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My problem is to show that $$ \left( \frac{\partial C_{V}}{\partial V} \right)_{T} = -T \left[ \frac{\partial (\alpha/\kappa_{T})}{\partial T} \right]_{V}$$ where $C_{V}$ is molar specific heat capacity of constant volume and $$\alpha := \frac{1}{V} \left(\frac{\partial V}{\partial T} \right)_{P}$$ which is Thermal Expansion coefficient and $$\kappa_{T}:=-\frac{1}{V} \left(\frac{\partial V}{\partial P} \right)_{T}$$ is isothermal compressibility.

My attempt: First, $\displaystyle{C_{V}=T\left( \frac{\partial S}{\partial T} \right)_{V}}$. So by chain rule, $$\left( \frac{\partial C_{V}}{\partial V} \right)_{T}=T\left( \frac{\partial}{\partial V} \left( \frac{\partial S}{\partial T} \right)_{V} \right)_{T}+\left( \frac{\partial T}{\partial V} \right)_{T} \left( \frac{\partial S}{\partial T} \right)_{V}=T\left( \frac{\partial}{\partial V} \left( \frac{\partial S}{\partial T} \right)_{V} \right)_{T}$$ We can interchange the order of differentiation so $\displaystyle{\left( \frac{\partial C_{V}}{\partial V} \right)_{T}=T\left( \frac{\partial}{\partial T} \left( \frac{\partial S}{\partial V} \right)_{T} \right)_{V}}$. Since $\displaystyle{\left( \frac{\partial S}{\partial V} \right)_{T}=\left( \frac{\partial P}{\partial T} \right)_{V}}$ by Maxwell's Relation and using the fact that $\displaystyle{\alpha/\kappa_{T}=-\left(\frac{\partial V}{\partial T} \right)_{P} \left(\frac{\partial P}{\partial V} \right)_{T}=\left( \frac{\partial P}{\partial T} \right)_{V}}$, I get the result without the minus sign. Where does minus sign come from??

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  • $\begingroup$ I can't see anything wrong with your derivation either - is this an exercise question? Are you sure there's not a typo in the problem or so? (According to Google, $\frac{dC_V}{dV_T} = T\frac{d^2P}{dT^2_V}$ is definitely correct, and then there's only one step missing...) $\endgroup$
    – Noiralef
    Apr 24, 2017 at 15:37

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I started with $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$ So, $$\frac{\partial ^2U}{\partial T \partial V}=\left(\frac{\partial C_v}{\partial V}\right)_T=-\left( \frac{\partial \left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]}{\partial T}\right)_V$$ The rest is easy.

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