Question

A vessel of volume $20\, L$ contains a mixture of hydrogen and helium at temperature of $27^{\circ} C$ and pressure $2\, atm$. The mass of mixture is $5\, g$. 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
$p=\frac{n_{1} R T}{V}+\frac{n_{2} R T}{V}=\left(n_{1}+n_{2}\right) \frac{R T}{V}$
$\Rightarrow 2 \times 101.3 \times 10^{3}$
$=\left(n_{1}+n_{2}\right) \times \frac{(8.3 \times 300)}{20 \times 10^{-3}}$
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)
$n_{1} \times 2+n_{2} \times 4=5$
$\Rightarrow \left(n_{1}+2 n_{2}\right)=2.5$ ...(2)
Solving the eqns. (1) and (2),
we get $n_{1}=0.74$ and $n_{2}=0.88$
Hence,$\frac{m_{H}}{m_{H e}=}=\frac{0.74 \times 2}{0.88 \times 4}$
$=\frac{1.48}{3.52}=\frac{2}{5}$