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24 December, 07:23

Water standing in the open at 33.0°C evaporates because of the escape of some of the surface molecules. The heat of vaporization (557 cal/g) is approximately equal to εn, where ε is the average energy of the escaping molecules and n is the number of molecules per gram. (a) Find ε. (b) What is the ratio of ε to the average kinetic energy of H2O molecules, assuming the latter is related to temperature in the same way as it is for gases?

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  1. 24 December, 07:40
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    a) ε = 6.961 * 10⁻²⁰Joules

    b) The ratio of ϵ to the average kinetic energy of H2O molecules = 10.642

    Explanation:

    a) The formula to be used is given below as:

    Heat of Vapourisation (Lv) = εn

    Where: ε = is the average energy of the escaping molecules and

    n = is the number of molecules per gram

    The first step is to convert 557 cal/g to joules/kilogram (j/kg)

    1 cal/g = 4186.8j/kg

    557cal/g = ?

    We cross multiply

    557cal/g * 4186.8j/kg

    = 2332047.6j/kg

    Therefore, 557 cal/g = 2332047.6j/kg

    ε = Lv/n

    ε = LvM/n

    Where Lv = 2332047.6j/kg

    M = 0.018kg/mol

    n = 6.03 * 10²³mol

    ε = (2332047.6j/kg * 0.018kg/mol) : 6.03 * 10²³mol

    = 6.961336119 * 10⁻²⁰Joules

    Approximately, ε = 6.961 * 10⁻²⁰Joules

    b) Kinetic energy = (3/2) KT

    The ratio of ε to Kinetic energy = ε / (3/2) kT = 2ε / 3kT

    Where ε = 6.961 * 10⁻²⁰Joules

    k = 1.38 * 10⁻²³ Joules/kelvin

    T = 33°C, which will be converted to kelvin as

    33°C + 273K

    = 306K

    The ratio of ε to Kinetic energy will be calculated as

    2ε / 3kT

    = (2*6.961 * 10⁻²⁰ Joules) : (3 * 1.38 * 10⁻²³Joules/kelvin * 306K)

    = 10.642

    Hence, The ratio of ϵ to the average kinetic energy of H2O molecules = 10.642
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