Question

vaporized at 100°C and 1 atmosphere pressure. Assuming ideal gas 1 g mole of water is behavior calculate the work done and compare this with the latent heat (40.57 kJ/mole). Why is the heat so much larger than the work?

Answer 1

q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.

Step-by-step 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.