Question

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

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

Answer 1

Answer: (C)

15 bar

Answer 2

The correct answer is C:15bar
Partial pressure of $N_2 =$ mole fraction $\times P_{Total}$ of $N_2$
$X_{N_2} = \frac{\text{moles of }N_2}{\text{ total moles}}$

moles of $N_2 = \frac{7}{28} = \frac{1}{4};$

moles of $Ar = \frac{8}{40} = \frac{1}{5}$

$X_{N_2} = \frac{(\frac{1}{4})}{\frac{1}{4} + \frac{1}{5}} = \frac{5}{9}$

$P_{N_2} = \frac{5}{9} \times 27 = 15$ bar