- #1
Gregg
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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.