University Physics with Modern Physics with Mastering Physics (11th Edition)

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20.58: a) For constant-volume processes for an ideal gas, the result of Example 20.10 may be used; the entropy changes are ). ln( and ) ln( d a V b c V T T nC T T nC b) The total entropy change for one cycle is the sum of the entropy changes found in part (a); the other processes in the cycle are adiabatic, with Q = 0 and . 0 = S The total is then . ln ln ln = + = d a a c V d a V b c V T T T T nC T T nC T T nC
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Unformatted text preview: S From the derivation of Eq. (20.6), , and 1 1 d γ c a γ b T r T T r T--= = and so the argument of the logarithm in the expression for the net entropy change is 1 identically, and the net entropy change is zero. c) The system is not isolated, and a zero change of entropy for an irreversible system is certainly possible....
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This document was uploaded on 02/06/2008.

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