(a) How much work is done on the steam when 3.89 mol of water at 100°C boils and becomes 3.89 mol of steam at 100°C at 1.00 atm pressure? (Assume the latent heat of vaporization of water is 2.26 106 J/kg.)
(b) Assume the steam to behave as an ideal gas. Determine the change in internal energy of the system of the water and steam as the water vaporizes.
Q + W = ΔE
Q = m * Lv
W = -P * ΔV = -P * (V2 - V1)
P * V = n * R * T
The Attempt at a Solution
Convert moles of water to volume (multiply by molar mass, then use density of water)
Volume of water: 0.07002 m^3.
Then find the volume of steam using ideal gas law: V = ((3.89 mol) * (8.31 J / mol K) * (373.15K)) / (101325 Pa)
Volume of steam: 0.1190 m^3
So work should be -P * (V2 - V1) -> -(101325 Pa) * ((.1190 m^3) - (.07002 m^3))
W = -4963 J = -4.963 kJ.
But my web assign homework tells me that I am over 10% away from the correct answer.
I can't do part (b) until I can complete part (a) because W + Q = ΔE, so without knowing the work, I can't find the change in internal energy (at least that's what my prof said when I asked him).
Thanks in advance for any help :)