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]

Question

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] (A) 9 bar (B) 12 bar (C) 15 bar (D) 18 bar

Tardigrade Admin · Accepted Answer

Partial presure of N2 = mole fraction × PTotal of N2 XN2 = ( textmoles of N2/ text total moles) moles of N2 = (7/28) = (1/4); moles of Ar = (8/40) = (1/5) XN2 = (((1/4))/(1)4 + (1)5 = (5/9) P/N2) = (5/9) × 27 = 15 bar

Q. A mixture of N2​ and Ar gases in a cylinder contains 7g of N2​ and 8g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27bar, the partial pressure of N2​ is : [Use atomic masses (in g mol−1) :N=14,Ar=40]

9 bar

12 bar

15 bar

18 bar

Partial presure of N2​= mole fraction ×PTotal​ of N2​ XN2​​= total molesmoles of N2​​ moles of N2​=287​=41​; moles of Ar=408​=51​ XN2​​=41​+51​(41​)​=95​ PN2​​=95​×27=15 bar

Q. A mixture of $N_{2}$ and $A r$ gases in a cylinder contains $7 g$ of $N_{2}$ and $8 g$ of $A r$ . If the total pressure of the mixture of the gases in the cylinder is $27 ba r$ , the partial pressure of $N_{2}$ is :
[Use atomic masses (in g $m o l^{- 1}$ ) : $N = 14, A r = 40$ ]