Solving dU < TdS − PextdV & Showing dQ = −TdSproduced

[subtletuna]

Homework Statement

(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?

Homework Equations

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!

 

[nate808]

Thank you for your post. I am a scientist specializing in thermodynamics and would be happy to help you with your questions.

For part (a), you are correct in using the first and second laws of thermodynamics to show that dU < TdS − PextdV for an infinitesimal spontaneous process in an open system. This inequality is known as the Gibbs inequality and is a fundamental concept in thermodynamics.

For part (b), you are correct in stating that d(bar)Q = TdS according to the second law of thermodynamics. However, in a constant S, V process, dS = 0, which means that d(bar)Q = 0. This indicates that no heat is being absorbed or lost in the process, as there is no change in entropy. Therefore, dU < 0 can be viewed as a criterion for spontaneity at constant S, V, as it indicates a decrease in internal energy without any heat transfer.

I hope this helps clarify your understanding. If you have any further questions, please don't hesitate to ask. Keep up the good work in your studies of thermodynamics!