# When $C_v=\left.\frac {\delta Q} {dT}\right|_v=\left.\frac {\partial E} {\partial T}\right|_v$?

The definition of specific heat at constant volume is $$C_v=\left.\frac {\delta Q} {dT}\right|_v$$ but sometimes I find this expression instead: $$C_v=\left.\frac {\partial E} {\partial T}\right|_v$$ . I guess that the reason why these two expression are both used is that they are equivalent, however I don't know how to demonstrate it.

I tried to demonstrate the equivalence by myself but I was able to do it only assuming $$\delta W=-PdV$$ indeed in this case: $$dE=\delta Q +\delta W=\delta Q - PdV \rightarrow \left.dE\right|_v=\delta Q$$ so it follow $$C_v=\left.\frac {\delta Q} {dT}\right|_v=\left.\frac {\partial E} {\partial T}\right|_v$$ .

However, when I don't assume $$\delta W=-PdV$$ it becomes impossible for me to show that $$C_v=\left.\frac {\delta Q} {dT}\right|_v=\left.\frac {\partial E} {\partial T}\right|_v$$. Consider for example the case of a paramagnetic materials, since it can be magnetized the expression for the work should be $$\delta W=-PdV+BdM$$ where $$M$$ is the magnetization and $$B$$ is the magnetic field. Even in this case i think the expression $$C_v=\left.\frac {\partial E} {\partial T}\right|_v$$ is valid. Why?

Note: in the last example I think that the independent variables are $$V$$ and $$T$$ while the number of particle and the magnetic field are taken constant and that $$M=M(V,T)$$

• That is the definition of Cv. Aug 17, 2022 at 13:32
• It is not clear where you get the last relation. If both $V$ and $M$ are present, you should specify what else is taken constant when partial derivatives are evaluated. Aug 17, 2022 at 14:24
• @GiorgioP thank you for the observation. I tried to edit the question to make it clear. About the $\delta W=-PdV+BdM$: i think the magnetic field is taken constant and also the number of particles while $M=M(V,T)$ Aug 18, 2022 at 9:17
• Then, in our last case, you should be working with a $C_{v,B}= \left.\frac{\partial{E}}{\partial{T}}\right|_{v,B}$. The independent variables are $V$, $T$, and $B$. Aug 18, 2022 at 9:22
• However, I do not see why the introduction of an additional variable should solve your starting problem. Aug 18, 2022 at 9:25

$$\delta Q$$ is a non-exact differential form. According to the First Law of Thermodynamics $$\delta Q=dU+\delta W$$ Where $$\delta W$$ is the work differential form and $$dU$$ is the differential of the internal, which is an exact differential form. For mechanical work only $$\delta W=pdV$$.

Let's consider a $$pVT$$ system. To define specific heat at constant volume, consider $$(T,V)$$ as independent variables, then the heat differential form is

$$\delta Q=\underbrace{\left(\frac{\partial U}{\partial V}\right)_TdV+\left(\frac{\partial U}{\partial T}\right)_VdT}_{dU}+p(T,V)dV$$ Specific heat at constant volume $$C_v$$ is defined as the component along $$dT$$ of the differential form $$\delta Q^1$$, i.e. $$C_v=\left(\frac{\partial U}{\partial T}\right)_V \tag{A}$$

To emphasize the definition, writing heat differential form more generally as $$\delta Q=f(T,V)dT+g(T,V)dV$$

Specific heat at constant volume is defined as the function $$f(T,V)$$. As you can see above, this coefficient is equal to $$(A)$$.

$$^1$$This is the reason why the other variable is $$V$$ and not $$p$$, so that the volume is fixed when differentiating with respect to T.

• Thank you, i found this answer useful but unfortunately it seems that you are also assuming $\delta W=-PdV$. My question is about when this is not true. I edited the question to make it clear, sorry for the confusion Aug 18, 2022 at 9:22

The formal definition of the specific heat at constant volume is

$$c_{v}=\biggl (\frac{\partial u}{\partial T}\biggr )_v$$

Your last expression does not show that the partial derivative is at constant volume.

Hope this helps.

• You formal definition of specific heat is different from the one i'm accustomed to. I edited the question to make it clear. Can you give me some extra details? Aug 18, 2022 at 9:23
• $c_v$ is a thermodynamic property and, as such, it is defined in terms of another thermodynamic property, $du$. A quantity of heat, $\delta q$ is not a thermodynamic property. The equation $\delta q=c_{v}dT$ calculates the amount of heat that causes a change in temperature at constant volume and is derived by applying the formal definition of $c_v$ along with the first law for a closed system as you already know. But it is not the definition of $c_v$. Aug 18, 2022 at 12:19