JEE Main 2022 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 28 June 2022, Shift 1, Physics Q58

Question 58 of 90 in this shift, Physics question 28 of 30, Section B.

The total internal energy of two mole monoatomic ideal gas at temperature T = 300 K will be ___________ J. (Given R = 8.31 J/mol.K)

Official answer

7479

NTA final key.

Chapter: Kinetic Theory
Topic: Kinetic Theory of Gases
Idea tested: Degrees of Freedom

Same idea in other shifts

Asked 4× in all

4 Apr 2024, Shift 2 · Q40The translational degrees of freedom ( $f_t$ ) and rotational degrees of freedom ( $f_r$ ) of $CH_4$ molecule are:EasySingle correct
10 Apr 2023, Shift 1 · Q32Match List I with List II : List I List II (A) 3 Translational degrees of freedom (I) Monoatomic gases (B) 3 Translational, 2 rotational…EasyMatch the list
27 Jul 2022, Shift 2 · Q39Which statements are correct about degrees of freedom ? (A) A molecule with n degrees of freedom has n $^2$ different ways of storing…MediumOther

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.