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Q.

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 vaporisation at this pressure is 540 cal/gm. Increase in internal
energy of water is (1 atmosphere = 1.01×10⁶dyne/cm²)

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a

4200J

b

8200J

c

1200J

d

2100 J

answer is D.

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Detailed Solution

$\begin{array}{l} m = 1 gm = 1 \times 10^{- 3} kg \\ p = 1 atm = 1 .01 \times 10^{6} dyne / {cm}^{2} \\ = 1 .01 \times 10^{5} N / m^{2} \\ dv = 1671 cc - 1 cc = 1670 cc = 1670 \times 10^{- 6} m^{3} \\ L = 540 cal / gm; dU = ? \\ dQ = mL = 1 \times 10^{- 3} \times 540 \times 4 .2 \times 10^{3} \\ (dQ)_{P} = 2268 J; dW = PdV = P (V_{2} - V_{1}) \\ dW = 1 .01 \times 10^{5} \times 1670 \times 10^{- 6} = 168 .67 J \\ du = (dQ)_{p} - dW = 2268 - 168 .67 = 2100 J \end{array}$

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