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
0.0091 joule
0.0182 joule
– 0.0091 joule
– 0.0182 joule
It is an isothermal process. Hence work done =P(V2−V1)
=1×105×(1.091−1)×10−6=0.0091J