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 $B=2$.