Question

One gram mole of an ideal gas $A$ with the ratio of constant pressure and constant volume specific heats $\gamma_{A} = 5 / 3$ is mixed with $n$ gram moles of another ideal gas B $y_{B} = 7/5. $If the $\gamma$ for the mixture is $19/13,$ then what will be the value of $n$?

• A. 0.75
• B. 2
• C. 1
• D. 3

Answer 1

Answer: (B)

By using formula,
$\gamma_{\text {mixture }}=\frac{\left(\frac{\mu_{1} C p_{1}+\mu_{2} C p_{2}}{\mu_{1}+\mu_{2}}\right)}{\left(\frac{\mu_{1} C v_{1}+\mu_{2} C v_{2}}{\mu_{1}+\mu_{2}}\right)} $
$\gamma_{\text {mixture }}=\frac{\mu_{1} C p_{1}+\mu_{2} C p_{2}}{\mu_{1} C v_{1}+\mu_{2} C v_{2}}$
$=\frac{\mu_{1}\left(\frac{\gamma_{1}}{\gamma_{1}-1}\right) R+\mu_{2}\left(\frac{\gamma_{2}}{\gamma_{2}-1}\right) R}{\mu_{1}\left(\frac{R}{\gamma_{1}-1}\right)+\mu_{2}\left(\frac{R}{\gamma_{2}-1}\right)}$
or $\gamma_{\text {mixture }}=\frac{\mu_{1} \gamma_{1}\left(\gamma_{2}-1\right)+\mu_{2} \gamma_{2}\left(\gamma_{1}-1\right)}{\mu_{1}\left(\gamma_{2}-1\right)+\mu_{2}\left(\gamma_{1}-1\right)}$
It is given, $\gamma_{\text {mixture }}=\frac{19}{13}$
$\gamma_{1}=\frac{5}{3}, \gamma_{2}=\frac{7}{5}$
$n_{1}=1, n_{2}=n$
Substituting values, we have
$\frac{19}{13}=\frac{\frac{5}{3}\left(\frac{7}{5}-1\right)+n \frac{7}{5}\left(\frac{5}{3}-1\right)}{\left(\frac{7}{5}-1\right)+n\left(\frac{5}{3}-1\right)}$
$\Rightarrow \frac{19}{13}=\frac{\frac{5}{3} \times \frac{2}{5}+n \frac{7}{5} \times \frac{2}{3}}{\frac{2}{5}+n \frac{2}{3}}$
$\Rightarrow 19\left(\frac{2}{5}+n \frac{2}{3}\right) $
$=13\left(\frac{2}{3}+n \frac{14}{15}\right) $
$\Rightarrow n=2$
$\therefore $ Number of moles of gas $B=2$.