NEET UG 2019 · Chemistry · Question 51
NEET 2019 Chemistry, question 51
Question 51 of the NEET 2019 paper (code P1) is a Chemistry question from Thermodynamics. Its idea, Work Against Constant External Pressure, was asked only this once in NEET 2017–2026. The NTA final answer key gives option 4.
Updated 2026-10-08
Q51 · Chemistry
Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is :
[Given that 1 L bar = 100 J]
- 1− 30 J
- 25 kJ
- 325 J
- 430 JOfficial answer
Official answer: option 4 · NTA final answer key
Same idea
The only question on Work Against Constant External Pressure
No other NEET paper from 2017–2026 tested this exact idea.
Work Against Constant External Pressure: 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.