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…

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]

MARKS Content Team · Accepted Answer

Correct answer: (B) 15 bar. Mass of N_2 = 7 g Moles of N_2=7/28=1/4 Mass of Ar=8 g Moles of Ar=8/40=1/5 Partial pressure_N2=P_T×mole fraction_N2 =27×(1/4)/(1/4+1/5) =27×(1/4)/(9/20) =27×20/(4×9) =3×5 =15 bar

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

Solution

Relevant Questions

A mixture of N2 and Ar gases in a cylinder contains 7 g of N2 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 N2 is:[Use atomic masses (in g mol–1): N=14,Ar=40]

Solution

Relevant Questions

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