Exercise 6 | Properties of Gases

Calculate the pressure (in bar) exerted by 55.0 g of propane C3H8 (g) confined to a 2.00 L container at 220°C, considering that propane is an ideal gas and then considering propane as a non-ideal gas.

Data: Van der Waals constants of propane: a = 9.392 L2.bar.mol-2 and b = 0.0905 L.mol-1

1) Ideal gas ⇒ Ideal-gas equation:


PV = nRT 

⇒ P = nRTV

⇒ P = mRTMV


m = mass in g

R = ideal gas constant = 8.314 m3.Pa.K-1.mol-1

T = temperature in K ⇒ 220°C = 493 K

M = molar mass ⇒ M = 3 x MC + 8 x MH = 44.1 g.mol-1

V = volume in m3 ⇒ 2.00 L = 2.00 dm3 = 2.00 x 10-3 m3


P = 55.0 × 8.314 × 49344.1 × 2.0 × 10-3= 2.56 x 106 Pa = 25.6 bar

 

2) Non ideal gas ⇒ Van der Waals equation:


(P + a n2V2)(V – nb) = nRT

⇒ (P + a n2V2) = nRTV - nb

⇒ P = nRTV - nb - a n2V2


n = number of moles = 55.044.1 = 1.25 mol

a = 9.392 L2.bar.mol-2 = 9.392 x 10-1 m6.Pa.mol-2

b = 0.0905 L.mol-1 = 9.05 x 10-5 m3.mol-1


P = 2.72 x 106 – 3.67 x 105 = 2.35 x 106 Pa = 23.5 bar