# 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 = $\frac{\mathrm{nRT}}{\mathrm{V}}$

⇒ P = $\frac{\mathrm{mRT}}{\mathrm{MV}}$

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 = = 2.56 x 106 Pa = 25.6 bar

2) Non ideal gas ⇒ Van der Waals equation:

(P + a $\frac{{\mathrm{n}}^{2}}{{\mathrm{V}}^{2}}$)(V – nb) = nRT

⇒ (P + a $\frac{{\mathrm{n}}^{2}}{{\mathrm{V}}^{2}}$) =

⇒ P =  - a $\frac{{\mathrm{n}}^{2}}{{\mathrm{V}}^{2}}$

n = number of moles = $\frac{55.0}{44.1}$ = 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