Thermodynamics

JEE Main level, single correct, medium

151 of 190

Question 151 of 190

One mole of an ideal monatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27oC{27}^{o}C . The magnitude of work done on the gas will be:

Show answer and solution

Answer: (B) 300Rln⁡2300R\ln 2

Work done on gas=nRTln⁡(PfPi)=R(300)ln⁡(2)=300Rln⁡2=nRT\ln \left(\dfrac{{P}_{f}}{{P}_{i}}\right)=R(300)\ln (2)=300R\ln 2.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermodynamics (Thermodynamic Processes), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.