Question

If 1.00 mol of argon is placed in a 0.500-L container at 20.0 ∘C , what is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation)?

For argon, a=1.345(L2⋅atm)/mol2 and b=0.03219L/mol.

Answer 1

Hey There!

ideal gas law :

PV = nRT

P = nRT/V

P = (1*0.082)(18+273)/(0.5) = 47.724 atm

For VDW :

(P + a(n/V)²)(V - nB) = nRT

P = nRT/(V - nB) - a(n/V)²

P = 1*0.082*(18+273) / (0.5-1*0.03219) - 1.345*( 1/0.5 )² = 45.62

P = 45.62 atm

Pdif = P2-P1 = 47.724 - 45.62 = 2.104 atm