Previous Year NEET Question- Solution L-6

Question: A mixture of $\mathrm{N}_2$ and $\mathrm{Ar}$ gases in a cylinder contains $7 \mathrm{~g}$ of $\mathrm{N}_2$ and $8 \mathrm{~g}$ of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of $\mathrm{N}_2$ is#

[Use atomic masses (in $\mathrm{g} \mathrm{mol}^{-1}$ ) : $\mathrm{N}=14, \mathrm{Ar}=40]$

A) 12 bar

B) 15 bar

C) 18 bar

D) 9 bar

Answer: 15 bar#

Solution:#

From Dalton’s law of partial pressure of gases. We know, $$ p_i=\chi_i \times p $$ where, $p_i=$ partial pressure of $i$ th component $\chi_i=$ mole-fraction of ith component $p=$ total pressure $=27$ bar (partial pressure) $){N_2}$ $$ \begin{aligned} & =(\text { mole-fraction }){\mathrm{N}2} \times p \ & =\frac{n{\mathrm{N}2}}{n{\mathrm{N}2}+n{\mathrm{Ar}}} \times p \ & =\frac{\frac{7}{28}}{\frac{7}{28}+\frac{8}{40}} \times 27 \text { bar }=15 \text { bar } \end{aligned} $$ $\therefore$ Partial pressure of $\mathrm{N}_2$ in the mixture is 15 bar.