Question

The volumetric fraction of the constituents of a gas mixture at a specified pressure and temperature are given. The mass fraction and partial pressure of each gas are to be determined:
Given: 65% N2 20% O2 15% CO2 at 290K, 250 KPa, The molar masses of N2, O2 and CO2 are 28.0 g/mol, 32.0 g/mol and 44.0 g/mol, respectively.

Answer 1

we know that

for ideal gases

mole fraction = volume fraction

so

mole fraction of N2 = 0.65

mole fraction of 02 = 0.20

mole fraction of CO2 = 0.15

now

Consider 100 kmol of mixture

then

mole of N2 = 0.65 x 100 = 65 kmol

moles of 02 = 20 kmol

moles of C02 = 15 kmol

we know that

mass = moles x molar mass

so

mass of N2 = 65 x 28 = 1820 kg

mass of 02 = 20 x 32 = 640 kg

mass of C02 = 15 x 44 = 660 kg

now

total mass = 1820 + 640 + 660 = 3120 kg

now

mass fraction = mass / total mass

so

mass fraction of N2 = 1820 / 3120 = 0.583

mass fraction of O2 = 640 / 3120 = 0.205

mass fraction of CO2 = 660 / 3120 = 0.212

now

we know that

partial pressure = mol fraction x total pressure

so

partial pressure of N2 = 0.65 x 250 = 162.5 kPa

partial pressure of O2 = 0.2 x 250 = 50 kPa

partial pressure of CO2 = 0.15 x 250 = 37.5 kPa