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Unformatted text preview: T and V shown above. (c) For a monatomic gas, = 5 / 3 (see the discussion in Chapter 20). Therefore, T f = T i V i V f 1 = T 1 2 2 / 3 = 0 . 63 T which implied process AF is adiabatic. (d) Since ln( x ) is positive for all x > 1, then Eq. 21-4 makes it clear that all processes (with the possible exception of AF ) have S > 0. We assume process AF to be reversibly adiabatic, in which case Eq. 21-1 gives S = 0 (since Q = 0 for the process, or any small portion of the process); in fact, if AF represented (in some sense) an irreversible process which generated entropy, then we would still end up with the overall conclusion that none of the processes shown are accompanied by an entropy decrease....
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- Fall '08