Calculate the work associated with the vaporization of 1 mol of water is `373 K` and 1 atm. Assume ideal gas behavior.
Strategy : As liquid water absorbs heat, it forms vapor 1 atm pressure. As the amount of vapor increases, the piston rises. The value of `Delta V` is the difference in volume between the initial and final states of the system. The final volume is the volume of one mole of water vapor at the secified conditions. We can find by using the ideal gas equaition `(PV = nRT)`. The initial volume of 1 mol of liquid water `[V = d. m = (1 g mL^(-1)) (18 g) = 18 mL]` is negligible relative to the volume of water vapor. In general the volume of liquid or solid can be neglected in the calculation of `Delta V` because they are so much smaller the the volume of the same amount of gas.

Explanation

`P = 1`
`V_(f) = (nRT)/(P) = ((1 mol) (0.082 (Latm)/(mol L)) (373 K))/((1 atm))`
`= 31 L`
`Delta V = V_(f) - v_(i) = v_(f)`
Thus, `w = - P Delta V`
`= - (1 atm) (31 L) ((101.3J)/(1 L atm)) = - 3140 J`