Question

15.0 moles of gas are in a 5.00 L tank at 23.1 ∘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/mol2and b=0.0430 L/mol.

Express your answer with the appropriate units.

Answer 1

Number of moles of gas = 15.0 , Volume = 5.0 L , Temperature = 23.1 + 273 = 296.1 K, P = ?

R= 0.08206 atm.L/mol K

Ideal gas condition:

PV = nRT

P = nRT/V = (15.0 moles * 0.08206 atm.L/mol K * 296.1 K)/5.0 L = 72.89 atm.

Vanderwals equation: a = 2.300L2⋅atm/mol2 , b=0.0430 L/mol.

[ P + a (n/V)²] [V/n - b] = RT

We have , a = 2.300L2⋅atm/mol2 , b=0.0430 L/mol.

[P + 2.300L2⋅atm/mol2 (15 mol/5 L)²] [ 5L/15 mol - 0.0430 L/mol] = 0.08206 atm.L/mol K * 296.1 K

[P + 20.7 atm] [0.29 L/mol] = 24.45 atm.L/mol

[P + 20.7 atm] = (24.45 atm.L/mol)/ [0.29 L/mol]

[P + 20.7 atm] = 84.31 atm

P = 84.31 atm - 20.7 atm = 63.61 atm.

DIfference in the pressures between the ideal gas and methane = 72.89 atm - 63.61 atm = 9.28 atm