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When 1 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 = $1.01 \times 10^{5} N / m^{2}$ )

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1200J

2100 J

8200J

4200J

answer is D.

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Detailed Solution

m = 1gm = 1 × $10^{- 3}$ kg P=1.01X $1.01 \times 10^{5} N / m^{2}$ $∆ v = 1671 c c - 1 c c = 1670 c c$

L = 540cal/gm; dU = ?

Q = mL = 1 × 10–3 × 540 × 4.2 × 103 J

Q = 2268J

W = PdV = P $(V_{2} - V_{1})$

dW = 1.01 × $10^{5}$ × 1670 × $10^{- 6}$ = 168.67J

ΔU = Q – W = 2268 – 168.67 = 2100J

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