Exercise 6 | Properties of Gases
General Chemistry 2 - Exercise 6
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 =
⇒ P =
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 )(V – nb) = nRT
⇒ (P + a ) =
⇒ P = - a
n = number of moles = = 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