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In the following image I have given the derivation of the formula for the derivation of the equation

$C_{v} = \frac{fR}{2}$

As you can see from the derivation below, we have assumed two times that the system we are considering is an ideal gas. So, does that mean that the formula $C_{v} = \frac{fR}{2}$ is applicable for ideal gases only? Is it not applicable for real gases? If it is applicable for real gases, then how? Is it applicable for all types of gases. Please explain.

!Derivation of formula for molar heat capacity at constant volume](https://i.sstatic.net/Vesyl.jpg)

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  • $\begingroup$ See my answer here $\endgroup$
    – Mark_Bell
    Commented Apr 18, 2021 at 13:07
  • $\begingroup$ If f is the number of degrees of freedom, we can derive this result when the energy is written as a sum of quadratic terms. If the potential that describes interations is quadratic than is also valid, and you have to include these degrees of freedom. $\endgroup$
    – Mark_Bell
    Commented Apr 18, 2021 at 13:14
  • $\begingroup$ Mark_Bell, I am asking whether $C_{v}=\frac{fR}{2}$ is applicable for ideal gases only or is it also applicable for real gases also. Please answer it. If it is applicable for real gases, also explain why is it applicable for real gases. $\endgroup$ Commented Apr 18, 2021 at 13:32
  • $\begingroup$ My answer includes this. What model of real gas would you use? What is the potential for the interactions? The term $\frac{1}{2} $ arise from quadratic form. $\endgroup$
    – Mark_Bell
    Commented Apr 18, 2021 at 13:48
  • $\begingroup$ Actually, I am just a Grade 12 student. I did not understand any of your derivation. I do not know hardcore maths. So, I am asking whether it is applicable for real gas or ideal gas? I am really sorry, but I do not know much maths. Just answer my question. $\endgroup$ Commented Apr 18, 2021 at 15:58

1 Answer 1

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From the comments, we have to address this question for high school.

The energy of a gas can be written as a sum of kinetic energy, describing the energy due to the motion, and potential energy, that describe the interactions between the molecules of the gas:

$$ U = U_{kin} + U_{pot} $$ For calculate the heat capacity at constant volume, we have to derive respect to the temperature, keeping the volume constant. $$ c_{m, V} = \frac{1}{n} \frac{d\langle U \rangle}{dT} = \frac{1}{n} (\frac{d\langle U_{kin} \rangle }{dT} + \frac{d\langle U_{pot}\rangle}{dT}) $$ Where $c_{m, V} $ is the molar heat capacity at constant volume, $\langle... \rangle$ is the mean and $n$ is the number of moles. The equipartition of the energy dictates that the mean kinetic energy is: $\langle U_{kin} \rangle = \frac{nfRT}{2} $, where $f$ is the number of degrees of freedom, $R$ the gas constant and $n$ the number of moles. Then: $$ c_{m, V} = \frac{1}{n} \frac{d\langle U \rangle}{dT} = \frac{fR}{2} + \frac{1}{n}\frac{d\langle U_{pot}\rangle}{dT} $$ So the point is the potential energy. The relation written in the question is valid for energy that is quadratic in the proper variables. The equipartition, clearly, not only applies on ideal gas (where the potential energy is zero), but also in all situations where the potential energy has the same form of the kinetic energy. The potential energy for real gases don't have the right form.

In general heat capacity depends on temperature, using the equipartition the final result is independent from the temperature, because the average kinetic energy is linear on temperature. The potential energy introduce this dependence in the heat capacity, because the mean potential energy can be higher order on $T$.

The potential energy can be quadratic and so the equipartition holds, but when is not quadratic as i said the dependence of the temperature is different. In real gases the potential is not quadratic.

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  • $\begingroup$ So, can we say that for real gases we cannot apply $C_{v}=\frac{fR}{2}$? For real gases we have to add that 'rate of change of potential energy w.r.t. time per no. of moles? Can we say that? $\endgroup$ Commented Apr 19, 2021 at 7:21
  • $\begingroup$ It's written in the answer. You cannot apply it to real gases. It's a derivative respect to temperature, divided by the number of moles $\endgroup$
    – Mark_Bell
    Commented Apr 19, 2021 at 14:19

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