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Unformatted text preview: Chemistry 131, Fall 2007 Solutions to Homework No. 3 Problem 1. See Levine, problem 3.41. (a) The second law cannot prove this statement because the system is not isolated. According to the second law, the position of maximum entropy corresponds to equilibrium of an isolated system (Levine p. 3.6). Since a closed system can interact with the surroundings, equilibrium is achieved at the position of maximum entropy for the combined system and surroundings. (b) The second law cannot prove that the entropy of an iso lated system must remain constant, unless the system is known to be at equilibrium (Levine p. 93). (c) Since the walls of the system are adiabatic and imper meable, neither heat nor matter can be extracted from the system so there is no way to decrease the entropy of this system. This system is thus equivalent to a closed system that can undergo only adiabatic processes. In this case, the second law can prove that Δ S syst > 0 for an irreversible pro cess (Levine eq. 3.37), while Δ S syst = 0 for reversible adia batic processes follows from the definition of dS (Levine eq. 3.23). Therefore, the second law implies that entropy can not decrease for this system, i.e., when this system achieves equilibrium with the surroundings the entropy of the system is maximized. (d) The second law does not prevent the entropy of a closed system, which is not isolated, from decreasing. Decreasing the entropy of a closed system does not violate the second law, as long as the entropy of the universe increases (Levine p. 93). (e) The second law can prove that the entropy of an iso lated system must increase in an irreversible process (Levine eq. 3.38). Also, the entropy of an isolated system cannot change in a reversible process because such a system can not exchange heat with the surroundings, so that Δ S = R 2 1 dq rev /T = 0. Hence, according to the second law the entropy of an isolated system cannot decrease. Problem 3. See Levine, problem 4.1. (a) True. The quantities U , H , A , and G all have the dimen sions of energy. (b) False. Since G = H TS (Levine eq. 4.17), then in general (Levine p. 124) Δ G = Δ H Δ( TS ) = Δ H ( T 2 S 2 T 1 S 1 ) = Δ H ( T 2 S 2 T 2 S 1 + T 2 S 1 T 1 S 1 ) = Δ H [ T 2 ( S 2 S 1 ) + ( T 2 T 1 ) S 1 ] = Δ H ( T 2 Δ S + S 1 Δ T ) = Δ H ( T 2 Δ S T 1 Δ S + T 1 Δ S + S 1 Δ T ) = Δ H [( T 2 T 1 )Δ S + T 1 Δ S + S 1 Δ T ] = Δ H [(Δ...
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This note was uploaded on 05/22/2008 for the course CHEM 131 taught by Professor Lindenberg during the Fall '08 term at UCSD.
 Fall '08
 Lindenberg
 Physical chemistry, Enthalpy, pH

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