The latent heat of evaporation of water is 80 kJ mol-1 . If 100 g wate

1.	The latent heat of evaporation of water is 80 kJ mol-1 . If 100 g water are evaporated at 100 °C and 1 atm, calculate W, ΔH, ΔU and Q.
Answer» Given : Latent heat of evaporation = ΔH = 80 kJ mol^-1 of water Temperature = T = 273 + 100 = 373 K Pressure = P = 1 atm Mass of water = m = 100 g Molar mass of water = 18 g mol^-1 W = ?, ΔH = 1, U = ?, Q = ? Number of moles of water = \(\frac{m}{M}\) = \(\frac{100}{8}\) = 5.556 mol H₂O_(I) → H₂O_(g) 5.556 mol Change in number of moles = Δn = 5.556 – 0 = 5.556 mol For evaporation of 1 mol H₂O, ΔH = 80 kJ For 5.556 mol H₂O, ΔH= 80 × 5.556 = 444.5 kJ In this reaction, the work will be of expansion. W= -ΔnRT = -5.556 × 8.314 × 373 = – 17230 J = -17.23 kJ Now, ΔH = ΔU + ΔnRT ΔU = ΔH – ΔnRT = 444.5 – 5.556 × 8.314 × 373 × 10 = 444.5 – 17.23 = 427.27 kJ In this, Q = Q_P = ΔH = 444.5 kJ ∴ W= -17.23 kJ; ΔH = 444.5 kJ ΔU= 427.21 kJ, Q = 444.5 kJ

The latent heat of evaporation of water is 80 kJ mol-1 . If 100 g water are evaporated at 100 °C and 1 atm, calculate W, ΔH, ΔU and Q.

Answer»

Given :

Latent heat of evaporation = ΔH

= 80 kJ mol^-1 of water

Temperature = T = 273 + 100 = 373 K

Pressure = P = 1 atm

Mass of water = m = 100 g

Molar mass of water = 18 g mol^-1

W = ?,

ΔH = 1,

U = ?,

Q = ?

Number of moles of water = \(\frac{m}{M}\) = \(\frac{100}{8}\) = 5.556 mol

H₂O_(I) → H₂O_(g)

5.556 mol

Change in number of moles = Δn

= 5.556 – 0

= 5.556 mol

For evaporation of 1 mol H₂O,

ΔH = 80 kJ

For 5.556 mol H₂O,

ΔH= 80 × 5.556

= 444.5 kJ

In this reaction, the work will be of expansion.

W= -ΔnRT

= -5.556 × 8.314 × 373

= – 17230 J

= -17.23 kJ

Now,

ΔH = ΔU + ΔnRT

ΔU = ΔH – ΔnRT

= 444.5 – 5.556 × 8.314 × 373 × 10

= 444.5 – 17.23

= 427.27 kJ

In this, Q = Q_P = ΔH = 444.5 kJ

∴ W= -17.23 kJ;

ΔH = 444.5 kJ

ΔU= 427.21 kJ,

Q = 444.5 kJ

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