Question

A mixture of $n_1$ moles of monoatomic gas and $n_2$ moles of diatomic gas has $\frac{C_p}{C_v}=\gamma =1.5$. What should be the ratio of $n_1:n_2$?

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $3:2$

Answer 1

Answer: (A)

$1:1$

If $n_1$ moles of a gas of ratio of specific heat ${\gamma }_1$ is mixed with $n_2$ moles of another gas of ratio of specific heat ${\gamma }_2$, the ratio of specific heats of the mixture $\gamma$ is given by,
$\frac{n_1+n_2}{\gamma -1}=\frac{n_1}{{\gamma }_1-1}+\frac{n_2}{{\gamma }_2-1}$
In the given problem,
$\gamma =\frac{3}{2}$
${\gamma }_1=\frac{5}{3}$ (mono atomic gas)
${\gamma }_2=\frac{7}{5}$ (diatomic gas)
$\frac{n_1+n_2}{\frac{1}{2}}=\frac{n_1}{\frac{2}{3}}+\frac{n_2}{\frac{2}{5}}$
$2n_1+2n_2=\frac{3n_1}{2}+\frac{5n_2}{2}$
$.\cdot .\frac{n_1}{2}=\frac{n_2}{2}$ or $n_1:n_2=1:1$