Question

A mixture of monoatomic gas ($n_1$ moles) and diatomic gas ($n_2$ moles) has $\frac{C_P}{C_V}$ equal to $\frac{11}{7}$. Then

• A. $n_1=3n_2$
• B. $n_2=3n_1$
• C. $n_1=n_2$
• D. $5n_1=2n_2$

Answer 1

The correct option is A $n_1=3n_2$

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 ${\gamma }_2$ , then the ratio of specific heats of the mixture $\gamma$ is given by,

$\frac{n_1+n_2}{\gamma -1}=\frac{n_1}{{\gamma }_1-1}+\frac{n_2}{{\gamma }_2-1}$

Given, $\gamma =\frac{11}{7}$

For monatomic gas, ${\gamma }_1=\frac{5}{3}$

For diatomic gas, ${\gamma }_2=\frac{7}{5}$

$\Rightarrow \frac{n_1+n_2}{\frac{11}{7}-1}=\frac{n_1}{\frac{5}{3}-1}+\frac{n_2}{\frac{7}{5}-1}$

$\Rightarrow \frac{7n_1+7n_2}{4}=\frac{3n_1}{2}+\frac{5n_2}{2}$

$\Rightarrow 7n_1+7n_2=6n_1+10n_2$

$\Rightarrow n_1=3n_2$

Hence, option $\left(a\right)$ is the correct answer.