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PHSX 446 EXAM 2 – SOLUTIONS Spring 2015 1 1. First, Some Basic

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PHSX 446 EXAM 2 – SOLUTIONS Spring 2015 1 1. First, Some Basic PHSX 446 EXAM 2 – SOLUTIONS Spring 2015 1 1. First, some basic knowledge questions. You need not show work here; just give the answer. More than one answer might apply. Don’t waste time transcribing answers; just write on this exam sheet. (a) An inexact differential is: a) inexpressible as a function, b) of no use in mathematics or physics, c) only approximately correct. (b) Maxwell relations express: a) conservation laws, b) the conditions under which phase transitions occur, c) relations among thermodynamic variables, d) the properties of ideal gases only. (c) Does equipartition always hold for a gas in equilibrium? Briefly explain. No. Equiparition holds for an ideal gas, and generally not for a non-ideal gas. (d) Give an example of an adiabatic process that is not reversible. Thermal diffusion in a closed system, mixing, free expansion. (e) Give an example of a reversible process that is not adiabatic. Heating of a system from one temperature to another by an outside heat source. The heat can always be removed (by putting the system in contact with a reservoir at the original temperature), which will restore the original state with unchanged entropy. The pressure, or volume, or both, could change. (f) What property must a gas have for it to cool in the Joule-Thompson (throt- tling) process? The gas must be non-ideal. (g) Write down the expression for the heat current in thermal conduction. ∂T j = −k one dimension. ∂x (h) Why are engines usually cyclical? Because we want to repeat the process to get some amount of work at each cycle, rather than do it only once. PHSX 446 EXAM 2 – SOLUTIONS Spring 2015 2 (i) Explain why the efficiency of even an ideal engine is less than one. The max- imum theoretical efficiency is the Carnot efficiency: Tc ecarnot = 1 − . Th The efficiency is unity only if the engine exhausts to a reservoir at absolute zero. Absolute zero is unattainable. (j) What is the maximum number of variables needed to express a thermodynamic state? 3. 2. This is a warm-up exercise. Recall that the Helmholtz free energy is F = E − TS. Use the first law to express dF in terms of S, p, µ, dT , dV , and dN. From your expression for dF , derive two Maxwell relations. Explain exactly how you are obtaining these relations. Take the differential of F : dF = dE − T dS − SdT. From the First Law, dE = T dS − pdV + µdN, giving dF = −SdT − pdV + µdN Since dF is an exact differential, we obtain ∂S ! ∂p ! ∂p ! ∂µ ! ∂S ! ∂µ ! = = − = − ∂V T,N ∂T V,N ∂N T,V ∂V T,N ∂N T,V ∂T V,N 3. One mole of an ideal gas at temperature Ti and volume Vi is held in a vessel by a piston. The gas has ν degrees of freedom. The adiabatic index of the gas is γ ≡ (ν + 2)/ν. (This expression had a typo on the exam, which did not affect anybody). (a) The piston adiabatically compresses the gas from volume Vi to volume Va. Find the temperature Ta after this process. Evaluate the work done by the piston on the gas. For an adiabatic process γ−1 γ−1 γ−1 Vi TaVa = TiVi → Ta = Ti . Va PHSX 446 EXAM 2 – SOLUTIONS Spring 2015 3 The work done by the piston on the gas is Z a W = pdV i For an adiabatic process γ γ pV = piVi , giving γ Z Va γ −γ piVi 1−γ 1−γ W = piVi dV V = Vi − Va . Vi γ − 1 (b) The piston is withdrawn so that the gas expands back to its original volume Vi. Just the right amount of heat is added to the gas from outside to ensure the expansion process occurs at constant pressure. Find the temperature Tb after this process, in terms of Ta. According to the ideal gas law, for one mole: Tb Ta Vi pV = RT → = → Tb = Ta . Vi Va Va (c) Heat Q is injected into the vessel. The piston is held fixed. Find the tempera- ture Tc after this process, in terms of Tb. One mole of an ideal gas has a heat capacity at constant pressure of CV = Rν/2. Since CV is independent of temperature: 2Q Q = C (T − T ) → T = T + . V c b c b Rν (d) The piston is quickly pulled out of the vessel, and the gas undergoes free expansion. Find the final temperature of the gas Tf in terms of the initial temperature Ti. Your answer should depend only on Ti, Vi, Va, γ, ν, Q, and the gas constant R. Do not introduce any new variables. Upon free expansion, the temperature of an ideal gas does not change, so in this case, the temperature is Tc. Putting together the results for Tb and Ta gives γ Vi 2Q Tf = Ti + . Va Rν Since Vi >> Va, the gas heats up. Heat was added in processes (b) and (c)..
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