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