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27 December, 04:56

vaporized at 100°C and 1 atmosphere pressure. Assuming ideal gas 1 g mole of water is behavior calculate the work done and compare this with the latent heat (40.57 kJ/mole). Why is the heat so much larger than the work?

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  1. 27 December, 06:19
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    q = 40.57 kJ; w = - 3.10 kJ; strong H-bonds must be broken.

    Explanation:

    1. Heat absorbed

    q = nΔH = 1 mol * (40.57 kJ/1 mol) = 40.57 kJ

    2. Change in volume

    V (water) = 0.018 L

    pV = nRT

    1 atm * V = 1 mol * 0.082 06 L·atm·K⁻¹mol⁻¹ * 373.15 K

    V = 30.62 L

    ΔV = V (steam) - V (water) = 30.62 L - 0.018 L = 30.60 L

    3. Work done

    w = - pΔV = - 1 atm * 30.60 L = - 30.60 L·atm

    w = - 30.60 L·atm * (101.325 J/1 L·atm) = - 3100 J = - 3.10 kJ

    4. Why the difference?

    Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.

    The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.
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