Kinetic Theory

JEE Main level, single correct, medium

93 of 168

Question 93 of 168

Molecules of an ideal gas are known to have three translational degrees of freedom. The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of γ=(CpCv)\gamma =\left(\dfrac{{C}_{p}}{{C}_{v}}\right) are given, respectively, by

Show answer and solution

Answer: (C) U=52RTU=\dfrac{5}{2}RT and γ=75\gamma =\dfrac{7}{5}

f=5f=5

γ=1+2f=1+25=75\gamma =1+\dfrac{2}{f}=1+\dfrac{2}{5}=\dfrac{7}{5}
CpCv=75\dfrac{{C}_{p}}{{C}_{v}}=\dfrac{7}{5}

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 Kinetic Theory (Degree of Freedom and Specific Heat), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.