Question

If 2.46 g of Ar are added to 2.75 atm of He in a 2.00 L cylinder at 27.0

Answer 1

Mass of Ar is = 2.46 g

Molar mass of Ar is = 40 g/mol

Number of moles of Ar is ,

Ideal gas equation is PV = nRT

Where n = no.of moles of Ar = 0.0615 mol

R = gas constant = 0.0821 L atm/mol

T = temperature in kelvin = 27.0 oC = 27.0 + 273 = 300 .0 K

V = volume of cylinder = 2.00 L

P = pressure of Ar = ?

Plug the values we have ,

So the partial pressure of Ar is = 0.76 atm

According to Dalton's law ,

Total pressure of the gaseous mixture = sum of the individual partial pressures

                                                       = partial pressure of Ar + partial pressure of He

                                                       = 0.76 + 2.75 atm

                                                      = 3.51 atm