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2 September, 13:08

a certain quantity of gas occupies a volume of 0.1 L when collected over water at 288 K and a pressure of 0.92 bar. the same volume of gas occupied a volume 0.85 L at S. T. P in dry conditions. Calculate the aqueous tension at 288K.

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  1. 2 September, 14:40
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    We know that

    Ptotal = Pdry gas + Pwater = Pdry gas + aqueous tension [Dalton's law]

    Ptotal = 0.92 bar

    So

    Aqueous tension = 0.92 - Pdry gas

    Now moles of gas is constant

    We can use ideal gas equation

    PV = nRT

    P1V1 / T1 = P2V2 / T2 [n and R constant]

    P1 X 0.1 / 288 = 1 X 0.085 / 273.15 = 0.896

    so aqueous tension = 0.92 - 0.896 = 0.024 bar
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