Pressure  of 1 gm 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 becomes 3 bar. Find a relationship between their molecular masses.

pV = nRT

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

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

Number of moles of A gas ;

${\mathrm{n}}_{\mathrm{A}}=\frac{1}{{\mathrm{M}}_{\mathrm{A}}}\left({\mathrm{M}}_{\mathrm{A}}=\text{ molar mass of }gas\mathrm{A}\right)$

$\text{ Number of moles of }\mathrm{B}\text{ gas; }$

${\mathrm{n}}_{\mathrm{B}}=\frac{2}{{\mathrm{M}}_{\mathrm{B}}}\left({\mathrm{M}}_{\mathrm{B}}=\text{ molar mass of gas }\mathrm{B}\right)$

Pressure of gas A, ${\mathrm{p}}_{\mathrm{A}}=2\text{ bar }$

Total pressure, ${\mathrm{p}}_{\text{total }}={\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}=3\text{ bar }$

Pressure of gas B, ${\mathrm{p}}_{\mathrm{B}}={\mathrm{p}}_{\text{total }}-{\mathrm{p}}_{\mathrm{A}}=3-2=1\mathrm{bar}$

V , R and T are same for both the gases.

Hence, from Eqs.(i) and (ii), we get 

$\frac{{\mathrm{p}}_{\mathrm{A}}}{{\mathrm{p}}_{\mathrm{B}}}=\frac{{\mathrm{n}}_{\mathrm{A}}}{{\mathrm{n}}_{\mathrm{B}}}=\frac{1\times {\mathrm{M}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{A}}\times 2}$

$\frac{{\mathrm{M}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{A}}}=\frac{2{\mathrm{p}}_{\mathrm{A}}}{{\mathrm{p}}_{\mathrm{B}}}=\frac{2\times 2}{1}$

${\mathrm{M}}_{\mathrm{B}}=4{\mathrm{M}}_{\mathrm{A}}$

Alternate method

(b) (i) $\because$${\mathrm{p}}_{\mathrm{T}}={\mathrm{p}}_{\mathrm{A}}+{\mathrm{p}}_{\mathrm{B}}$

Given, ${\mathrm{p}}_{\mathrm{T}}=3\text{ bar and }{\mathrm{p}}_{\mathrm{A}}=2\text{ bar }$

$\therefore {\mathrm{p}}_{\mathrm{B}}=3-2=1\mathrm{bar}$

$\text{ (ii) }\because \mathrm{pV}=\mathrm{nRT}=\frac{\mathrm{w}}{\mathrm{M}}\mathrm{RT}$

$\therefore \mathrm{For} \left(\mathrm{A}\right)$

$2=\frac{1}{{\mathrm{M}}_{\mathrm{A}}}$---------(I)

$\left(\because {\mathrm{p}}_{\mathrm{A}}=2\text{ bar and }\mathrm{V},\mathrm{T}\right.\text{ are constant. }$

$\left.{\mathrm{w}}_{\mathrm{A}}=1 \mathrm{g},{\mathrm{M}}_{\mathrm{A}}=\text{ molar mass of }\mathrm{A}\right)$

Similarly for (B)

$1=\frac{2}{{\mathrm{M}}_{\mathrm{B}}}$ --------(ii)

$\left(\because {\mathrm{p}}_{\mathrm{B}}=1\mathrm{bar}\text{ and }\mathrm{V},\mathrm{T}\right.\text{ are constant }$

${\mathrm{w}}_{\mathrm{B}}=2 \mathrm{g},{\mathrm{M}}_{\mathrm{B}}=\text{ molar mass of }\mathrm{MB}\text{ ) }$

Dividing (i) by (ii)

$\frac{2}{1}=\frac{\frac{1}{{\mathrm{M}}_{\mathrm{A}}}}{\frac{2}{{\mathrm{M}}_{\mathrm{B}}}}=\frac{{\mathrm{M}}_{\mathrm{B}}}{2{\mathrm{M}}_{\mathrm{A}}},\text{ i.e. }{\mathrm{M}}_{\mathrm{B}}=4{\mathrm{M}}_{\mathrm{A}}$