Q158 / 170 · Manipal (MET) 2020 · Physics
If one mole of a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole diatomic gas $\left(\gamma=\frac{7}{5}\right)$, the value of $\gamma$ for the mixture is
- A$1.40$
- Correct answerB$1.50$
- C$1.53$
- D$3.07$
Solution
For monoatomic gas,
$\gamma_1=\frac{C_{p_1}}{C_{V_1}}=\frac{5}{3}$ and $C_{p_1}-C_{V_1}=R$
this gives $C_{\rho_1}=\frac{5}{2} R$ and $C_{V_1}=\frac{3}{2} R$
For diatomic gas,
$\gamma_2=\frac{C_{\rho_2}}{C_{V_2}}=7 / 5$ and $C_{p_2}-C_{V_2}=R$
$\Rightarrow \quad C_{P_2}=\frac{7}{2} R \quad$ and $\quad C_{V_2}=\frac{5}{2} R$
Now, we can write,
$\gamma_{\text {mix }}=\frac{n_1 C_{p_1}+n_2 C_{\rho_2}}{n_1 C_{V_1}+n_2 C_{V_2}}$
$=\frac{C_{\rho_1}+C_{\rho_2}}{C_{V_1}+C_{V_2}}=\frac{\frac{5}{2} R+\frac{7}{2} R}{\frac{3}{2} R+\frac{5}{2} R}=\frac{12}{8}=\frac{3}{2}=1.5$
Hence, the correct answer is option: 'B'.