91Ó°ÊÓ

In the microsystem, the potential energy between the bonds is zero, and the kinetic energy of a monatomic gas is translational. If the temperature varies by a certain amount, $∆T$, then the thermal energy for the system is given by equation in the form

$\mathrm{Δ}{E}_{\mathrm{th}}=\frac{i}{2}nR\mathrm{Δ}T$

$\mathrm{Δ}{E}_{\mathrm{th}}=\frac{3}{2}{n}_{1}R\mathrm{Δ}T$

The diatomic gas is

$\mathrm{Δ}{E}_{\mathrm{th}}=\frac{5}{2}{n}_{2}R\mathrm{Δ}T$

Hence, the sum of the energy is

$\mathrm{Δ}{E}_{\mathrm{th}}=\frac{3}{2}{n}_{1}R\mathrm{Δ}T+\frac{5}{2}{n}_{2}R\mathrm{Δ}T$

The molar specific heat is specified by

$\mathrm{Δ}{E}_{\mathrm{th}}=\left(n_1+n_2\right)C_V\mathrm{Δ}T.$

The molar specific heat $C_V$by

$\begin{array}{c}\frac{3}{2}{n}_{1}R\mathrm{Δ}T+\frac{5}{2}{n}_{2}R\mathrm{Δ}T=\left(n_1+n_2\right)C_V\mathrm{Δ}T\\ \left(\frac{3}{2}{n}_1+\frac{5}{2}{n}_2\right)R=\left(n_1+n_2\right)C_V\\ C_V=\frac{\frac{3}{2}{n}_1+\frac{5}{2}{n}_{2}R}{n_1+n_2}.\end{array}$

When $n_1→0$,there is no monatomic gas.

$\begin{array}{r}{C}_V=\frac{\frac{3}{2}\left(0\right)+\frac{5}{2}{n}_{2}R}{0+n_2}\\ =\frac{5}{2}R\end{array}$

This is the molar specific heat for diatomic gas.

When $n_2→0$,there is no monatomic gas.

$\begin{array}{r}{C}_V=\frac{\frac{3}{2}{n}_1+\frac{5}{2}\left(0\right)R}{n_1+0}\\ =\frac{3}{2}R\end{array}$

For monatomic gas, this is the molar specific heat.