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Thermodynamics question

2025 · 4 Apr · Shift 1 · Q9
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Thermodynamics question

2025 · 4 Apr · Shift 1 · Q9

JEE MainChemistryThermodynamicsMCQ+4 / −1
One mole of an ideal gas expands isothermally and reversibly from 10dm310 \mathrm{dm}^310dm3 to 20dm320 \mathrm{dm}^320dm3 at 300 K . ΔU,q\Delta \mathrm{U}, \mathrm{q}ΔU,q and work done in the process respectively are Given: R=8.3 J K−1 mol−1ln⁡10=2.3log⁡2=0.30log⁡3=0.48\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\ln 10=2.3\log 2=0.30\log 3=0.48R=8.3 J K−1 mol−1ln10=2.3log2=0.30log3=0.48
  1. A
    0,21.84 kJ,−1.726 J0,21.84 \mathrm{~kJ},-1.726 \mathrm{~J}0,21.84 kJ,−1.726 J
  2. B
    0,21.84 kJ,21.84 kJ0,21.84 \mathrm{~kJ}, 21.84 \mathrm{~kJ}0,21.84 kJ,21.84 kJ
  3. C
    0,1.718 kJ,−1.718 kJ0,1.718 \mathrm{~kJ},-1.718 \mathrm{~kJ}0,1.718 kJ,−1.718 kJ
  4. D
    0,−17.18 kJ,1.718 J0,-17.18 \mathrm{~kJ}, 1.718 \mathrm{~J}0,−17.18 kJ,1.718 J
View written solutionFree

Correct answer: C

  1. Given
  • One mole of ideal gas: n=1n=1n=1
  • Isothermal reversible expansion
  • V1=10 dm3V_1=10\,\mathrm{dm}^3V1​=10dm3
  • V2=20 dm3V_2=20\,\mathrm{dm}^3V2​=20dm3
  • T=300 KT=300\,\mathrm{K}T=300K
  • R=8.3 J K−1 mol−1R=8.3\,\mathrm{J\,K^{-1}\,mol^{-1}}R=8.3JK−1mol−1
  1. Find ΔU\Delta UΔU

For an ideal gas, internal energy depends only on temperature. Since the process is isothermal, ΔT=0\Delta T=0ΔT=0.

Therefore,

ΔU=0\Delta U = 0ΔU=0
  1. Calculate reversible work

For reversible isothermal expansion of an ideal gas,

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

Substitute values:

w=−(1)(8.3)(300)ln⁡(2010)w = -(1)(8.3)(300)\ln\left(\frac{20}{10}\right)w=−(1)(8.3)(300)ln(1020​) w=−2490ln⁡2w = -2490\ln 2w=−2490ln2

Now,

log⁡2=0.30,ln⁡10=2.3\log 2 = 0.30,\quad \ln 10 = 2.3log2=0.30,ln10=2.3

so

ln⁡2=2.3log⁡2=2.3×0.30=0.69\ln 2 = 2.3\log 2 = 2.3\times 0.30 = 0.69ln2=2.3log2=2.3×0.30=0.69

Thus,

w=−2490×0.69=−1718.1 Jw = -2490\times 0.69 = -1718.1\,\mathrm{J}w=−2490×0.69=−1718.1J w≈−1.718 kJw \approx -1.718\,\mathrm{kJ}w≈−1.718kJ
  1. Calculate heat qqq

From the first law:

ΔU=q+w\Delta U = q+wΔU=q+w

Since ΔU=0\Delta U=0ΔU=0,

q=−wq=-wq=−w

Hence,

q=+1.718 kJq = +1.718\,\mathrm{kJ}q=+1.718kJ
  1. Final values

Therefore, respectively ΔU,q,\Delta U, q,ΔU,q, and www are:

0,  1.718 kJ,  −1.718 kJ0,\; 1.718\,\mathrm{kJ},\; -1.718\,\mathrm{kJ}0,1.718kJ,−1.718kJ
  1. Check options
  • A: Incorrect
  • B: Incorrect
  • C: Correct
  • D: Incorrect

So the correct option is C.

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