Pressure of the 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at Same temperature, the pressure become 3 bar. Relationship between their molecular masses is M_B = xMA. Find the value of x.

pV = nRT

For A gas, ${\mathrm{p}}_{\mathrm{A}}\mathrm{V}={\mathrm{n}}_{\mathrm{A}}\mathrm{RT} -----\left(\mathrm{i}\right)$

$\text{ Similarly for }\mathrm{B}\text{ gas }{\mathrm{p}}_{\mathrm{B}}\mathrm{V}={\mathrm{n}}_{\mathrm{B}}\mathrm{RT}-------\left(\mathrm{ii}\right)$

(V is same because the gases are taken in the same flask)

${\mathrm{n}}_{\mathrm{A}}=\frac{1}{{\mathrm{M}}_{\mathrm{A}}},\text{ similarly }{\mathrm{n}}_{\mathrm{B}}=\frac{2}{{\mathrm{M}}_{\mathrm{B}}}\text{ (where, }{\mathrm{M}}_{\mathrm{A}}\text{ and }{\mathrm{M}}_{\mathrm{B}}\text{ are the molar masses of gases A and B respectively)}$

$$\begin{gathered} {\mathrm{p}}_{\mathrm{A}}=2\text{ bar, }\mathrm{p}={\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}=3\text{ bar } \\ \therefore {\mathrm{p}}_{\mathrm{B}}=3-2=1\mathrm{bar} \\ \frac{{\mathrm{p}}_{\mathrm{A}}}{{\mathrm{p}}_{\mathrm{B}}}=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{B}}}=\frac{1\times {\mathrm{M}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{A}}\times 2}\Rightarrow \frac{{\mathrm{M}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{A}}}=\frac{2{\mathrm{p}}_{\mathrm{A}}}{{\mathrm{p}}_{\mathrm{B}}}=\frac{2\times 2}{1} \\ \Rightarrow {\mathrm{M}}_{\mathrm{B}}=4{\mathrm{M}}_{\mathrm{A}} \\ \mathrm{Value} \mathrm{of} \mathrm{x} =4 \end{gathered}$$