For water at 100 ° C and 1 bar, Δ _ vap H - Δ _ vap U = _ ¯ × 10 ^ 2 J mol ^ - 1 (Round off to the Nearest Integer) [Use : R = 8 . 31J mol ^ - 1 K ^ - 1 ] [Assume volume of H _ 2 O ( l ) is much smaller than volume of H

Correct answer: 0 H _ 2 O _ ( ℓ ) ⇌ H _ 2 O _ ( v ) Δ H = Δ U + Δ n _ g RT for 1 mole waters ; Δ n _ g = 1 ∴ Δ n _ g RT = 1mol × 8 . 31J / mol - k × 373K = 3099 . 63J ≅ 31 × 10 ^ 2 J

JEE Main 202127 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 July
Shift: Morning Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Main 2021 · 27 July · Morning Shift · Chemistry · Thermodynamics (C)

For water at 100 ° C and 1 bar, Δ _ vap H - Δ _ vap U = _ ¯ × 10 ^...

For water at

100 ° C and 1

bar,

{\Delta}_{vap}H - {\Delta}_{vap}U = \mathop{}_{¯} \times {10}^{2}J{mol}^{- 1}

(Round off to the Nearest Integer)
[Use : \leftR = 8 . 31J{mol}^{- 1}{K}^{- 1}]

<br />[Assume volume of

{H}_{2}O ( l )

is much smaller than volume of

{H}_{2}O ( g )

. Assume

{H}_{2}O ( g )$ treated as an ideal gas]

Correct answer

Step-by-step solution

{H}_{2}{O}_{( ℓ )} ⇌ {H}_{2}{O}_{( v )}

\Delta H = \Delta U + {\Delta n}_{g}RT

for

1

mole waters ;

{\Delta n}_{g} = 1

∴ {\Delta n}_{g}RT = 1mol \times 8 . 31J / mol - k \times 373K

= 3099 . 63J ≅ 31 \times {10}^{2}J

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