Question

A gas mixture contains each of the following gases at the indicated partial pressures: N2, 211 torr ; O2, 135 torr ; and He, 129 torr .

1. What is the total pressure of the mixture?

2. What mass of each gas is present in a 1.35 −L sample of this mixture at 25.0 ∘C?

Answer 1

1)

Ptotal = sum of all partial pressures

Ptotal = 211+135+129 = 475 torr = 475/760= 0.625 atm

2)

total mass of gas in V = 1.35 L T = 25° = 298 K

apply ideal gas law

PV = nRT

n = PV/(RT)

n = (0.625)(1.35)/(0.082*298) = 0.03452

total mol = 0.03452

we need to find out each compound mass so

x N2 = PN2 / PT =211/475 = 0.444210

x O2 = PO2 / PT =135/475 = 0.28421

x He= PHe / PT =129/475 = 0.2715789

then

mol of N2 = x*mol = 0.444210*0.03452 = 0.015325245

mass of N2 = mol*MW = 0.015325245*28 = 0.429106 g

mol of O2 = x*¨mol = 0.28421*0.03452 = 0.0098109292

mass of O2 = mol*MW = 0.0098109292*32 = 0.31394 g

mol of H2 = x*mol = 0.2715789*0.03452 = 0.00937490

mass of He = mol*MW = 0.00937490*0.03452 = 0.000323621 g