If the mixture of $7 g$ of $N_2$ and $8 g$ of $\mathrm{Ar}$ in a cylinder has the total pressure $27…

If the mixture of $7 g$ of $N_2$ and $8 g$ of $\mathrm{Ar}$ in a cylinder has the total pressure $27 \mathrm{bar}$. What is the partial pressure of $N_2$ ? (Atomic mass of $N=14 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{Ar}=40 \mathrm{~g} \mathrm{~mol}^{-1}$ )
  1. 18 bar
  2. 12 bar
  3. 15 bar
  4. 9 bar

Solution

Moles of $N_2=\frac{7}{28}=\frac{1}{4}$ Moles of $\mathrm{Ar}=\frac{8}{40}=\frac{1}{5}$ Partial pressure $=X_{N_2} \times P_T$ Partial pressure of $\begin{aligned} N_2 & =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{5}} \times 27 \\ & =\frac{\frac{1}{4}}{\frac{9}{20}} \times 27 \\ & =\frac{1}{4} \times \frac{20}{9} \times 27 \\ & =15 \mathrm{bar} \end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 1)

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