3
$\begingroup$

I came across this formula in thermodynamics. Please give me a rigorous proof to this formula. My teacher did not even give any proof neither do any of my books. The formula is : $C_{V}=\frac{fR}{2}$ where $C_{V}$ is the molar heat capacity at constant volume, $f$ is the total number of degrees of freedom .

$$f=f_{\rm translational}+f_{\rm rotational}+f_{\rm vibrational}$$

And $R$ is the universal gas constant.

$\endgroup$
0

3 Answers 3

1
$\begingroup$

Within thermodynamics there is no way to deduce that formula. It is something one should look at as directly provided by experiments.

Using statistical mechanics it is possible to derive that results as a consequence of the description of an ideal gas (i.e. a gas in conditions where the effect of interactions is negligible). It can be obtained from a theorem valid only for systems described by classical mechanics (i.e. no quantum systems): the equipartition theorem, stating that the average kinetic energy per degree of freedom is $\frac12 k_BT$

$\endgroup$
1
  • $\begingroup$ For sake of completeness, I think you should add where one could read of the stat mech derivation of the equations mentioned $\endgroup$ Apr 16, 2021 at 9:57
1
$\begingroup$

Not really sure if this amounts to a "proof", but, from the equipartition theorem, we know that each degree of freedom has an average KE = $KT/2$ associated with it.

So if you have a system of $N$ particles, you have an average total KE of $fNKT/2=fN(R/N_{a})T/2 =fnRT/2$. (Where $n=N/N_{a}$ is the number of moles, and $R$ is the gas constant)

Now, if you have other forms of (Temperature independent) forms of energy, that along with the average kinetic energy accounts for the internal energy, $U$, of the system.

$$U(T)=U(0) + fnRT/2$$

Now, consider the First law of thermodynamics:

$$\delta{q}=dU + \delta{w}$$ $$\dfrac {\delta q}{ndT}=\dfrac{dU}{ndT}+\dfrac{\delta w}{ndT}$$ $$\dfrac {\delta q}{ndT}=\dfrac{fR}{2}+\dfrac{\delta w}{ndT}$$

Consider a situation at constant volume. The reason for this is that $\delta{w}$ will vanish.

$$\dfrac {\delta q}{ndT}_{(const V)}=\dfrac{fR}{2}$$

The LHS of this equation is precisely how $Cv$ is defined. We thus finally end up with, $$\dfrac {\delta q}{ndT}_{(constV)}\equiv C_{v} =\dfrac{fR}{2}$$

$\endgroup$
7
  • $\begingroup$ How did you write $ \frac{dU}{ndT}= \frac{fR}{2}$? $\endgroup$ Apr 16, 2021 at 12:00
  • $\begingroup$ based on how I defined $$U(T)= U(0) + nfRT/2$$ earlier $\endgroup$
    – satan 29
    Apr 16, 2021 at 12:43
  • $\begingroup$ What is $U(0)$? $\endgroup$ Apr 16, 2021 at 13:23
  • $\begingroup$ some constant term. This contains the energy which is not T dependent. .This term is usually omitted (since we are mostly concerned about the change) in internal energy, so this term cancels out.. another (identical) way to see it as it is the U at T=0. $\endgroup$
    – satan 29
    Apr 16, 2021 at 13:26
  • $\begingroup$ Can you give some examples as to what this constant term $U(0)$ can be? What type of energies can it be? $\endgroup$ Apr 16, 2021 at 16:11
1
$\begingroup$

This is the so called equipartition theorem, that says that each degree of freedom contributes with $\frac{k_BT} {2} $ to the average energy, or using mol $\frac{RT} {2} $. More precisely, each degree of freedom in the Hamiltonian that is quadratic contributes for $\frac{k_BT} {2} $.

If we can write the energy:$ E = \sum_n \alpha_n x^2_n$ like a sum of quadratic degrees of freedom, we can compute the average, using Boltzmann distribution: $$ \langle E \rangle =\frac{\int E \exp(-\beta E) d^3 x_1...d^3 x_N}{\int \exp(-\beta E) d^3 x_1...d^3 x_N} = \frac{\int \sum_n \alpha_n x^2_n\exp(-\beta \sum_n \alpha_n x^2_n) d^3 x_1...d^3 x_N} {\int \exp(-\beta \sum_n \alpha_n x^2_n) d^3 x_1...d^3 x_N} $$ We can factorize, using the property of the exponential and Gaussian integrals: $$ \frac{\int \alpha x^2\exp(-\beta \alpha x^2) d^3 x} {\int \exp(-\beta \alpha x^2) d^3 x} = \frac{\frac{\alpha \sqrt{\pi} }{2{(\alpha \beta})^{\frac{3}{2} } }} {\frac{\sqrt{\pi}}{{(\alpha \beta})^{\frac{1}{2} } }} = \frac{1} { 2 \beta} $$ Knowing that: $\beta = \frac{1} {k_B T} $, we have: $$ \langle E \rangle = N f \frac{k_B T} {2} $$ Where $N$ is the number of molecules and $f$ is the number of degrees of freedom per molecule. Or writing this for a mole of gas: $\langle E \rangle = f \frac{R T} {2}$, where $R$ is the constant of gas.

Now, we know that the heat capacity at constant volume is: $C_V = \frac{d\langle E \rangle}{dT}$ then we have: $$ C_V = f \frac{R} {2} $$

For a perfect monoatomic gas, where we neglect interactions, we have only the kinetic term: $$ E = \sum_n^N \frac{p^2_n}{2m}$$ Where the sum is over the number of molecules. Using the equipartition theorem we have: $\langle E \rangle = N(\frac{3}{2})k_B T$ or $\langle E \rangle = \frac{3} {2}RT$. The factor of $3N$ arise from the number of molecules and the fact that we have three dimensional space, then we have three degrees of freedom from the momentum of each molecule. Clearly the degrees of freedom include translational, rotational and vibrational terms, depending on the system.

$\endgroup$

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.