Question

A vessel of volume V = 20 litre contains a mixture of hydrogen and helium at a temperature $t={20}^{\circ }$ C and Pressure P = 2.0 atm. The mass of the mixture is equal to m = 5.0g. Find the ratio of the mass of hydrogen to that of helium in the given mixture.

• A. 0.961
• B. 2.61
• C. 0.61
• D. 1.062

Answer 1

The correct option is C 0.61

Suppose the number of moles of hydrogen and helium in the mixture are $n_1$ and $n_2$ respectively. If molecular weight of $H_2$ and He are $M_1$ and $M_2$ respectively, then mass of hydrogen in the mixture, $m_1=n_1{M}_1$ and mass of helium in the mixture, $m_2=n_2{M}_2.$
Therefore, total mass of the mixture is
$m=m_1+m_2=n_1{M}_1+n_2{M}_2$ ...(1)
If n is the total number of moles of the mixture in the vessel, then $n=n_1+n_2$ ...(2)
Solving for $n_1$ and $n_2$ from (1) and (2), we get
$n_1=\frac{(n{M}_2-m)}{M_2-M_1}$ and $n_2=\frac{(m-n{M}_1)}{M_2-M_1}$
Therefore, we get
$m_1=M_1\frac{(n{M}_2-m)}{M_2-M_1}$ and $m_2=M_2\frac{(m-n{M}_1)}{M_2-M_1}$
$\therefore \frac{m_1}{m_2}=\frac{M_1}{M_2}.\frac{(n{M}_2-m)}{m-n{M}_1}$ ....(3)
Now, according to gas equation, effective molecular weight of the mixture
$M=\frac{m}{n}=m\frac{RT}{PV}$
Thus, $\frac{m_1}{m_2}=\frac{M_1}{M_2}.\frac{M_2-M}{M-M_1}=\frac{1-\left(\frac{M}{M_2}\right)}{\left(\frac{M}{M_1}\right)-1}$
Given that m = 0.005 Kg,
P = 2 atm = $2.1\times {10}^{5}N/m^2,V=0.02m^3,$
T = 293 K, $M_2=4g,M_1=2g$
Therefore,
$M=m\frac{RT}{PV}=\frac{0.005\times 8.31\times 293}{2.1\times {10}^5\times 0.02}=2.9g$
$\therefore \frac{m_1}{m_2}=\frac{1-\left(\frac{2.9}{4}\right)}{\left(\frac{2.9}{2}\right)-1}=0.61$