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dQ = −TdS(produced)

 
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Apr9-10, 01:56 AM   #1
 

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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