Question

Given that the density of air at 0.987 bar and 27°C is 1.146 kg m 3, calculate the mole fraction and partial pressure of nitrogen and oxygen assuming that a) air consists only of these two gases b) air also contains 1.0 mole per cent Ar. Please show step by step of calculations.

Answer 1

(a) Calculate the absolute temperature, T = (27 + 273) K = 300 K.

Assume the volume of air to be V = 1.000 m³.

Given the density of air is ρ = 1.146 kg/m³, the mass of air is

m = (1.146 kg/m³)*(1.000 m³) = 1.146 kg.

The pressure of the gas is P = 0.987 bar; calculate the total number of moles of gases in air as

n =P*V/RT = (0.987 bar)*(1.000 m³)/(8.314*10^-5 bar.m³/mol.K).(300 K) = 39.5718 mole.

Use the set of equations n = n(N₂) + n(O₂) where n(N₂) and n(O₂) are the number of moles of N₂ and O₂. Therefore,

n(N₂) + n(O₂) = 39.5718

===> n(N₂) = 39.5718 – n(O₂)

Again, m = m(N₂) + m(O₂)

===> MW(N₂)*n(N₂) + MW(O₂)*n(O₂) = 1.146 [MW denotes the molecular mass; MW*n = mass of the gas taken].

Therefore,

(0.02802 kg)*n(N₂) + (0.0319988 kg)*n(O₂) = 1.146 kg

===> (0.02802)*[39.5718 – n(O₂)] + (0.0319988)*n(O₂) = 1.146

===> 1.10880 – 0.02802*n(O₂) + 0.0319988*n(O₂) = 1.146

===> 0.0039188*n(O₂) = 0.0372

===> n(O₂) = 0.0372/0.0039188 = 9.4927

The number of moles of O₂ is 9.4927 mole and hence, the number of moles of N₂ = (39.5718 – 9.4927) mole = 30.0791 mole.

Mole fraction of N₂ = (moles of N₂)/(total number of moles of air) = (30.0791 mole)/(39.5718 mole) = 0.7601 (ans).

Mole fraction of O₂ = (moles of O₂)/(total number of moles of air) = (9.4927 mole)/(39.5718 mole) = 0.2399 (ans).

Partial pressure of N₂ = (mole fraction of N₂)*(air pressure) = (0.7601)*(0.987 bar) = 0.7502 bar (ans).

Partial pressure of O₂ = (mole fraction of O₂)*(air pressure) = (0.2399)*(0.987 bar) = 0.2368 bar (ans).

(b) It is given than air contains 1 mole% Ar; hence, n(Ar)/n*100 = 1

===> n(Ar) = 1/100*n = 0.010*n = 0.010*(39.5718 mole) = 0.395718 mole.

Use the constraint

n = n(N₂) + n(O₂) + n(Ar)

===> 39.5718 = n(N₂) + n(O₂) + 0.395718

===> n(N₂) + n(O₂) = 39.5718 – 0.395718 = 39.176082

===> n(N₂) = 39.176082 – n(O₂) ……(3)

Again, we have

0.02802*n(N₂) + 0.0319988*n(O₂) + MW*n(Ar) = 1.146

===> (0.02802)*[39.176082 – n(O₂)] + 0.0319988*n(O₂) + 0.039948*0.395718 = 1.146

===> 1.09771 – 0.02808*n(O₂) + 0.0319988*n(O₂) + 0.015808 = 1.146

===> 0.0039188*n(O₂) = 0.032482

===> n(O₂) = 0.032482/0.0039188 = 8.2888

Therefore, n(N₂) = 39.176082 – n(O₂) = 39.176082 – 8.2888 = 30.887282.

Mole fraction of N₂ = (moles of N₂)/(total number of moles of air) = (30.887282 mole)/(39.5718 mole) = 0.7805 (ans).

Mole fraction of O₂ = (moles of O₂)/(total number of moles of air) = (8.2888 mole)/(39.5718 mole) = 0.2095 (ans).

Partial pressure of N₂ = (mole fraction of N₂)*(air pressure) = (0.7805)*(0.987 bar) = 0.7703 bar (ans).

Partial pressure of O₂ = (mole fraction of O₂)*(air pressure) = (0.2095)*(0.987 bar) = 0.2068 bar (ans).