Question

$1\, g$ of water at atmospheric pressure has volume of $1\, c c$ and when boiled it becomes $1681\, c c$ of steam. The heat of vaporisation of water is $540\, cal / g$. Then the change in internal energy in this process is

• A. 540 cal
• B. 500 cal
• C. 1681 cal
• D. None of the above

Answer 1

Answer: (B)

Given heat, $Q=m L=1 \times 540\, cal $
$=540 \times 4.18 \times 107\, J$
Work done, $W=P \Delta V$
$P=1\, atm =1 \times 10^{5} N / m ^{2}=10^{6} dyne / cm ^{2}$
$W=10^{6} \times 1680\, J$
Change in internal energy
$\Delta U=Q-W$
$=540 \times 4.18 \times 10^{7}-10^{6} \times 1680$
$=\frac{540 \times 4.18 \times 10^{7}-10^{6} \times 1680}{4.18 \times 10^{7}} cal$
$=500\, cal$