Question

12.0 moles of gas are in a 4.00 L tank at 21.6 ∘C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L2⋅atm/mol2 and b=0.0430 L/mol.

Answer 1

Answer – We are given, the mole = 12 , volume = 4.0 L ,

T = 21.6^oC + 273.15 = 294.75 K

Now we need to calculate the pressure using the ideal gas law

PV = nRT

P = nRT/V

   = 12.0 moles * 0.0821 L.atm.mol^-1.K^-1*294.75 K / 4.0 L

   = 72.60 atm

a=2.300L2⋅atm/mol2 and b=0.0430 L/mol.

van der Waals equation –

(P + n²a / V²) (V-nb) = nRT

(P + 12²*2.300 / 4.00²) (4.00 -12*0.0430) = 12*0.0821 *294.75

(P +20.7) (3.484) = 290.39

P+20.7 = 290.39/ 3.484

            = 83.35

P = 83.35-20.7

   = 62.65 atm

So the difference in pressure between methane and an ideal gas under these conditions

= 72.60 – 62.65

= 9.95 atm