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Thermodynamics question

2023 · 12 Apr · Shift 1 · Q22
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Thermodynamics question

2023 · 12 Apr · Shift 1 · Q22

JEE MainChemistryThermodynamicsNumerical+4 / −1
One mole of an ideal gas at 350 K350 \mathrm{~K}350 K is in a 2.0 L2.0 \mathrm{~L}2.0 L vessel of thermally conducting walls, which are in contact with the surroundings. It undergoes isothermal reversible expansion from 2.0 L to 3.0 L3.0 \mathrm{~L}3.0 L against a constant pressure of 4 atm4 \mathrm{~atm}4 atm. The change in entropy of the surroundings ( ΔS)\Delta \mathrm{S})ΔS) is ‾\underline{\hspace{2cm}}​JK−1\mathrm{J} \mathrm{K}^{-1}JK−1(Nearest integer) Given: R=8.314 J K−1 mol−1\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}R=8.314 J K−1 mol−1.
Numerical answer
View written solutionFree

Correct answer: -1

  1. Given data
  • n=1n=1n=1 mol
  • T=350 KT=350\,\mathrm{K}T=350K
  • Isothermal expansion: V1=2.0 LV_1=2.0\,\mathrm{L}V1​=2.0L to V2=3.0 LV_2=3.0\,\mathrm{L}V2​=3.0L
  • External constant pressure: Pext=4 atmP_{\text{ext}}=4\,\mathrm{atm}Pext​=4atm
  • Ideal gas, thermally conducting walls, so temperature remains constant.

We need entropy change of the surroundings.

  1. Use first law for isothermal process of ideal gas

For an ideal gas in an isothermal process,

ΔU=0\Delta U=0ΔU=0

So,

q=−wq=-wq=−w

Since the gas expands against a constant external pressure, the work done by the system is

w=−PextΔVw=-P_{\text{ext}}\Delta Vw=−Pext​ΔV

(using chemistry sign convention).

Now,

ΔV=V2−V1=3.0−2.0=1.0 L\Delta V=V_2-V_1=3.0-2.0=1.0\,\mathrm{L}ΔV=V2​−V1​=3.0−2.0=1.0L

Hence,

w=−(4 atm)(1.0 L)=−4 L atmw=-(4\,\mathrm{atm})(1.0\,\mathrm{L})=-4\,\mathrm{L\,atm}w=−(4atm)(1.0L)=−4Latm

Convert to joules:

1 L atm=101.325 J1\,\mathrm{L\,atm}=101.325\,\mathrm{J}1Latm=101.325J

So,

w=−4×101.325=−405.3 Jw=-4\times 101.325=-405.3\,\mathrm{J}w=−4×101.325=−405.3J

Therefore,

q=−w=405.3 Jq= -w = 405.3\,\mathrm{J}q=−w=405.3J

This is heat absorbed by the gas, so the surroundings lose this heat:

qsurr=−405.3 Jq_{\text{surr}}=-405.3\,\mathrm{J}qsurr​=−405.3J
  1. Entropy change of surroundings

The surroundings are effectively a thermal reservoir at 350 K350\,\mathrm{K}350K, so

ΔSsurr=qsurrT\Delta S_{\text{surr}}=\frac{q_{\text{surr}}}{T}ΔSsurr​=Tqsurr​​

Thus,

ΔSsurr=−405.3350=−1.158 J K−1\Delta S_{\text{surr}}=\frac{-405.3}{350}=-1.158\,\mathrm{J\,K^{-1}}ΔSsurr​=350−405.3​=−1.158JK−1

Nearest integer:

ΔSsurr≈−1 J K−1\Delta S_{\text{surr}}\approx -1\,\mathrm{J\,K^{-1}}ΔSsurr​≈−1JK−1
  1. Check against stored answer

Stored correct answer = 333

My derived answer is −1-1−1, which does not match the stored answer.

A value near 333 would arise if one incorrectly used the reversible heat of the gas:

qrev=nRTln⁡V2V1q_{\mathrm{rev}}=nRT\ln\frac{V_2}{V_1}qrev​=nRTlnV1​V2​​

leading to

ΔSgas=nRln⁡32≈3.37 J K−1\Delta S_{\text{gas}}=nR\ln\frac{3}{2}\approx 3.37\,\mathrm{J\,K^{-1}}ΔSgas​=nRln23​≈3.37JK−1

which is the entropy change of the gas/system, not of the surroundings.

Since the question explicitly asks for entropy change of the surroundings, we must use the actual heat exchanged with surroundings:

ΔSsurr=−405.3350≈−1.16 J K−1\Delta S_{\text{surr}}=\frac{-405.3}{350}\approx -1.16\,\mathrm{J\,K^{-1}}ΔSsurr​=350−405.3​≈−1.16JK−1

So the nearest integer is −1-1−1.

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