Solve First Law of Thermo Homework: (DT/DV)s = -(DP/DS)v

AI Thread Summary
The discussion revolves around demonstrating the relationship (DT/DV)s = -(DP/DS)v using the first law of thermodynamics expressed as dU = TdS - PdV. The participants note that this is a Maxwell relation, which connects partial derivatives of thermodynamic variables. The equation involves holding entropy (S) and volume (V) constant while differentiating temperature (T) and pressure (P). The approach involves applying the equality of mixed partial derivatives to establish the relationship. Understanding these concepts is crucial for solving the homework problem effectively.
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Homework Statement



First law can be written dU = TdS - PdV where the internal energy U may be written in terms of any two of T,P,V,S.

I have to show that (DT/DV)s = -(DP/DS)v

where D is partial d, and the subscripts s and v mean hold those constant..

Homework Equations





The Attempt at a Solution



Not really sure how to proceed at all?
 
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This is one of the Maxwell relations.
If you have a function
f(x,y)=\left( \frac{\partial f}{\partial x} \right)_y dx + \left( \frac{\partial f}{\partial y} \right)_x dy = A dx + B dy
then, according to
\frac{\partial}{y} \left( \frac{\partial f}{\partial x} \right) = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right)
you have
\left( \frac{\partial A}{\partial y} \right)_x = \left( \frac{\partial B}{\partial x} \right)_y
 
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