[NOTES/SM-04010] Classical Theory of Specific Heat of Gases

The law of equipartition of energy
Assume that the energy \(E\) is quadratic in some variable, coordinate or momentum. So we can write $E=Cs^2+E^\prime, C>0$, for some variable \(s\) and assume that $E^\prime$ does not depend on $s$. Then
$$
Z=\int_V\cdots\int_{-\infty}^\infty \exp(-\beta
E(\vec{r}_1\cdots\vec{v}_N)d^3\vec{r}_1\cdots d^3v_N
$$

\begin{align*}
Z=&\int_{-\infty}^\infty ds \exp(-\beta Cs^2)
\underbrace{\int\cdots\int} \exp(-\beta E^\prime)\\
Z=&Z^\prime \int_{-\infty}^\infty ds\exp(-\beta Cs^2)\\
= &Z^\prime\left(\frac{\pi}{\beta C}\right)^{1/2}
\end{align*}
Internal energy is given by
\begin{align*}
U = &\frac{1}{2}\frac{\partial}{\partial\beta}\ln\left(\frac{\beta
C}{\pi}\right) = \frac{\partial}{\partial\beta} \ln Z^\prime\\
=&\frac{1}{2} kT+U^\prime
\end{align*}
$N$ such quadratic terms will contribute $\frac{1}{2}(NkT)$ to
internal energy.

Average energy associated with every quadratic variables is
$\frac{1}{2}kT$. For vibrational modes, the energy is has the form of that for harmonic oscillators.
$$
\epsilon = \frac{1}{2} mv^2_x +\frac{1}{2} kx^2
$$
such a mode will contribute a term \(\frac{1}{2}kT + \frac{1}{2}kT = kT\) to the energy.

Perfect monoatomic gases
\begin{align*}
E = & \frac{1}{2}\sum_{i=1}^N m\vec{v}_i^2.
\end{align*}
For perfect gases, the equipartition law gives \(U=\frac{3}{2}NkT\). Therefore
\begin{align*}
C_V = & \left(\frac{\partial U}{\partial T}\right)_V =
\frac{3NK}{2}
=\frac{3}{2}nR
\end{align*}
and
$$
C_p=C_v+Rn=\frac{5}{2} nR
$$
$\gamma=C_p/C_v=\frac{5}{3}=1.667$

Perfect Diatomic Gases

We consider classical model as a rotating rigid dumb bell model and vibrating molecule model.

Rigid dumb bell model
$$
\epsilon=\frac{1}{2}M\vec{V}^2+\frac{1}{2}I(\omega_1^2+\omega^2_2)
$$
In this case the classical predictions are as follows.
\begin{align*}
C_v =&\left(\frac{\partial V}{\partial T}\right)_V = \frac{5}{2}
NK\\
C_p =&C_v+NK =\frac{7}{2} Nk\\
C_p/C_v =&\frac{7}{5}=14
\end{align*}

Specific heat due to vibrations and rotations

For gas consisting of diatomic molecules, the rigid dumbbell model is only an approximation. A better model is the the molecules can rotate as well as vibrate. In such a case the prediction of specific heat will be as follows. $$
\epsilon=\frac{1}{2}M\vec{V}^2+\frac{1}{2}I(\omega_1^2+\omega^2_2) +
\frac{1}{2} \mu \dot{x}^2+\frac{1}{2}Kx^2
$$

$$
C_p/C_v = \frac{9}{2}\times\frac{2}{7}=\frac{9}{7} = 1.286
$$
Comparison with experiments

It should be noted that the classical theory predicts the specific heat to be independent of temperature. This does not agree with experiments.
\underline{$T=15^\circ$}

$H_2\qquad r=1.408$

$O_2\qquad r=1.400$

$N_2 \qquad r=1.404$

$Cl_2\qquad r=1.340$

$Co\qquad r=1.404$

In fact the heat capacities depend on temperature

For hydrogen, the experimental values
are as follows.

$T<75 \quad C_p\cong \frac{5}{2}$

$T \approx {250-1100} \quad C_p \approx \frac{7}{2}$

$> 5000 \quad C_p=\frac{9}{2}$

It may be noted that below 75K
only translational degree of freedom contribute.

• Between 250-1000K the translational and rotational degrees of freedom contribute to the specific heat.
• Above 5000K all the three translational, rotational and vibrational modes contribute to the energy.
• This feature cannot be understood in classical terms. It appears due to quantization of energy.