Q142 / 180 · COMEDK 2020 · Physics
A gaseous mixture consists of $16 \mathrm{~g}$ of helium and $16 \mathrm{~g}$ of oxygen. The ratio $C_{p} / C_{V}$ of the mixture is
- A$1.4$
- B$1.54$
- C$1.59$
- Correct answerD$1.62$
Solution
Mass of helium, $m_{1}=16 \mathrm{~g}$
Mass of oxygen, $m_{2}=16 \mathrm{~g}$
$\therefore$ Number of moles of He gas,
$$
n_{1}=\frac{m_{1}}{4}=\frac{16}{4}=4
$$
Number of moles of $\mathrm{O}_{2} \mathrm{gas}$,
$$
n_{2}=\frac{m_{2}}{32}=\frac{16}{32}=\frac{1}{2}
$$
We know that for the mixture,
$\begin{aligned} C_{V} &=\frac{n_{1} C_{V}+n_{2} C_{V_{2}}}{n_{1}+n_{2}}, \text { where } C_{V}=\frac{f R}{2} \\ \text { and } C_{p} &=\frac{n_{1} C_{p_{1}}+n_{2} C_{p_{2}}}{n_{1}+n_{2}}, \text { where } C_{p}=\left(\frac{f}{2}+1\right) R \end{aligned}$
For He gas, $f=3, n_{1}=4$
For $\mathrm{O}_{2}$ gas, $f=5, n_{2}=\frac{1}{2}$
$$
\begin{aligned}
\therefore \quad \frac{C_{p}}{C_{V}} &=\frac{n_{1} C_{p_{1}}+n_{2} C_{p_{2}}}{n_{1} C_{V_{1}}+n_{2} C_{V_{2}}} \\
&=\frac{4 \times\left(\frac{3}{2}+1\right) R+\frac{1}{2}\left(\frac{5}{2}+1\right) R}{4 \times \frac{3}{2} R+\frac{1}{2} \times \frac{5}{2} R} \\
&=\frac{\left(4 \times \frac{5}{2} R\right)+\left(\frac{1}{2} \times \frac{7}{2} R\right)}{\left(4 \times \frac{3}{2} R\right)+\left(\frac{1}{2} \times \frac{5}{2} R\right)}=\frac{47}{29} \\
&=1.62
\end{aligned}
$$
Hence, the correct answer is option: 'D'.