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Thermodynamics question

2021 · 27 Jul · Shift 1 · Q20
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Thermodynamics question

2021 · 27 Jul · Shift 1 · Q20

JEE MainChemistryThermodynamicsNumerical+4 / −1
For water at 100 ∘^\circ∘ C and 1 bar, Δ\DeltaΔ vap H −Δ-\Delta−Δ vap U = ‾×\underline{\hspace{2cm}}\times​× 102 J mol −-− 1. (Round off to the Nearest Integer) [Use : R = 8.31 J mol −-− 1 K −-− 1] [Assume volume of H2O(l)H_2O(l)H2​O(l) is much smaller than volume of H2O(g)H_2O(g)H2​O(g). Assume H2O(g)H_2O(g)H2​O(g) treated as an ideal gas]
Numerical answer
View written solutionFree

Correct answer: 31

  1. We use the relation

H=U+PVH = U + PVH=U+PV

So for vaporization,

ΔvapH−ΔvapU=Δ(PV)\Delta_{\text{vap}} H - \Delta_{\text{vap}} U = \Delta(PV)Δvap​H−Δvap​U=Δ(PV)

  1. For the process

H2O(l)→H2O(g)\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g)H2​O(l)→H2​O(g)

we have

Δ(PV)=PVg−PVl\Delta(PV) = P V_{g} - P V_{l}Δ(PV)=PVg​−PVl​

Given that the volume of liquid water is negligible compared to steam,

Δ(PV)≈PVg\Delta(PV) \approx P V_gΔ(PV)≈PVg​

  1. Steam is treated as an ideal gas, so for 1 mol:

PVg=RTPV_g = RTPVg​=RT

Hence,

ΔvapH−ΔvapU=RT\Delta_{\text{vap}} H - \Delta_{\text{vap}} U = RTΔvap​H−Δvap​U=RT

  1. Now substitute the values:
  • R=8.31 J mol−1K−1R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}R=8.31 J mol−1K−1
  • T=100∘C=373 KT = 100^\circ C = 373\ \text{K}T=100∘C=373 K

So,

RT=8.31×373=3099.63 J mol−1RT = 8.31 \times 373 = 3099.63\ \text{J mol}^{-1}RT=8.31×373=3099.63 J mol−1

  1. The question asks in the form:

‾×102 J mol−1\underline{\hspace{2cm}} \times 10^2\ \text{J mol}^{-1}​×102 J mol−1

Thus,

3099.63 J mol−1=30.9963×102 J mol−13099.63\ \text{J mol}^{-1} = 30.9963 \times 10^2\ \text{J mol}^{-1}3099.63 J mol−1=30.9963×102 J mol−1

Rounding to the nearest integer:

313131

Therefore,

31\boxed{31}31​

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