1
$\begingroup$

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 $M=M(V,T)$

$\endgroup$
9
  • $\begingroup$ That is the definition of Cv. $\endgroup$ Aug 17, 2022 at 13:32
  • 1
    $\begingroup$ 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

2
$\begingroup$

$\delta Q$ 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.

$\endgroup$
1
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.