Question

$1$ mole of a gas with $\gamma =\frac{7}{5}$ is mixed with $1$ mole of gas with $\gamma =\frac{5}{3}$, the value of $\gamma$ of the resulting mixture of.

• A. $\frac{7}{5}$
• B. $\frac{2}{5}$
• C. $\frac{3}{2}$
• D. $\frac{12}{7}$

Answer 1

The correct option is C $\frac{3}{2}$

$Y_{mis}=\frac{n_{1}C{\rho }_1+n_{2}C{\rho }_2}{n_{1}C{\gamma }_1+n_{2}C{\gamma }_2}$

${C\rho }_1=\frac{5}{2}R$ then its $C{v}_1=\frac{3}{2}R$

Because $C_{\rho }-C_v=R$

for diatomic gas ${C\rho }_2=\frac{7R}{2}$ then ${Cv}_2=\frac{5}{2}R$

$Y_{mis}=\frac{n_1\times \frac{5}{2}R+n_2\times \frac{7}{2}R}{n_1\times \frac{3}{2}R+n_2\times \frac{5}{2}R}$

Here $n_1=n_2=1$

$Y_{mis}=\frac{3}{2}$