Question

A mixture of $N_2$ and $Ar$ gases in a cylinder contains $7g$ of $N_2$ and $8g$ of $Ar$. If the total pressure of the mixture of the gases in the cylinder is $27bar$, the partial pressure of $N_2$ is:
[Use atomic masses (in $gmo{l}^{-1}$ $N=14,Ar=40$]

• A. 9 bar
• B. 15 bar
• C. 12 bar
• D. 18 bar

Answer 1

The correct option is B 15 bar

$\text{Partial presure of}{N}_2=\text{mole fraction}\times P_{Total}\text{of}{N}_2$
${\chi }_{N_2}=\frac{n_{N_2}}{n_T}$
$\text{moles of}{N}_2=\frac{7}{28}=\frac{1}{4}=0.25$
$\text{moles of}Ar=\frac{8}{40}=\frac{1}{5}=0.2$
${\chi }_{N_2}=\frac{.25}{0.25+0.2}=\frac{5}{9}$
$P_{N_2}=\frac{5}{9}\times 27=15bar$