When `1g` of water at `0^(@)C` and `1 xx 10^(4)Nm^(-2)` pressure is converted into ice of volume `1.091 cm^(3)`, find the work done by water? `(rho_(w) = 1 gm//cm^(3))`

Answer

`0.0091J`

Explanation

`m = 1 gm rho_(w) = 1 gm//cm^(3)`
`V_(i) = 1 cm^(3) V_(F) = 1.091 cm^(3)`
`W = PdV`
`= 1xx 10^(5) (1.091-1) xx 10^(-8)`
`= 0.091 xx 10^(5) xx 10^(-6) , W = 0091 J`