The temperature change of an ideal gas, Joule Kelvin expansion (const. enthelpy)

[Gregg]

Homework Statement

This is the last part of the question. So far have been made to derive:

$\mu _{\text{JK}}=\left(\frac{\partial T}{\partial P}\right)_H=-\frac{1}{C_P}\left(\frac{\partial H}{\partial P}\right)_T$

Then

$\left(\frac{\partial H}{\partial P}\right)_T=V - T \left(\frac{\partial V}{\partial T}\right)_P$

It says you need to derive an expression for the temperature change as an integral over pressure.

The Attempt at a Solution

$dT=\left(\frac{\partial T}{\partial P}\right)_HdP+\left(\frac{\partial T}{\partial H}\right)_PdH$

At constant enthalpy $dH = 0$.

$dT=\left(\frac{\partial T}{\partial P}\right)_HdP\text{=}-\frac{1}{C_P}\left(V-T\left(\frac{\partial V}{\partial T}\right)_P\right)$

So I think the change in temperature will be:

$\text{\Delta T} =-\frac{1}{C_P} \int_{P_1}^{P_2} \left(V-T\left(\frac{\partial V}{\partial T}\right)_P\right) \, dP$

Then it says derive an expression for the temperature change for an ideal gas.

$p V = n R T$

$V = \frac{ n R T}{P}$

$T\left(\frac{\partial V}{\partial T}\right)_P = \frac{ n R T}{p}$

So it would seem the integral vanishes, and $\Delta T = 0$

I don't think this is right.

 

[rude man]

Homework Statement

I don't think this is right.

Oh, but it is. The Joule-Kelvin coefficient μ_JK for an ideal gas is identically zero.