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vaporized at 100Ā°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?

Respuesta :

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Answer:

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

Explanation:

1. Heat absorbed

q = nĪ”H = 1 mol Ɨ (40.57 kJ/1 mol) = 40.57 kJ

2. Change in volume

V(water) = 0.018 L

pV = nRT

1 atm Ɨ V = 1 mol Ɨ 0.082 06 LĀ·atmĀ·Kā»Ā¹molā»Ā¹ Ɨ 373.15 K

V = 30.62 L

Ī”V = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L

3. Work done

w = -pĪ”V = - 1 atm Ɨ 30.60 L = -30.60 LĀ·atm

w = -30.60 LĀ·atm Ɨ (101.325 J/1 LĀ·atm) = -3100 J = -3.10 kJ

4. Why the difference?

Every gas does 3.10 kJ of work when it expands at 100 Ā°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.

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