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?

Question

Wade Adams

Chemistry

12 December, 20:02

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?

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Answers (1)

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Ryann · Answer 1

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.