Question

Two moles of helium are mixed with $n$ moles of hydrogen. If $\frac{C_P}{C_V} = \frac{3}{2}$ for the mixture, then the value of n is :

• A. 1
• B. 3
• C. 2
• D. 44622

Answer 1

Answer: (C)

2

Answer 2

For mixture of gases, we have $\frac{\left(n_{1}+n_{2}\right)}{(\gamma-1)}=\frac{n_{1}}{\left(\gamma_{1}-1\right)}+\frac{n_{2}}{\left(\gamma_{2}-1\right)}$ ...(1) Given: $n_{1}=2$ (for helium); $\gamma=\frac{3}{2}$. Since, $f=3$ for monatomic gas and $f=5$ for diatomic gas. Therefore, $\gamma_{1}=\left(1+\frac{2}{f}\right)=\left(1+\frac{2}{3}\right)=\frac{5}{3}$ $\Rightarrow \gamma_{2}=\left(1+\frac{2}{5}\right)=\frac{7}{5}$ where $f$ is number of degrees of freedom. Now, substitute $n_{2}=n$ in E (1), we get $\frac{(2+n)}{\left(\frac{3}{2}-1\right)}=\frac{2}{\left(\frac{5}{3}-1\right)}+\frac{n}{\left(\frac{7}{5}-1\right)}$ $\frac{(2+n)}{1 / 2}=\frac{2}{2 / 3}+\frac{n}{2 / 5}$ $\Rightarrow 2(2+n) =3+\frac{5}{2} n$ $4+2 n=3+2.5 n$ $\Rightarrow 0.5 n=1$ $\Rightarrow n=2$