How much work is done on the steam when 100 mol of water at

How much work is done on the steam when 1.00 mol
of water at 100°C boils and becomes 1.00 mol of steam at
100°C at 1.00 atm pressure.

So far I know that the the mass is from the molar mass of water as we have 1 mol.
Also i solved the amount of energy needed to convert that mass to steam using:

Q = m * L

Q 40,680 Joules

Is it as simple as subtracting the Q of steam @ 1 atm by the Q above, making the difference the work done?

Solution

a) W = -PV = -P(Vs-Vw)= - P(Nrt)/P+P( 18g/(1g/cm3)(10exp6 cm3/m3)

   W = -(1.00mol)(8.314 J/K.mol)(373K)+(1.013 x1 0exp5 N/m2)(18g/10 exp6 g/m3)

       = -3.10 Kj

b) Q = mLw= 0.018Kg(2.26x10exp6 J/kg) = 40.7 Kj

c) Eint = Q + W = 37.6 Kj