Question

a chemical engineer places 2.12 mol Ar, 1.37 mol N2 and 0.419 mol Kr in a 6.00 L piton-cylinder assemply at 25C and 0.825 atm. what is the partial pressure of Ar

Answer 1

According to Raoult's law ,

The partial pressure of a gas = mole fraction x total pressure of the gaseous mixture

The number of moles of Ar , n = 2.12 mol

The number of moles of N₂ , n' = 1.37 mol

The number of moles of Kr , n'' = 0.419 mol

Total number of moles , N = n + n' + n''

                                      = 2.12+1.37+0.419

                                     = 3.909 mol

Mole fraction of Ar , X = n / N

                                = 2.12 / 3.909

                               = 0.542

Partial pressure of Ar , pAr = X * P                   P = total pressure = 0.825 atm

                                       = 0.542 * 0.825 atm

                                       = 0.447 atm