Why heat capacity at constant volume relates to change in volume

Question

I have been dealing with the establishment of a relationship between $w$, $q$, $\Delta U$ and $\Delta H$. I think the entire derivation is not necessary here so I will elaborate until the concerned part.

The process here is adiabatic. So $$\delta q=0$$ From the first law of thermodynamics, $$dU=\delta q + \delta w$$ which in this case simply becomes $$dU=\delta w \tag{1}$$ For only the work of expansion $$\delta w=-PdV \tag{2}$$ Putting this value in (1), we get $$dU=-PdV \tag{3}$$ As is known that for an ideal gas, $$C_v= \frac{dU}{dT}$$ or $$dU=C_vdT$$ Putting this value in (3), we get $$C_vdT=-PdV$$ I am wondering why in this last equation are we relating $C_v$ and $dV$ while $dV$ is 0 as the given specific heat is at constant volume. How to justify that in the last equation, if process is at constant volume, that $C_v$ is being related to $dV$ anyways?

My best guess is that we could have better used $C_p$ instead of $C_v$ as that would have led us to $\frac{dH}{dT}$ which would account for $\Delta U$ as well as $-PdV$. Or maybe it has something to do with the process being quasi-static.

Answer 1

Your confusion comes because you think that you can only write $\Delta U = C_V \Delta T$ only for isochoric processes.

The specific heat at constant volume is defined as $$ C_V := \left(\dfrac{\partial U}{\partial T}\right)_V \tag{1} $$ Eq.(1), as it is written, denotes $C_V$ as a state function.

The changes in internal energy are always given by $$ \int_{U_1}^{U_2} \mathrm{d}U = \int_{T_1}^{T_2} C_V(T) \mathrm{d}T \tag{2} $$ if

1. The fluid is an ideal gas, or
2. The process is really isochoric, independently if the fluid is an ideal gas or not.

To make sure that the formula works even though the process is not isochoric, take a look at the following image, where you are interested in the $\color{purple}{\text{path}}$ in purple: Since for an ideal gas $U$ is only a function of $T$, if we plot $U$ vs $V$ we get straight lines. Lets calculate then, by virtue that $U$ is a state function, the change between those two arbitrary points by the $\color{red}{\text{path}}$ in red \begin{align} U(T_2,V_2) - U(T_1,V_1) &= [U(T_2,V_2) - U(T_1,V_2)] + \underbrace{[U(T_1,V_2) - U(T_1,V_1)]}_ {=0 \; (\text{constant $T$}))} \\ U(T_2,V_2) - U(T_1,V_1) &= \underbrace{[U(T_2,V_2) - U(T_1,V_2)]}_{\text{real isochoric process}} \\ U(T_2,V_2) - U(T_1,V_1) &= \int_{T_1}^{T_2} C_V(T) \mathrm{d}T \tag{3} \end{align} In the last derivation we "assumed" that we could only replace $C_V \; \mathrm{d}T$ if the process was isochoric. In Eq. (2) we didn't assume anything, and the result was the same.

I am wondering why in this last equation are we relating $C_V$ and $\mathrm{d}V$ while $\mathrm{d}V$ is $0$ as the given specific heat is at constant volume. How to justify that in the last equation, if process is at constant volume, that $C_V$ is being related to $\mathrm{d}V$ anyways?

If you add that the process is indeed isochoric, in addition to that it was also adiabatic, the first law will tell you $\mathrm{d}U = \delta q + \delta w = 0 + 0 = 0$. For a simple fluid that has two degrees of freedom, asking for two restrictions leads to this behaviour: nothing has happened to the fluid. You will arrive to the same conclusion if you demand an isentropic and isenthalpic process, or a constant pressure and constant volume process, etc. Ask for two restrictions and you will get always $U_2 = U_1$.

Answer 2

For a real gas, U=U(T,V) and $$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV=nC_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$Sp, for a real gas you get:$$nC_vdT+\left(\frac{\partial U}{\partial V}\right)_TdV=-PdV$$or $$nC_vdT=-\left(P+\left(\frac{\partial U}{\partial V}\right)_T\right)dV$$ For an ideal gas where U=U(T) , this reduces to w$$nC_vdT=-PdV$$where $C_v=C_v(T)$