When $1\, gm$ of water at ${0^o}C$ and $1 \times {10^5}\;N/{m^2}$ pressure is converted into ice of volume $1.091\;c{m^2}$, the external work done will be
A$0.0091\, joule$
B$0.0182 \,joule$
C$-0.0091 \,joule$
D$-0.0182\, joule$
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A$0.0091\, joule$
a (a) It is an isothermal process. Hence work done $ = P({V_2} - {V_1})$
$ = 1 \times {10^5} \times (1.091 - 1) \times {10^{ - 6}} = 0.0091\;J$
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