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dQ = −TdS(produced) |
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| Apr9-10, 01:56 AM | #1 |
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dQ = −TdS(produced)
1. The problem statement, all variables and given/known data
(a) Show that dU < TdS − PextdV , using the first and second laws, for an infinitasimal spontaneous process in an open system. (b) Thus dU < 0 could be viewed as a criterion for spontaneity at constant S, V . Show that dQ = −TdSproduced in such a process. Is heat absorbed or lost in a constant S, V process? 2. Relevant equations 3. The attempt at a solution I did part (a) easily enough. I just included it for background information. I'm having troubles with the idea that d(bar)Q = -TdS(produced) since the 2nd law of thermodynamics says d(bar)Q = TdS. Any help with this would be great. Thanks! |
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