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When 1 gm of water at $0^{o} C$ and $1 \times 10^{5} N / m^{2}$ pressure is converted into ice of volume $1 .091 {cm}^{2}$ , the external work done will be

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0.0091 joule

0.0182 joule

– 0.0091 joule

– 0.0182 joule

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detailed solution

Correct option is A

It is an isothermal process. Hence work done =P(V2−V1)=1×105×(1.091−1)×10−6=0.0091J

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When 1 gm of water at 0oC and 1×105 N/m2 pressure is converted into ice of volume 1.091 cm2 , the external work done will be

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When 1 gm of water at $0^{o} C$ and $1 \times 10^{5} N / m^{2}$ pressure is converted into ice of volume $1 .091 {cm}^{2}$ , the external work done will be