Q$4.(12$ points$)$ Show that

$(C_P-C_V)\frac{dT}{T}=VdP+\frac{}{}dV$

$(\frac{\text{d}\text{e}\text{l}\text{C}\text{V}}{\text{d}\text{e}\text{l}\text{P}}{)}_T=T(\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{P}})T(\frac{de{l}^{2}P}{del{T}^2})V$

Useful Information

$[$:$[dU=TdS-PdV],[dH=TdS+VdP],[dA=-PdV-SdT],[dG=VdP-SdT]\}$

$$ If $dF=$

$(\frac{\text{d}\text{e}\text{l}\text{M}}{\text{d}\text{e}\text{l}\text{y}}{)}_x=(\frac{\text{d}\text{e}\text{l}\text{N}}{\text{d}\text{e}\text{l}\text{x}}{)}_y$