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Q.

While boiling 1 gm of water at pressure 1.013×105 N/m3, its volume 1471 cm3 from 1 cm3, then work done by the system is

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a

148.911 J

b

150 J

c

130.24 J

d

120.57 J

answer is A.

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Detailed Solution

dQ=dU+dW=dU+pdvdW=pdv=1.013×105×(1470)×10−6=148.91 joule
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