NAME and Section No. b). A refrigerator is a Carnot cycle run backwards. That is, heat is now withdrawn from the cold reservoir at T cold

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1 Chemistry 391 Fall 007 Exam II KEY 1. (30 Points) ***Do 5 out of 7***(If 6 or 7 are done only the first 5 will be graded)*** a). How does the efficiency of a reversible engine compare with that of an irreversible engine? How can you improve the efficiency of an irreversible engine? he efficiency of an irreversible engine must be less than that of a reversible, with equality only when both are reversible. he only way to improve the efficiency of an irreversible engine is to make it operate closer to reversibly, e.g., by making it operate more slowly. b). A refrigerator is a Carnot cycle run backwards. hat is, heat is now withdrawn from the cold reservoir at cold and is deposited in the hot reservoir at hot. he coefficient of q performance of a refrigerator η is defined as η w hot. Since du = 0 = wcycle + qcold + qhot, and since qcold qhot = cold hot qcold qcold cold η = = = w q + q ( ) cycle cold hot hot cold cold cycle. Express η in terms of cold and c). You are told that ΔS sys = 0 for a process in which the system is coupled to its surroundings. Can you conclude that the process is reversible? Justify your answer. No. he criterion for reversibility is ΔS sys + ΔS surroundings = 0. o decide if this criterion is satisfied, ΔS surroundings must be known. 1

2 d). How and why is it possible to calculate Δ S in an irreversible transformation (since it only obeys the Clausius inequality)? It can be done by finding a reversible path that connects the same initial and final states as for the irreversible transformation. his is possible because entropy is a state function. e). Calculate At fixed pressure Δ G for melting ice reversibly at fixed pressure. Δ H = q and qp Δ G =ΔH Δ S = qp = 0. p q p Δ S = so f). From the definition of the enthalpy obtain the Gibbs differential expression for dh from the Gibbs differential expression for du. ( ) dh = du + d p = du + pd + dp = ds pd + pd + dp dh = ds + dp g) Is Δ A positive, negative or zero for an adiabatic expansion of an ideal gas into a vacuum? he expansion is free ( 0 ext p = ) so no work and adiabatic so no heat. hus Δ U( ) = 0 and is fixed. For a free, irreversible expansion Δ A=ΔU Δ S = Δ S is negative. Δ S is positive and

3 . (5 points). Show that for du. U p = α + pκ by starting from the Gibbs differential du = ds pd. hen taking the indicated derivative U S p = p p p Use the Maxwell from the dg Gibbs differential S = p p to obtain U p = p p p he definitions of α and κ on the equation sheet gives the answer U = α + pκ. p 3

4 3. (0 points) A general expression for the entropy change in the transformation from state 1 = 1, p1 to state =, p is (don t assume ideal gas) (, p) C P Δ S = d (, p) α (, p) dp 1 1 A. Define a path that simplifies this so it can be evaluated using ordinary integrals and carefully indicate what the integrals to be done are: You can for example use 1 = 1, p1 >, p1 >, p = (, p ) C p P 1 Δ S = d (, p) α (, p) dp 1 p1 B. Assuming that the and p dependencies of the various coefficients can be neglected, generate an expression to evaluate Δ S. Δ = 1 = ( ) p S Cp d α dp Cpln p p1 1 α p1 1 4

5 4. ( 5 points) A. hinking of S S(, ) =, write down the partial differential expression for ds. S S ds = d + d S B. Noting that C = and that you want to use equation of state data in your expression, express ds suitably. S p Need the Maxwell from Helmholtz = and with C definition C p ds = d + d C. Assume that we are dealing with an ideal gas, show that the expression for ds is indeed exact (show equality of mixed partials). For ideal gas exact. p nr =. he mixed derivative is ( nr / ) C / v = 0 = so 5

6 D. Assuming that C is constant, evaluate a finite change in entropy gas. Δ S for this ideal. ( ) ln ( ) Δ S = C ln + nr 1 1 E. If > 1 for fixed what sign is the entropy change? If > 1 for fixed what sign is the entropy change? Discuss the result for the volume increase. For both situations the entropy change is positive. A larger volume means the gas has more room to occupy and so more ways of arrangement; hence, a larger entropy. 6

7 Mathematical expressions: n ( n+ 1) xdx x n n = /( + 1); = 0,1,,... 1 dx = ln x x ( n 1) ( ) n x dx= x / n 1 ; n=,3,... x y z Euler: = 1 y zx xy z x y Inverter: = 1 y x z z 7

8 i n dμ + Sd dp = 0 (Gibbs-Duhem) i i νμ i i = 0 (reaction equilibrium). i μ = μ + Rln a ; ai = γ ixi; 0 i i i 8