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Thermodynamics question

2022 · 27 Jun · Shift 2 · Q17
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Thermodynamics question

2022 · 27 Jun · Shift 2 · Q17

JEE MainChemistryThermodynamicsNumerical+4 / −1
When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is ‾\underline{\hspace{2cm}}​ J. [nearest integer] (Given : R = 8.3 J K −-− 1 mol −-− 1 and log 2 = 0.3010)
Numerical answer
View written solutionFree

Correct answer: 8630

  1. For isothermal reversible expansion of an ideal gas, the maximum work obtained is:
∣wmax⁡∣=nRTln⁡(V2V1)|w_{\max}| = nRT \ln\left(\frac{V_2}{V_1}\right)∣wmax​∣=nRTln(V1​V2​​)

Here,

  • n=5n = 5n=5 mol
  • R=8.3 J mol−1K−1R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}R=8.3 J mol−1K−1
  • T=300 KT = 300\ \text{K}T=300 K
  • V1=10V_1 = 10V1​=10 L
  • V2=20V_2 = 20V2​=20 L
  1. Substitute the values:
∣wmax⁡∣=5×8.3×300×ln⁡(2010)|w_{\max}| = 5 \times 8.3 \times 300 \times \ln\left(\frac{20}{10}\right)∣wmax​∣=5×8.3×300×ln(1020​) =5×8.3×300×ln⁡2= 5 \times 8.3 \times 300 \times \ln 2=5×8.3×300×ln2
  1. Convert log⁡2\log 2log2 to ln⁡2\ln 2ln2 using
ln⁡2=2.303log⁡2\ln 2 = 2.303\log 2ln2=2.303log2

Given log⁡2=0.3010\log 2 = 0.3010log2=0.3010,

ln⁡2=2.303×0.3010≈0.6932\ln 2 = 2.303 \times 0.3010 \approx 0.6932ln2=2.303×0.3010≈0.6932
  1. Now calculate:
5×8.3×300=124505 \times 8.3 \times 300 = 124505×8.3×300=12450

So,

∣wmax⁡∣=12450×0.6932|w_{\max}| = 12450 \times 0.6932∣wmax​∣=12450×0.6932 ≈8630.34 J\approx 8630.34\ \text{J}≈8630.34 J
  1. Nearest integer:
8630\boxed{8630}8630​

Thus, the magnitude of maximum work obtained is 8630 J.

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