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In this lecture we derive law of equipartition of energy under the assumption that the energy is quadratic function of some variable such as coordinates and momenta. As an application, the classical theory of specific heat of gases is given.
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
$$
C_p=C_v+Rn=\frac{5}{2} nR
$$
$\gamma=C_p/C_v=\frac{5}{3}=1.667$
He | Ne | Ar | Kr | Xe |
1.66 | 1.64 | 1.67 | 1.63 | 1.66 |
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.