This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: CHEM 131 K. Lindenberg Fall 2007 Solutions to Homework No. 2 Posted Monday October 15 These solutions to Homework No. 2 were written partly by Dario and partly by Jim. Note: they chose how to divide up the task, so the solutions are not in the order in which they were assigned. However, the solutions to all the problems are there. Chemistry 131, Fall 2007 Solutions to Homework No. 2 Problem 1. See Levine, problem 3.1. (a) True. The statement may be obvious from e rev = 1- T C T H (Levine eq. 3.15), but a more general approach to test the validity of the statement to look at the sign of the partial derivative of e rev with respect to T H : e rev T H T C = T C T 2 H , which is positive for all non-zero temperatures. Therefore, increasing T H must cause e rev to increase. (b) True. Reapplying the approach of part (a), e rev T C T H =- 1 T H < , for all non-zero temperatures, so increasing T C must cause e rev to decrease. (c) True. See Levine p. 81. (d) False. Work is not a state function so the work done in a Carnot cycle need not be zero. The work would be zero if the efficiency of the engine was zero, since- w = e rev q H , which would imply that H = C and would thus contradic the assumption that H > C (Levine p. 79). So the work done by one Carnot cycle cannot be zero. Problem 3. See Levine, problem 3.21. (a) False. S can be negative for a closed system, because such a system is allowed to exchange heat with the surround- ings. So it is possible to have dS = dq rev /T < 0, and hence S < 0 for some process involving a closed system. (b) False. S = 0 for the universe in any reversible process (Levine eq. 3.35). But the entropy of a closed system can change in a reversible process. For example, in a reversible isothermal process the entropy change is S = q rev /T (Levine eq. 3.27). (c) True. S univ = 0 for any reversible process (Levine eq. 3.35). (d) True. For S to be negative heat would have to flow out of the system, since S = R 2 1 dq rev /T (Levine eq. 3.21). But such heat flow is not possible during an adiabatic process. (e) True. Follows from part (d), since an isolated system is also a closed system and any process in an isolated system is also an adiabatic process....
View Full Document