Q104 / 125 · BITSAT 2015 · Physics
A vessel of volume 20L contains a mixture of hydrogen and helium at temperature of $27^{\circ} \mathrm{C}$ and pressure 2 atm. The mass of mixture is $5 \mathrm{~g}$. Assuming the gases to be ideal, the ratio of mass of hydrogen to that of helium in the given mixture will be
- A1: 2
- B2: 3
- C2: 1
- Correct answerD2: 5
Solution
Let there are $\mathrm{n}_{1}$ moles of hydrogen and $\mathrm{n}_{2}$ moles of helium in the given mixture. As $\mathrm{Pv}$ $=\mathrm{nRT}$
Then the pressure of the mixture
$$
\begin{array}{r}
P=\frac{\mathrm{n}_{1} \mathrm{RT}}{\mathrm{V}}+\frac{\mathrm{n}_{2} \mathrm{RT}}{\mathrm{V}}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right) \frac{\mathrm{RT}}{\mathrm{V}} \\
2 \times 101.3 \times 10^{3}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right) \times \frac{(8.3 \times 300)}{20 \times 10^{-3}}
\end{array}
$$
or,
$$
\begin{array}{l}
\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)=\frac{2 \times 101.3 \times 10^{3} \times 20 \times 10^{-3}}{(8.3)(300)} \\
\text { or, } \mathrm{n}_{1}+\mathrm{n}_{2}=1.62
\end{array}
$$
The mass of the mixture is (in grams)
$$
\begin{array}{l}
\mathrm{n}_{1} \times 2+\mathrm{n}_{2} \times 4=5 \\
\Rightarrow \quad\left(\mathrm{n}_{1}+2 \mathrm{n}_{2}\right)=2.5 \quad \ldots
\end{array}
$$
Solving the eqns. (1) and (2), we get $\mathrm{n}_{1}=0.74$ and $\mathrm{n}_{2}=0.88$
Hence, $\frac{\mathrm{m}_{\mathrm{H}}}{\mathrm{m}_{\mathrm{He}}}=\frac{0.74 \times 2}{0.88 \times 4}=\frac{1.48}{3.52}=\frac{2}{5}$
Hence, the correct answer is option: 'D'.