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 the helium in the given mixture will be

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

Answer 1

Answer: (D)

$2:5$

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
$P=\frac{n_{1}RT}{V}+\frac{n_{2}RT}{V}=(n_1+n_2)\frac{RT}{V}$
$\Rightarrow 2\times 101.3\times {10}^3=(n_1+n_2)\times \frac{(8.3\times 300)}{20\times {10}^{-3}}$
or, $(n_1+n_2)=\frac{2\times 101.30\times {10}^3\times 20\times {10}^{-3}}{\left(8.3\right)\left(300\right)}$
or, $n_1+n_2=1.62....\left(1\right)$
The mass of the mixture is (in grams)
$n_1\times 2+n_2\times 4=5$
$\Rightarrow (n_1+2n_2)=2.5.....\left(2\right)$
Solving the eqns. $\left(1\right)$ and $\left(2\right)$, we get
$n_1=0.74$ and $n_2=0.88$
Hence, $\frac{m_H}{m_{He}}=\frac{0.74\times 2}{0.88\times 4}=\frac{1.48}{3.52}=\frac{2}{5}$.