Question

A closed bulb of capacity $200$ ml containing $C{H}_4,H_2$ and He at $300$K. The ratio of partial pressures of $C{H}_4,H_2$ and He, respectively is $2:3:5$. Calculate the ratio of their weights present in the container.

Answer 1

For mixture of gases under identical condition of Temperature and Volume

$p_i={\chi }_i.P$

$p_i$= partial pressure of the gas

and ${\chi }_i$=mole fraction of the gas

P=total pressure of the gas mixture

also ${\chi }_i=\frac{n_i}{n_T}=\frac{m_i}{M_i\times n_T}$

$n_i=$moles of ith gas

$M_i=$molar mass of ith gas

$n_T=$total number of moles of gases in the mixture

$\therefore p_i=P.\frac{m_i}{M_i\times n_T}$

or $p_i\propto \frac{m_i}{M_i}$ (for a given mixture)

Given that $p_{C{H}_4}:p_{H_2}:p_{He}::2:3:5$

thus $\frac{m_{C{H}_4}}{M_{C{H}_4}}:\frac{m_{H_2}}{M_{H_2}}:\frac{m_{He}}{M_{He}}::2:3:5$

or $\frac{m_{C{H}_4}}{16}:\frac{m_{H_2}}{2}:\frac{m_{He}}{4}::2:3:5$

$m_{C{H}_4}:m_{H_2}:m_{He}::2\times 16:3\times 2:5\times 4$

$m_{C{H}_4}:m_{H_2}:m_{He}::16:3:10$