Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

64 of 276

Question 64 of 276

The temperature of 3.00mol3.00mol of an ideal diatomic gas is increased by 40.0∘C40.0^{\circ}C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is x10.\dfrac{x}{10}. Then the value of xx (round off to the nearest integer) is _________.
(( GivenR=8.31Jmol−1K−1)\left.R=8.31J{mol}^{-1}{K}^{-1}\right)

Show answer and solution

Answer: 25

First, recall the formula of change in internal energy in terms of molar specific heat at constant volume and change in temperature, ∆U=nCVΔT∆U=n{C}_{V}\Delta T, for diatomic gas, CV=f2R=52R{C}_{V}=\dfrac{f}{2}R=\dfrac{5}{2}R, ⇒∆U=n×52×RΔT\Rightarrow ∆U=n\times \dfrac{5}{2}\times R\Delta T
⇒ΔUW=52nRΔTnRΔT=52=2510\Rightarrow \dfrac{\Delta U}{W}=\dfrac{\dfrac{5}{2}nR\Delta T}{nR\Delta T}=\dfrac{5}{2}=\dfrac{25}{10}
So, x=25x=25

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 Magnetic Effects & Magnetism (Magnetic Field), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.