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

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

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

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

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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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120.57 J

148.911 J

150 J

130.24 J

answer is A.

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Detailed Solution

$\begin{array}{l} dQ = dU + dW = dU + pdv \\ dW = pdv = 1 .013 \times 10^{5} \times (1470) \times 10^{- 6} \\ = 148 .91 joule \end{array}$

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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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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