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Thermodynamics question

2011 · Shift 0 · Q6
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Thermodynamics question

2011 · Shift 0 · Q6

JEE MainChemistryThermodynamicsMCQ+4 / −1
The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27oC is :
  1. A
    35.8 J mol-1 K−1
  2. B
    32.3 J mol-1 K−1
  3. C
    42.3 J mol-1 K−1
  4. D
    38.3 J mol-1 K−1
View written solutionFree

Correct answer: D

  1. Formula for entropy change in isothermal reversible expansion

For an ideal gas undergoing isothermal reversible expansion,

ΔS=nRln⁡(V2V1)\Delta S = nR \ln\left(\frac{V_2}{V_1}\right)ΔS=nRln(V1​V2​​)
  1. Given data
  • Number of moles: n=2n = 2n=2
  • Initial volume: V1=10 dm3V_1 = 10\,\text{dm}^3V1​=10dm3
  • Final volume: V2=100 dm3V_2 = 100\,\text{dm}^3V2​=100dm3
  • Gas constant: R=8.314 J mol−1K−1R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}R=8.314J mol−1K−1
  1. Substitute the values
ΔS=2×8.314×ln⁡(10010)\Delta S = 2 \times 8.314 \times \ln\left(\frac{100}{10}\right)ΔS=2×8.314×ln(10100​) ΔS=16.628×ln⁡(10)\Delta S = 16.628 \times \ln(10)ΔS=16.628×ln(10)

Using

ln⁡(10)=2.303\ln(10) = 2.303ln(10)=2.303

so,

ΔS=16.628×2.303\Delta S = 16.628 \times 2.303ΔS=16.628×2.303 ΔS≈38.3 J K−1\Delta S \approx 38.3\,\text{J K}^{-1}ΔS≈38.3J K−1
  1. Match with the options

The calculated entropy change is

38.3 J K−1\boxed{38.3\,\text{J K}^{-1}}38.3J K−1​

So the correct option is D.

  1. Note on units

The entropy change for the whole sample is 38.3 J K−138.3\,\text{J K}^{-1}38.3J K−1. Although the options mention J mol−1K−1\text{J mol}^{-1}\text{K}^{-1}J mol−1K−1, the numerical value clearly matches the total entropy change for 2 moles.

  1. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.

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