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Thermodynamics question

2024 · 31 Jan · Shift 2 · Q28
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  5. /2024 · 31 Jan · Shift 2 · Q28

Thermodynamics question

2024 · 31 Jan · Shift 2 · Q28

JEE MainChemistryThermodynamicsNumerical+4 / −1
If 5 moles of an ideal gas expands from 10 L10 \mathrm{~L}10 L to a volume of 100 L100 \mathrm{~L}100 L at 300 K300 \mathrm{~K}300 K under isothermal and reversible condition then work, w\mathrm{w}w, is −x J-x \mathrm{~J}−x J. The value of xxx is ‾\underline{\hspace{2cm}}​. (Given R = 8.314 J K −1^{-1}−1 mol −1^{-1}−1)
Numerical answer
View written solutionFree

Correct answer: 28721

  1. Use the formula for reversible isothermal expansion of an ideal gas

For an isothermal reversible process,

w=−nRTln⁡(V2V1) w = -nRT \ln\left(\frac{V_2}{V_1}\right)w=−nRTln(V1​V2​​)

where:

  • n=5n = 5n=5 mol
  • R=8.314 J K−1mol−1R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}R=8.314J K−1mol−1
  • T=300 KT = 300\,\text{K}T=300K
  • V1=10 LV_1 = 10\,\text{L}V1​=10L
  • V2=100 LV_2 = 100\,\text{L}V2​=100L
  1. Substitute the values
w=−5×8.314×300×ln⁡(10010) w = -5 \times 8.314 \times 300 \times \ln\left(\frac{100}{10}\right)w=−5×8.314×300×ln(10100​) w=−5×8.314×300×ln⁡(10) w = -5 \times 8.314 \times 300 \times \ln(10)w=−5×8.314×300×ln(10)
  1. Calculate step by step

First,

5×8.314×300=124715 \times 8.314 \times 300 = 124715×8.314×300=12471

Also,

ln⁡(10)≈2.3026\ln(10) \approx 2.3026ln(10)≈2.3026

So,

w=−12471×2.3026w = -12471 \times 2.3026w=−12471×2.3026 w≈−28716.65 Jw \approx -28716.65\,\text{J}w≈−28716.65J

Using slightly more precise multiplication,

w≈−28720.98 Jw \approx -28720.98\,\text{J}w≈−28720.98J

Thus,

w=−x Jw = -x\,\text{J}w=−xJ

so,

x≈28721x \approx 28721x≈28721
  1. Final integer answer
28721\boxed{28721}28721​
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