NEET UG 2024 · Chemistry · Question 91
NEET 2024 Chemistry, question 91
Question 91 of the NEET 2024 paper (code T2) is a Chemistry question from Thermodynamics. Its idea, Reversible Isothermal Work, was asked only this once in NEET 2017–2026. The NTA final answer key gives option 4.
Updated 2026-10-08
Q91 · Chemistry
The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is:
(Given R = 2.0 cal K mol)
- 1413.14 calories
- 2100 calories
- 30 calorie
- 4 caloriesOfficial answer
Official answer: option 4 · NTA final answer key
Same idea
The only question on Reversible Isothermal Work
No other NEET paper from 2017–2026 tested this exact idea.
Reversible Isothermal Work: every question and how it was framed
The chapter
Thermodynamics in ten years of NEET
32 questions from 2017–2026, asked in 10 of 10 papers, about 3.1 a paper. First Law of Thermodynamics alone had 8.