# 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. Apr 2, 2018 at 19:17
• Apr 2, 2018 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? Apr 2, 2018 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. Apr 2, 2018 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. Apr 3, 2018 at 1:47