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Standard enthalpy of vaporisation Δvap H⁰ for water at 100⁰C is 40.66 kJ mol⁻¹. The internal energy of vaporisation of water at 100⁰C (in kJ mol⁻¹) is
Options
(a) 37.56
(b) -43.76
(c) 43.76
(d) 40.66
Correct Answer:
37.56
Explanation:
ΔvapH⁰ = 40.66 kJ mol⁻¹,
T = 100 + 273 = 373K,
ΔE = ? , ΔH = ΔE + Δn(g)RT ⇒
ΔE = ΔH – Δn(g) RT,
Δn(g) = number of gaseous moles of products – number of gaseous moles of reactants, H₂O(l) ⇌ H₂O(g), Δn(g) = 1 – 0 = 1, ΔE = ΔH -RT , ΔE = ( 40.66 X 10³) – (8.314 X 373), ΔE = 37559 J/mol or 37.56 kJ/mol.
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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