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2 June, 04:02

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.

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  1. 2 June, 06:47
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    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
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