How much work is done on the steam when 1.00 mol of water at `100^(@)C` boils and becomes 1.00 mole of steam at `100^(@)C` at 1.00 atm pressure? Assume the steam to behave as an ideal gas , detemine the change in the internal energy of the material as it vapourizes.

Explanation

Consider the liquid water vapourizing inside a cylinder with a piston. It expands greatly to lift the piston and do the positive work. Many hundreds of joules of negative works is done on the material. Even more positive energy is put into it be heating, to produce a net increase in its internal energy.
We use the idea of the negative integal of PdV to find the work, identifying the volumes of the liquid and the gas as we do so. Then ml will tell us the heat input and the first law of thermodynamics will tell us the change in internal energy.
For a constant pressure process.
`W = - P Delta V = - P (V_(s) - V_(w))`
Where `V_(s)` is the volume of the steam and `V_(W)` is the volume of the liquid water.
We can PV, and `PV_(w)`, find, respectively, from `PV_(s) = vRT`
and `V_(w) = m//rho = nM//rho`
Calculating each work term term, we get
`PV_(s) = (1.00 mol) (8.314 (J)/(K mol)) (373 K) = 3101 J`
`PV_(w) = (1.00 mol) (18.0 g//mol) ((1.013 xx 10^(5) N//m^(2))/(1.00 xx 10^(6) g//m^(3))) = 1.82 J`
Thus the work done is `W = - 3.10 kJ`.
The energy input by heat is
`Q = mL_(v) = (18.0 g) (2.26 xx 106 J//kg) = 40.7 kJ`.
So the change in internal energy is `Delta U = Q + W = 37.6kJ`.
Steam at `100^(@)` is on the point of liquefaction, so it probably behaves differently from the ideal gas we were told to assume. Still, the volume increase by more than a thousand times means that a significant amount of the 'heat of vapourization' is coming out of the sample as it boils, as the work done on the environment in its expansion. Remember that we start with 18 cubic centimeters of liquid water. Visualize it!