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**

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