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5 March, 07:25

A closed, rigid 2.5 L flask contains a mixture of Ne (g) and F2 (g) with a total pressure of 3.32 atm at 0oC. The system is heated to 15oC. The change in entropy for this process was 0.345 J/K. What is the XNe in the flask

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  1. 5 March, 10:21
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    The heat capacities of Ne and F₂ are:

    Cv (F₂) ≅ (5/2) R

    Cv (Ne) ≅ (3/2) R.

    Using ideal gas equation,

    PV = nRT

    n (total) = PV/RT

    = (3.32 * 2.5) / (0.08206 * 273.15)

    = 0.3703 mol

    For one mole heated at constant volume, the entropy change is:

    ∆S = ∫dq/T

    = ∫ (Cv/T) dT from T1 to T2 = Cvln (T2/T1) = Cvln (288.15/273.15)

    = 0.05346•Cv

    So, for 0.3703 mol,

    ∆S = (0.3703 * 0.05346) Cv

    (0.3703 * 0.05346) Cv = 0.345

    Cv = 17.43 J/mol-K for the Ne/F₂ mixture.

    For pure Ne, Cv = (3/2) R

    = 2.5 * 8.314 J/mol•K

    = 12.471 J/mol•K

    For pure F₂, Cv = (5/2) R

    = 2.5 • 8.314 J/mol•K

    = 20.785 J/mol•K

    Cv (Ne/F₂) = Cv (Ne) + Cv (Ne/F₂)

    17.43 = X * 12.471 + (1 - X) * 20.785

    (20.875 - 8.314) * X = 17.43

    X = 0.415

    1 - X = 0.585

    moles Ne = (0.415 * 0.3703 mol)

    = 0.154 mol
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