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JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Chemical Thermodynamics

Δ_{vap} H^{⊖}

for water is

+ 40.49 kJ mol^{- 1}

at 1 bar and

10 0^{\circ} C

. Change in internal energy for this vapourisation under same condition is ____ kJ

mol^{- 1}

. (Integer answer)
(Given

R = 8.3 JK^{- 1} mol^{- 1}

)

Answer: 38

Solution

Δ_{v a p} H^{⊖} = 40.79

kJ/mol at

373

K (liquid

\to

gas,

Δ n_{g} = 1

).

Δ_{v a p} H = Δ_{v a p} U + Δ n_{g} R T ⟹ Δ_{v a p} U = Δ_{v a p} H - R T

R T = \frac{8.3 \times 373.15}{1000} \approx 3.10 kJ/mol

Δ_{v a p} U = 40.79 - 3.10 \approx 37.7 \approx 38 kJ/mol

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Integer answer

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