Question

A vessel of volume $20L$ contains a mixture of hydrogen and helium at a temperature of $27{}^{\circ }C$ and pressure $2.0atm$ . The mass of the mixture is $5g$ . Assuming the gases to be ideal, the ratio of the 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)

Suppose there are n₁ moles of hydrogen and n₂ moles of helium in the given mixture. Then the pressure of the mixture will be
​​​​ $\text{P}=\frac{{\left(\text{n}\right)}_1\text{RT}}{\text{V}}+\frac{{\left(\text{n}\right)}_2\text{RT}}{\text{V}}=\left({\left(\text{n}\right)}_1+{\left(\text{n}\right)}_2\right)\frac{\text{RT}}{\text{V}}$
$2\times \text{101.3}\times {\left(10\right)}^3=\left({\left(\text{n}\right)}_1+{\left(\text{n}\right)}_2\right)\frac{\left(\text{8.3}\right)\left(300\right)}{20\times {\left(10\right)}^{-3}}$
$\left({\left(\text{n}\right)}_1+{\left(\text{n}\right)}_2\right)=\frac{2\times \text{101.3}\times 10^3\times 20\times 10^{-3}}{\left(\text{8.3}\right)\left(300\right)}=\text{1.62}$ ...(i)
The mass of the mixture is (in $g$ )
${\text{n}}_1\times 2+{\text{n}}_2\times 4=5$
$\left({\left(\text{n}\right)}_1+2{\left(\text{n}\right)}_2\right)=\text{2.5}$ ......(ii)
Solving Eqs. (i) and (ii), $n_1=0.74$ , $n_2=0.88$
Hence $\frac{{\text{m}}_{\text{H}}}{{\text{m}}_{\text{He}}}=\frac{\text{0.74}\times 2}{\text{0.88}\times 4}=\frac{\text{1.48}}{\text{3.52}}\to \text{2 : 5}$