Pressure of $1 \mathrm{~g}$ of an ideal gas $\mathrm{A}$ at $27^{\circ} \mathrm{C}$ is found to be 2 bar. When $2 \mathrm{~g}$ of another ideal gas $B$ is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.

Question

Pressure of $1 \mathrm{~g}$ of an ideal gas $\mathrm{A}$ at $27^{\circ} \mathrm{C}$ is found to be 2 bar. When $2 \mathrm{~g}$ of another ideal gas $B$ is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.,

MARKS App · Accepted Answer

Suppose molecular masses of $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{M}_{\mathrm{A}}$ and $\mathrm{M}_B$ respectively. Then their number of moles will be $\mathrm{n}_{\mathrm{A}}=\frac{1}{\mathrm{M}_{\mathrm{A}}}, \mathrm{n}_{\mathrm{B}}=\frac{2}{\mathrm{M}_{\mathrm{B}}} \quad\left(\because \mathrm{n}=\frac{\text { given mass }}{\text { molar mass }}\right)$ $P_A=2$ bar, $P_A+P_B=3$ bar, i.e., $P_B=1$ bar Applying the relation $P V=n R T$ $\begin{aligned} &\mathrm{P}_{\mathrm{A}} \mathrm{V}=\mathrm{n}_{\mathrm{A}} \mathrm{RT}, \quad \mathrm{P}_{\mathrm{B}} \mathrm{V}=\mathrm{n}_{\mathrm{B}} \mathrm{RT} \ &\therefore \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}}}=\frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=\frac{1 / \mathrm{M}_{\mathrm{A}}}{2 / \mathrm{M}_{\mathrm{B}}} \ &\text { or } \frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{A}}}=2 \times \frac{\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{B}}}=2 \times \frac{2}{1}=4 \text { or } \mathrm{M}_{\mathrm{B}}=4 \mathrm{M}_{\mathrm{A}} \end{aligned}$

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