Question

A gas mixture consists of $0.4mole$ of ${N_2}$ , $0.6mole$of ${O_2}$ , and $0.2mole$ of argon at a total pressure of $300kPa$ . Calculate the partial pressure of each gas in the mixture?

Answer 1

Given that the mixture consists of three gases and moles of each gas were given. From the moles, the mole fraction can be calculated. By substituting the mole fraction of each gas, and total pressure in Rouault’s law gives the partial pressure of each gas in the mixture.

Complete answer:
Given that A gas mixture consists of $0.4mole$ of ${N_2}$ , $0.6mole$of ${O_2}$ , and $0.2mole$ of argon at a total pressure of $300kPa$ .
According to Rouault’s law, the partial pressure of a gas is equal to the mole fraction of a gas and the total pressure of a mixture. Which was given by ${P_{partial}} = \chi .{P_{total}}$
${P_{partial}}$ is the partial pressure
$\chi$ is mole fraction which is the ratio of moles to the total moles
${P_{total}}$ is total pressure of a mixture
The total number of moles of a mixture is the sum of the moles of ${N_2}$ , ${O_2}$ , and argon gas which will be equal to $0.4 + 0.6 + 0.6 = 1.2moles$
Given that total pressure of a mixture is $300kPa$
Partial pressure of ${N_2}$ gas will be $\dfrac{{0.4}}{{1.2}} \times 300 = 100kPa$
Partial pressure of ${O_2}$ gas will be $\dfrac{{0.6}}{{1.2}} \times 300 = 150kPa$
Partial pressure of argon gas will be $\dfrac{{0.2}}{{1.2}} \times 300 = 50kPa$
Thus, partial pressure of ${N_2}$ is $100kPa$ , partial pressure of ${O_2}$ is $150kPa$ , and partial pressure of argon gas is $50kPa$ .

Note:
By adding the obtained partial pressure of each gas in the mixture the value obtained should be equal to total pressure of a mixture, according to Dalton’s law of partial pressure. In the above mixture, the partial pressure of ${N_2}$ , ${O_2}$ , and argon gas is equal to total pressure.