Question

$n_1$moles of a monatomic gas and $n_2$moles of a diatomic gas are mixed together in a container.
a. Derive an expression for the molar specific heat at constant volume of the mixture.
b. Show that your expression has the expected behavior if $n_1\to 0$or$n_2\to 0$.

Short answer

Part$\left(a\right)$

$\left(a\right)$The molar specific heat at constant volume of mixture is, $C_V=\frac{\frac{3}{2}{n}_1+\frac{5}{2}{n}_{2}R}{n_1+n_2}$.

Part$\left(b\right)$

$\left(b\right)$localid="1648535844362" $\text{At}{n}_1\to 0,C_V=\frac{5}{2}R;$

$\text{At}{n}_2\to 0,C_V=\frac{3}{2}R$.

Step-by-step solution

Step: 1  Molar specific heat: (part a)

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

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

The monatomic gas is

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

The diatomic gas is

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

Hence, the sum of the energy is

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

The molar specific heat is specified by

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

Step: 2 Finding the value of CV: 

The molar specific heat $C_V$by

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

Step: 3  Molar specific heat for monatomic gas and diatomic gas: (part b)

When $n_1\to 0$,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 $n_2\to 0$,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.