Question

Nitrogen gas at 300K, 60 atm, has a volume of 418cm^3/mol. Compare this to the value calculated using Van Der Waals equation

Answer 1

The Van Der Waals equation for one mole is given by (P+a/V²)*(V-b)= RT   (1)

a= 1.370 Lit2bar/mol2 and b=0.0387 L/mol           

1 bar= 0.9869 atm a=1.37*0.9869 Lit²atm/mol²

Therefore a =1.352 Lit²atm/mol^2, R=0.08206 L.atm/mole.K

T= 300K RT (RHS of eq.1)= 300*0.08206=24.62

V has to be chosen such that LHS of equation (1) should also be equal to 24.62

Let initial trial be 418cc/mol =0.418L/mol (1000CC= 1L)

LHS of Eq.1= (60+1.352/0.418*0.418)*(0.418-0.0387)=25.69 LHS > RHS

Assume V= 0.5 L LHS of Eq.1=30.17 > RHS

So assume V=0.4L LHS =24.73 L,

V=0.39L LHS= 24.2

Assumed V= 0.395 LHS=24.5

Assumed V= 0.397, LHS= 24.57

Assumed V =0.398, V=24.62

Hence the volume of gas based on Vandewall equation is 0.397L/mol or 397 cm³/mol