must be equal. This means that A, being at a higher temperature, will have a lower density of nitrogen and oxygen. (b) The room-temperature heat capacity of 1 kg of iron (Fe) is about 450 J/K, while the heat capacity of 1 kg of lead (Pb) is only about 125 J/K. Explain the difference (qualitatively). SOLUTION: What counts for the heat capacity is the number of degrees of freedom. Lead atoms are heavier, implying a smaller number of them per kg. Thus, the heat capacity is expected to be lower per kg. 2. Starting from the Clausius-Clapeyron equation (5.47), consider the case of a liquid-gas phase transition, in which the latent heat L is constant and the volume V g of the gas phase is much larger than the volume V l of the liquid. By approximating V = V g V l V g and assuming that the gas is ideal, derive a differential equation for the vapor pressure P as a function of T, and show that the solution is P (T ) = const. e l/kt, where l is the latent heat per molecule. SOLUTION: This was essentially one of your homework problems. See the solutions to problem 5.35.

2 3. The following is a simple model of an electrically polarizable solid. Consider N independent, identical charges q that are each bound to lattice sites by identical springs with spring constant K. In the presence of an electric field E, the energy of a charge is U = 1 2 Kx2 Eqx, where x is the displacement of the charge. (a) Apart from an overall constant, what is the probability as a function of x of an individual charge at fixed temperature T and applied field E? SOLUTION: The probability where x 0 = Eq/K. P(x) e β( 1 2 Kx2 Eqx) = const. e βk(x x0)2 /2, (b) Find the average displacement x for one charge, and the total polarization of the solid P = Nq x. SOLUTION: Due to the symmetry of the distribution above, the average x = x 0, which also corresponds to the maximum of the distribution. Since the charges are assumed to be independent of each other, the total polarization simply adds, as indicated above. (c) Calculate the partition function Z(T, E) and find the average energy U. Z(T, E) = e 1 2 βkx2 0 e 1 2 βk(x x0)2 dx = e 1 2 βkx2 0 2πkT K. From this, we find U = β log Z = 1 2 Kx kt. 2

3 4. The following is a somewhat more realistic model of a rubber band than you considered in a homework problem. Here, the entropy S as a function of energy U and end-to-end distance x is given by x2 S(U, x) = k + Nk log U + constant. 2Na2 Here, k is Boltzmann s constant, N represents the number of links in the chain-like molecules making up the rubber band, and a is a molecular size. (a) Find the energy U as a function of temperature T. SOLUTION: 1 T = S U = Nk U, or U = NkT. (b) Consider a small, adiabatic (dq = 0) expansion of the rubber band from x to x + dx, were x > 0 and dx > 0. Does the temperature increase or decrease? By how much? SOLUTION: From the first law, du = dq + fdx = fdx for an adiabatic process (see solutions to problem 3.34). Thus, du = NkdT = fdx = kt Na xdx, or 2 dt = T xdx N 2 a 2 > 0, where I have used f = T S x = kt Na 2 x (again, see problem 3.34). Thus, the system heats up. We could have derived this another way. An adiabatic expansion is reversible, and S remains constant. Thus, Using this, we find x2 k = Nk log U + constant. 2Na2 kx Na Nk dx = 2 U du = Nk T dt, which is equivalent to the expression for dt above. 3

4 5. (a) In the nineteenth century, using Maxwell s equations, it was already known that the electromagnetic pressure exerted on the walls of an evacuated container of volume V was related to the energy U by the formula P = 1 3 U/V. Assuming that the entropy S(U, V ) is a function of the energy and volume, and that the energy density U V = AT λ, where both A and λ are constants, show that λ must be 4. (This is a result first derived by Boltzmann.) SOLUTION: If S is a function only of U and V (no N), then we can invert this to obtain U as a function only of S and V (again, no N). Then, from the first law, du = T ds P dv represents a complete and general expression for du. But, by the assumption above, U = AV T λ, and du = AV λt λ 1 dt + AT λ dv = T ds P dv. Solving this for ds gives ds = AV λt λ 2 dt AT λ 1 dv. Thus, S/ T = AV λt λ 2 and S/ V = 4 3 AλT λ 1. By equating 2 S/ V T and 2 S/ T V, we have AλT λ 2 = 4 3 A(λ 1)T λ 2, the only solution of which is λ = 4. 4

5 (b) Photons in one dimension. Consider a one-dimensional box of size L at temperature T. In equilibrium, this box will be filled with a gas of photons characterized by energy ɛ = hc 2L j, where j = 1, 2,.... Show that the photon energy in the box is U = A L T λ, where A and λ are constants. Find λ. (You do not need to give a precise value for A. You only need to show that it is finite and that it depends on only mathematical and physical constants.) SOLUTION: Here, U = j=1 ( ) hc 2L j 1 e βhcj/2l 1 )2 2(kT L hc 0 x e x 1 dx. I am neglecting a possible number of polarization states (in 1D?), but I am only asking for an answer up to a possible multiplicative constant. The integrand above is finite near x = 0, so the integral is certainly finite. Thus, U = ALT λ and λ = 2. 5