# Why is $C = C_v + \frac{P}{n}\left(\frac{dV}{dT}\right)$?

I found this equation for the molar heat capacity of a process as $C = C_v + \frac{P}{n}\left(\frac{dV}{dT}\right)$. I cannot find such an equation anywhere else. What sort of process is this equation applicable for? I know that for a polytropic process $C = \frac{R}{γ-1} + \frac{R}{1-n}$. Is there some sort of relation between these two equations?

• The equation isn't complete; there must be additional assumptions. For example, the condition(s) for the heat capacity on the left and the differential term on the right aren't specified. See here. – Chemomechanics Apr 2 '18 at 19:17
• – Bert Barrois Apr 2 '18 at 19:19

For an arbitrary reversible process on a closed system (involving only P-V work), you have (from the first law of thermodynamics) $dU=nC_vdT=dQ-PdV$. So, solving for dQ, you have $$dQ=n\left(C_v+\frac{P}{n}\frac{dV}{dT}\right)dT$$where dV/dT is the derivative of volume with respect to temperature along the specified process path. If you are willing to define heat capacity C (which should really be a function of state) in terms of Q the amount of heat transferred (which is really a function of path), then you have for the "molar heat capacity" along the path $$C=C_v+\frac{P}{n}\frac{dV}{dT}$$
• Sorry, why isn't it $dU=nC_vdT+(\partial U/\partial V)_TdV(=dQ-PdV)$ for a closed system, if you're leaving open the possibility of changes in volume? – Chemomechanics Apr 2 '18 at 21:34
• Alternatively, I calculate $dU=\frac{nPC_v}{\alpha T K}dT+\left(T-\frac{P}{\alpha K}\right)dS=\frac{nPC_v}{\alpha T K}dT$ for a closed-system reversible process for a general material, but this still doesn't lead to the original poster's result. – Chemomechanics Apr 2 '18 at 22:11
• By writing $U = nC_v T$, you assumed that $(\partial U/\partial V)_T=0$. That said, this assumption is needed to derive the relation OP cited, meaning that the relation doesn't hold generally. – higgsss Apr 3 '18 at 1:47