$1 \mathrm{cm}^3$ of water at its boiling point absorbs 540 calories of heat to become steam with a volume of $1671 \mathrm{cm}^3$ .If the atmospheric pressure = $1.013 \times 10^{5} \mathrm{N}/\mathrm{m}^{2}$ and the mechanical equivalent of heat = 4.19J/calorie , the energy spent in this process in overcoming intermolecular forces is
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$\Delta Q = \Delta U + \Delta W$
$\therefore \Delta U = \Delta Q - \Delta W = 540 - \frac{\mathrm{P}(V_2 / V_1)}{\mathrm{J}}$ $= 540 - \frac{1.013 \times 10^5 \times [(1671 - 1) \times 10^{-6}]}{4.2}$
= 540 - 39.7 = 500 calories
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