Gujarati
11.Thermodynamics
medium

$1c{m^3}$ 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 ..... $cal$

A

$540 $

B

$40 $

C

$500$

D

$0$

Solution

(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$

Standard 11
Physics

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