Question

A vessel of volume $20L$ contains a mixture of hydrogen and helium at temperature of $27^{\circ }C$ and pressure $2atm$. The mass of mixture is $5g$. Assuming the gases to be ideal, the ratio of mass of hydrogen to that of helium in the given mixture will be

• A. 1 : 2
• B. 2 : 3
• C. 2 : 1
• D. 2 : 5

Answer 1

Answer: (D)

Let there are $n_1$ moles of hydrogen and $n_2$ moles of helium in the given mixture.
As $PV=nRT$ Then the pressure of the mixture
$p=\frac{n_{1}RT}{V}+\frac{n_{2}RT}{V}=\left(n_1+n_2\right)\frac{RT}{V}$
$\Rightarrow 2\times 101.3\times 10^3$
$=\left(n_1+n_2\right)\times \frac{(8.3\times 300)}{20\times 10^{-3}}$
or $\left(n_1+n_2\right)=\frac{2\times 101.3\times 10^{3}20\times 10^{-3}}{(8.3)(300)}$
or, $n_1+n_2=1.62$ ...(1)
The mass of the mixture is (in grams)
$n_1\times 2+n_2\times 4=5$
$\Rightarrow \left(n_1+2n_2\right)=2.5$ ...(2)
Solving the eqns. (1) and (2),
we get $n_1=0.74$ and $n_2=0.88$
Hence,$\frac{m_H}{m_{He}=}=\frac{0.74\times 2}{0.88\times 4}$
$=\frac{1.48}{3.52}=\frac{2}{5}$