JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Chemical Thermodynamics

For water

Δ_{vap} H = 41 kJ mol^{- 1}

at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is

____

kJ mol^{- 1}

. [Use:

R = 8.3 J mol^{- 1} K^{- 1}

]

Answer: 38

Solution

Evaporation turns one mole of liquid into one mole of vapour, and the liquid volume is negligible beside the vapour volume, so

Δ n_{g} = 1

and

Δ U = Δ H - Δ n_{g} R T

Substituting the data,

Δ U = 41 - \frac{8.3 \times 373}{1000} = 41 - 3.10 = 37.90 kJ mo l^{- 1}

The internal energy change on evaporation is

37.9 kJ mo l^{- 1}

, which to the nearest integer is

38 kJ mo l^{- 1}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Numerical answer

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