Question

A vessel of volume $20 \, L$ contains a mixture of hydrogen and helium at a temperature of $27 \,{}^\circ C$ and pressure $2.0 \, atm$ . The mass of the mixture is $5 \, g$ . 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
​​​​ $\text{P}=\frac{\left(\text{n}\right)_{1} \text{RT}}{\text{V}}+\frac{\left(\text{n}\right)_{2} \text{RT}}{\text{V}}=\left(\left(\text{n}\right)_{1} + \left(\text{n}\right)_{2}\right)\frac{\text{RT}}{\text{V}}$
$2\times \text{101.3}\times \left(10\right)^{3}=\left(\left(\text{n}\right)_{1} + \left(\text{n}\right)_{2}\right)\frac{\left(\text{8.3}\right) \left(300\right)}{20 \times \left(10\right)^{- 3}}$
$\left(\left(\text{n}\right)_{1} + \left(\text{n}\right)_{2}\right)=\frac{2 \times \text{101.3} \times 1 0^{3} \times 2 0 \times 1 0^{- 3}}{\left(\text{8.3}\right) \left(3 0 0\right)}=\text{1.62}$ ...(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 $\frac{\text{m}_{\text{H}}}{\text{m}_{\text{He}}}=\frac{\text{0.74} \times 2}{\text{0.88} \times 4}=\frac{\text{1.48}}{\text{3.52}} \rightarrow \text{2 : 5}$