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Kelvi

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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 $^{-1}$ ): $N=14, Ar=40]$

medium

Equilibrium

2020

chemistry

$12$ bar

$15$ bar

$18$ bar

$9$ bar

Explanation

To find the partial pressure of $N_2,$ we will use the concept of partial pressures and mole fractions. $\\$ Given: $\\$ • Mass of $N_2 = 7$

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g

\\

• Mass of Ar

= 8

g

\\

• Total pressure

P_{total} = 27

bar

\\

Atomic masses:

\\

•

N = 14

g/mol

\\

•

Ar = 40

g/mol

\\

First, calculate the number of moles of each gas:

\\

Moles of

N_2 = \frac{mass \ of \ N_2}{molar \ mass \ of N_2} = \frac{7}{28} = 0.25

mol

\\

Moles of

Ar = \frac{mass \ of \ Ar}{molar \ mass \ of \ Ar} = \frac{8}{40} = 0.2

mol

\\

Total moles of gas in the mixture:

\\ n_{total} = n_{N_2} + n_{Ar} = 0.25 + 0.2 = 0.45

mol

\\

Mole fraction of

N_2: \\ \chi_{N_2} = \frac{n_{N_2}}{n_{total}} = \frac{0.25}{0.45} \\

Calculate

\chi_{N_2}: \\ \chi_{N_2} = \frac{0.25}{0.45} = \frac{5}{9} \\

Partial pressure of

N_2

is given by:

\\ P_{N_2} = \chi_{N_2} \times P_{total} \\ P_{N_2} = \frac{5}{9} \times 27 \\ P_{N_2} = 15

bar

\\

Therefore, the partial pressure of

N_2

is

15

bar, which corresponds to Option 2.