Physics · Thermodynamics
JEE Main 2026 — 23 January, Evening Shift — Question 36
One mole of an ideal diatomic gas expands from volume V to 2 V isothermally at a temperature and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from doing the same amount of work W, then its final temperature will be (close to) .
- Option A:
- Option B:Correct
- Option C:
- Option D:
Answer: B
Step-by-step solution
For Isothermal process
\mathrm{W}_{\text {isothermal }} & =\mathrm{nRT} \ell \mathrm{n}\left(\frac{\mathrm{~V}_{2}}{\mathrm{~V}_{1}}\right) = & 1 . \mathrm{R} 300 . \ell \mathrm{n}(2) = & 300 \mathrm{R}(0.693) \ldots . . \end{aligned}$$ Now for adiabatic process, It is given work done in Isothermal = work done in adiabatic $$\begin{gathered} \mathrm{W}_{\text {adiabatic }}=\frac{\mathrm{nR}\left(\mathrm{~T}_{1}-\mathrm{T}_{2}\right)}{\gamma-1} \end{gathered}$$ $(1)=(2)$ $\frac{\mathrm{nR}\left(300-\mathrm{T}_{\text {final }}\right)}{1.4-1}=300 \mathrm{R}(0.693)$ $\mathrm{T}_{\text {Final }}=216.84 \mathrm{~K}$ $=-56.3^{\circ} \mathrm{C}$Answer key and solution verified before publishing.
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- Exam
- JEE Main 2026
- Subject
- Physics
- Chapter
- Thermodynamics
- Topic
- Different Thermodynamic Processes