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kntsy
Sep21-10, 04:37 AM
Hi, how can i derive this fundamental identity "without using entropy"?
\left(\frac {\partial U}{\partial V}\right)_T = T\left(\frac {\partial P}{\partial T}\right)_V - P

I believe the above equation is purely thermal and has nothing to do with entropy and statistical mechanics but unfortunately the below identity is the key to this derivation:

\left(\frac {\partial P}{\partial T}\right)_V = \left(\frac {\partial S}{\partial V}\right)_T

of course:

dU=TdS-PdV

Thaakisfox
Sep21-10, 09:11 AM
What do you mean "without using entropy"? what do you mean by "purely thermal"?
Entropy is a fundamental quantity of thermodynamics. I dont see why you should want to leave it out of the game.

Btw. your last equation is wrong. (At least compared to the sign convention you use in the first two formulas, then the last one should be different.)

kntsy
Sep21-10, 11:07 AM
What do you mean "without using entropy"? what do you mean by "purely thermal"?
Entropy is a fundamental quantity of thermodynamics. I dont see why you should want to leave it out of the game.

Btw. your last equation is wrong. (At least compared to the sign convention you use in the first two formulas, then the last one should be different.)

sorry for the mistakes. In other words, can we derive the first equation just using P,T,V; without using 2nd law(which defines entropy and shows that it is a "state" function)?