From Wikipedia:

$$ \left(\frac{\partial U}{\partial T}\right)_V = \left(\frac{\partial Q}{\partial T}\right)_V = C_V, $$

$C_V$ is what known to be constant volume heat capacity. I don't really get the exact meaning of 'constant volume'. Does it mean that (a) the whole process takes place under a constant volume, i.e, it can be any volume as long as the volume does not change during the process, the volume is maintained at its initial value throughout the process, or (b) the $C$ is measured under a fixed constant volume $V$?

If (a) is correct, are the rates of change of $Q$ with respect to $T$ same at, for example, $V = \pu{1 dm3}$ and $V = \pu{2 dm3}$? We are not considering only perfect gases.

If (b) is correct, what is the value of $V$ for standard $C_V$?

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    $\begingroup$ @Aditya Garg For a real gas, Cv depends on specific volume. $\endgroup$ – Chet Miller Jan 7 '19 at 13:25

(a) is correct. (b) is wrong. $U$ depends on $V$ only in the combination $v=V/N$. $v$ is the specific volume, an intensive quantity that does not depend on the size of the system.

We can derive this from first principles. Take $U = U(T,V,N)$. Because $U$ is extensive, $$U(T,\lambda V,\lambda N) = \lambda U(T,V,N) \quad \Longleftrightarrow \quad U(T,V,N) = \lambda^{-1}U(T,\lambda V,\lambda N),$$ by taking $\lambda = N^{-1}$, we can write this expression as $$U(T,V,N) = NU(T,V/N,1) =: Nu(T,v),$$ which produces the desired functional relationship.


You need to put consideration of any process out of your mind. Cv and U are functions of state, that depend only on the characteristics of the particular thermodynamic equilibrium state that the system is in. How it gets from one thermodynamic equilibrium state to another is irrelevant. If you have two closely neighboring thermodynamic equilibrium states that are at the same volume, the difference in internal energy between these two states is $dU=C_vdT$. In the case of an ideal gas (where internal energy is independent of volume), even if the two neighboring states differ by both dT and dV, the difference between these two states is still $dU=C_vdT$.

Furthermore, I should add that, because Q is a function of process path, it should not really be part of the definition of Cv. So Wiki is wrong to include this. It only confuses the student.

As far as the title question is concerned, the exact definition of molar Cv is: $$C_v=\frac{1}{n}\left(\frac{\partial U}{\partial T}\right)_{V,n}$$


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