The volume of 1 g of water `(1 cm^3)` becomes `1671 cm^(3)` on being converted to steam at standard atmosphere pressure. Find out the work done and rise in internal energy. Given, latent heat of vaporisation of water =`540 cal cdot g^(-1)`, the standard atmosphere pressure = `1.013 xx 10^(5) N cdot m^(-2)`.

Explanation

`W = int_(V_1)^(V_2) pdV = p int_(V_1)^(V_2) dV = p(V_2 - V_1)`
Here, `V_(2) - V_(1) = 1671 - 1`
`= 1670 cm^(3)`
`=1670 xx 10^(-6) m^(3) = 1.67 xx 10^(-3)m^(3)`
`:. W = (1.013 xx 10^(5)) xx (1.67 xx 10^(-3))`
`= 169.17 J`.
`:. U_(2) - U_(1) = Q - W = 540 xx 4.2 - 169.17`
`[540 cal = 540 xx 4.2J]`
`=2099 J`.