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When 2 gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, the volume increases from 1cc to 1671cc. The heat of vaporization at this pressure is 540 cal/gm. Increase in internal energy of water is (1 atmosphere $= 1 .0 1 \times 1 0^{6} dyne / {cm}^{2}$ )

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4369 J

1200J

4200J

8200J

answer is D.

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Detailed Solution

Given that, $\to d Q = m L = 2 \times 540 \times 4.2 = 4536 J$

$\Rightarrow d w = P (v_{2} - v_{1})$

$= 10^{5} (1671 - 1) \times 10^{- 6}$

$= 167 J$

From the first law of thermodynamics, $d U = d Q - d W = 4536 - 167$

$d U = 4369 J$

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