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. 3/2

Answer 1

Answer: (C)

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 Eq. (1), we get
$\frac{(2+n)}{\left(\frac{3}{2}-1\right)}=\frac{2}{\left(\frac{5}{3}-1\right)}+\frac{n}{\left(\frac{7}{5}-1\right)}$
$\frac{(2+n)}{1/2}=\frac{2}{2/3}+\frac{n}{2/5}$
$\Rightarrow 2(2+n)=3+\frac{5}{2}n$
$4+2n=3+2.5n$
$\Rightarrow 0.5n=1$
$\Rightarrow n=2$