The molar volume of oxygen, O2, is 3.90 dm3 mol-1 at 10.0 bar and 200 degree centigrade. Assuming that the expansion may be truncated after the second term, calculate the second virial coefficient of oxygen at this temperature.

B = - 0.0326 dm³/mol

Explanation:

virial eq until second term:

PVm = RT [ 1 + B/Vm ]

∴ P = 10 bar * (atm / 1.01325 bar) = 9.869 atm

∴ T = 200°C = 473 K

∴ Vm = 3.90 dm³/mol

∴ R = 0.08206 dm³. atm/K. mol

⇒ PVm / RT = 1 + B/Vm

⇒ ((9.869 atm) * (3.90 dm³/mol)) / ((0.08206 dm³. atm/mol. K) * (473K)) = 1 + B/Vm

⇒ 0.99164 = 1 + B/Vm

⇒ B/Vm = - 8.357 E-3

⇒ B = (3.90 dm³/mol) * ( - 8.357 E-3)

⇒ B = - 0.0326 dm³/mol