# Joule Coefficient Derivation

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1. Oct 10, 2015

### lee403

The Joule coefficient, μJ, is defined as μ = (∂T/∂V)U. Show that μJ CV = p – αT/κ

Relevant equations
Cv=(∂U/∂T)v
κT=-1/V*(∂V/∂p)T
α=1/V*(∂V/∂T)p
dV(T,p)=(∂V/∂T)pdt+(∂V/∂p)Tdp

What I have attempted was to make μ = (∂T/∂V)U=(∂T/∂U)V(∂U/∂V)T=(∂U/∂V)T*Cv-1.

Then α/κT=(∂V/∂T)p/(∂V/∂p)T because the 1/V cancels.

Since (∂V/∂T)p=(∂V/∂p)T *(∂p/∂T)V
p – αT/κ becomes p-T(∂p/∂T)V

so μJ*Cv = (∂U/∂V)T (because Cv and Cv-1 cancel)
(∂U/∂V)T= p-T(∂p/∂T)V

I try to make the assumption that it is an ideal gas so (∂p/∂T)V becomes nR/V
so then p-T(nR/V)= p-p= 0. But that would mean (∂U/∂V)T = 0.

I don't know that this statement is true or if it is safe to assume an ideal gas. I am afraid that maybe I missed a negative or that I used properties of partial differentials incorrectly. Can anyone provide assistance?

2. Oct 11, 2015

### TSny

Hello. Welcome to PF.

You don't want to assume an ideal gas.

You've gotten to μ = (∂U/∂V)T*Cv-1. So, you need to deal with the factor (∂U/∂V)T. The first law will be helpful. Then, maybe use a Maxwell relation.