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Thermodynamics question

2012 · Q118
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Thermodynamics question

2012 · Q118

NEETChemistryThermodynamicsMCQ+4 / −1
Standard enthalpy of vaporisation Δ\DeltaΔvapHo for water at 100oC is 40.66 kJ mol−-−1. The internal energy of vaporisation of water at 100oC (in kJ mol−-−1) is
  1. A
    +37.56
  2. B
    −-−43.76
  3. C
    + 43.76
  4. D
    + 40.66
View written solutionFree

Correct answer: A

H2O(l) →100∘ C\xrightarrow{100^\circ \,C} 100∘C​ H2O(g)

∆Ho = 40.66kJ mol–1

∆Ho = ∆uo + Δ\Delta Δng RT

Δ\Delta Δng = 1, R = 8.314 × 10–3 kJ mol–1 k–1

T = 100 + 273 = 373 K

⇒\Rightarrow⇒ 40.66 = ∆uo + (1) (8.314 × 10–3) × 373

⇒\Rightarrow⇒ ∆uo = 37.56 kJ mol–1

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