Answer:
q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.
Explanation:
1. Heat absorbed
q = nΔH = 1 mol × (40.57 kJ/1 mol) = 40.57 kJ
2. Change in volume
V(water) = 0.018 L
pV = nRT
1 atm × V = 1 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 373.15 K
V = 30.62 L
ΔV = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L
3. Work done
w = -pΔV = - 1 atm × 30.60 L = -30.60 L·atm
w = -30.60 L·atm × (101.325 J/1 L·atm) = -3100 J = -3.10 kJ
4. Why the difference?
Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.
The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.
