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the heat of vaporization of liquid oxygen at 1.01325bar is 6820JMol at its boiling point,-183C, at that pressure. For the reversible vaporization of liquid oxygen calculate (a)q per mole (b) change in U (c) change in S (d) change in G

  • chemistry -

    q per one mole at constant pressure = ∆H = enthalpy of vapourization

    = 6820 J

    ∆U = q + w

    = ∆H - P∆V\

    ∆V will be equal to volume of one mole gas because volume of liquid one mole at that temperature will be negligible compared to volume of gas

    volume of one mole of gas = 1*8.314*90/101325 m3

    = 7.4 x10-3 m3

    = 6280 - 101315 Pa*7.4 x10-3 m3

    = 6280 - 750

    = 5530 J

    As the liquid and gas are in equilibrium at the boiling point ΔG = 0

    ∆G = ∆H -T∆S

    ∆S = ∆H/T

    = 6820/90

    = 75.8 J/K

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