Question

An experiment you're designing needs a gas with $\gamma =1.50$. You recall from your physics class that no individual gas has this value, but it occurs to you that you could produce a gas with $\gamma =1.50$by mixing together a monatomic gas and a diatomic gas. What fraction of the molecules need to be monatomic?

Short answer

The Fraction of molecules needs at monatomic$=\frac{1}{2}$.

Step-by-step solution

Step: 1  Change in temperature:

In the micro-system, 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 system is given by equation in the form

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

The monatomic gas equation is,

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

The diatomic gas equation is,

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

The sum of 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 calculated as follows:

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

Step: 2 Value of CV:

We are give $\gamma =1.5,\text{so}{C}_V\text{is}$

$\begin{array}{c}{C}_p=\gamma C_V\\ C_p=1.5C_V\\ C_V+R=1.5C_v\\ C_V=2R.\end{array}$

Step: 3 Finding fraction of molecules:

The molar specific heat of the the mixture is

$\begin{array}{r}\frac{3}{2}{n}_{1}R\mathrm{\Delta }T+\frac{5}{2}{n}_{2}R\mathrm{\Delta }T=\left(n_1+n_2\right)C_V\mathrm{\Delta }T\\ \left(\frac{3}{2}{n}_1+\frac{5}{2}{n}_2\right)R=\left(n_1+n_2\right)2R\\ \frac{3}{2}{n}_1-2n_1=-\frac{5}{2}{n}_2+2n_2\\ -0.5n_1=-0.5n_2\\ n_1=n_2\end{array}$

The molarity in each group is the same. The total mass is ${\raisebox{1ex}{$N$}\!\left/ \!\raisebox{-1ex}{$N$}\right.}_A$where $N_A$is the number calculated by Avogadro. The monatomic molecule's number is multiplied. $2$to exceed the diatomic's number of molecules As a result, the diatomic gas's fraction of monatomic gas is$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.