JEE Main 2022ChemistryThermodynamics (C)HardNumerical

JEE Main 2022Thermodynamics (C) Question with Solution

JEE Main 2022 (29 Jun Shift 2)

Question

2.2 g of nitrous oxide N2O gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, U is '-x'J. The value of 'x' is _____.

[nearest integer]

(Given: atomic mass of N=14 g mol-1 and of O=16 g mol-1. Molar heat capacity of N2O is 100 JK-1 mol-1)

Enter your answer

Show full solutionCorrect answer: 72
Correct answer
72

Step-by-step explanation

Molecular weight of nitrous oxide Number of moles of Specific heat capacity of Heat release on cooling from to Negative sign indicates heat released. Work done by surrounding on system to compares the gas from to at constant pressure. It means work done by surrounding is irreversible. Using 1st Law of thermodynamics .

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Thermodynamics (C) chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, covering the Thermodynamics (C) chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.