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 $M=M(V,T)$

  • $\begingroup$ That is the definition of Cv. $\endgroup$ Aug 17, 2022 at 13:32
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    $\begingroup$ 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. $\endgroup$ Aug 17, 2022 at 14:24
  • $\begingroup$ @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 $M=M(V,T)$ $\endgroup$
    – SimoBartz
    Aug 18, 2022 at 9:17
  • $\begingroup$ 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 $B$. $\endgroup$ Aug 18, 2022 at 9:22
  • $\begingroup$ However, I do not see why the introduction of an additional variable should solve your starting problem. $\endgroup$ Aug 18, 2022 at 9:25

2 Answers 2


$\delta Q$ 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.

  • $\begingroup$ 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 $\endgroup$
    – SimoBartz
    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.

  • $\begingroup$ 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? $\endgroup$
    – SimoBartz
    Aug 18, 2022 at 9:23
  • $\begingroup$ $c_v$ 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 $c_v$. $\endgroup$
    – Bob D
    Aug 18, 2022 at 12:19

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