Question

Suppose that 11.1 mol C₂H₆(g) is confined to 4.860 dm³ at 27°C. Predict the pressure exerted by the ethane from the following.
(a) the perfect gas equation of state for which I get 56.3 atm (correct)
(b) the van der Waals equations of state for which I get 37.4 atm (correct)
Calculate the compression factor based on these calculations. For ethane, a = 5.507 dm⁶ atm mol⁻², b = 0.0651 dm³ mol⁻¹.
Z = 1 is incorrect for which I used the formula Vm/Vm (ideal)

Answer 1

a) Note 1 dm³ = 1 litre ; 27 ⁰C = 300 K

Using the ideal gas equation, i.e P*V = n*R*T ; where P = pressure in atm ; V = volume in litres; R = universal gas constant = 0.0821 atm-L/mol/K ; n = number of moles of the gas ; T = temperature in Kelvin

Thus, P = (n*R*T)/V = (11.1*0.0821*300)/4.86 = 56.254 atm = 56.3 atm (approx.)

b) The Vander Waal's Equation is :-

{P + (n²a/V²)}*(V - nb) = n*R*T

Thus, [P + {(11.1²*5.507)/4.86²}]*{4.86 - (11.1*0.0651)} = 11.1*0.0821*300

or, (P + 28.727)*(4.137) = 273.293

or, P = 37.4 atm (approx.)

c) Compression factor = V_real/V_ideal = (p*V)/(R*T) = (37.4*4.86)/(0.0821*300) = 7.38