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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2026 · 23 January · Evening Shift · Physics · Thermodynamics

One mole of an ideal diatomic gas expands from volume V to 2 V...

One mole of an ideal diatomic gas expands from volume VV to 2 V isothermally at a temperature 27∘C27^{\circ} \mathrm{C} and does WW joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27∘C27^{\circ} \mathrm{C} doing the same amount of work WW, then its final temperature will be (close to) ____∘C\_\_\_\_ { }^{\circ} \mathrm{C}.
(log⁡e2=0.693)\left(\log _{\mathrm{e}} 2=0.693\right)

Options

  1. A-30
  2. B-189
  3. C-117
  4. D-56

Correct answer

D. -56

Step-by-step solution

For isothermal expansion of 1 mole diatomic gas from V to 2V at 300 K:
W=RTln⁡2=R(300)ln⁡2W = RT\ln 2 = R(300)\ln 2
For adiabatic expansion with same volume change and same work output:
W=nCVΔT=CV(Ti−Tf)W = nC_V\Delta T = C_V(T_i - T_f)
For diatomic gas, CV=52RC_V = \frac{5}{2}R. Setting work equal:
R(300)ln⁡2=52R(300−Tf)R(300)\ln 2 = \frac{5}{2}R(300 - T_f)
Simplifying:
300ln⁡2=2.5(300−Tf)300\ln 2 = 2.5(300 - T_f), so 300(0.693)=2.5(300−Tf)300(0.693) = 2.5(300 - T_f)
207.9=750−2.5Tf207.9 = 750 - 2.5T_f, giving Tf=216.84T_f = 216.84 K ≈\approx -56°C.
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