Thermodynamics Proofs: Showing Relationship to Ideal Gas

[IanPTerry]

Making use of the fact that dH = C_pdT + (V - T($$\delta$$V/$$\delta$$T)_P ) dP is an exact differential expression, show that: ($$\deltaCp$$/$$\delta$$P)_T = -T($$\delta$$²V/$$\delta$$T²)_P

What is the result of application of this equation to an ideal gas?So I literally have no idea what is going on. I've asked for help from the TA, but there's not much to it apparently besides just guessing?

 

[NateD]

Using the fact that dH = CpdT + (V - T(\deltaV/\deltaT)P ) dP is an exact differential expression, we can rearrange it to the following: CpdT + (V - T(\deltaV/\deltaT)P ) dP = dH We can then take the partial derivatives with respect to pressure (P) and temperature (T):\frac{\delta Cp}{\delta P}T + \frac{\delta V}{\delta P} - \frac{\delta^2 V}{\delta T^2}TP = \frac{\delta H}{\delta P}Rearranging the terms, we get: \frac{\delta Cp}{\delta P}T = -T\frac{\delta^2 V}{\delta T^2}P + \frac{\delta H}{\delta P}This equation is valid for any system, including ideal gases. If we apply this equation to an ideal gas, we get: \frac{\delta Cp}{\delta P}T = -T\frac{\delta^2 V}{\delta T^2}P + \frac{\delta H}{\delta P}Since ideal gases obey the ideal gas law, we can substitute PV = RT into the equation, which yields:\frac{\delta Cp}{\delta P}T = -T\frac{\delta^2 V}{\delta T^2}P + \frac{\delta H}{\delta P} = -T\frac{\delta^2 V}{\delta T^2}P + \frac{C_p}{R}\frac{\delta R}{\delta P} Therefore, the result of application of this equation to an ideal gas is: \frac{\delta Cp}{\delta P}T = -T\frac{\delta^2 V}{\delta T^2}P + \frac{C_p}{R}\frac{\delta R}{\delta P}