Thermodynamics

JEE Main level, single correct, medium

103 of 190

Question 103 of 190

Heat energy of 735J735J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be:

Show answer and solution

Answer: (A) 525J525J

For isobaric process and diatomic gas we can write, ΔQ=nCPΔT=735J\Delta Q=n{C}_{P}\Delta T=735J
⇒n7R2ΔT=735J⇒nRΔT=210\Rightarrow n\dfrac{7R}{2}\Delta T=735J \Rightarrow nR\Delta T=210
Now internal energy will be,
ΔU=nCVΔT=n5R2ΔT=52×210\Delta U=n{C}_{V}\Delta T =n\dfrac{5R}{2}\Delta T =\dfrac{5}{2}\times 210
=525J=525J

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.