Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2016-Paper-2 A gas is enclosed in a cylinder with a movab…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2016-Paper-2 Single Correct MCQ
Published on: August 14, 2026

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure P t = 10 5 Pa and volume V t = 10 -3 m 3 changes to a final state at P f = (1/32) x 10 5 Pa and V f = 8 x 10 -3 m 3 in an adiabatic quasi-static process, such that P 3 V 5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at P i followed by an isochoric (isovolumetric) process at volume V f . The amount of heat supplied to the system in the two-step process is approximately

A
112J
B
294J
C
588J
D
813J

Share this question

For Instagram sharing, use “Apps” on mobile or copy the link.

Text Solution

Verified by Experts
The correct answer is:
C

monoatomic gas

From first law of thermodynamics

H = W + U

W = P i V

= 700J

U = nC v T

So, H = W + U =588 J

──────────────────────────────────────────────────────────────────────────────────────────

Prepare Smarter with CGP Edu

Get practice questions, solutions, and test series in one place.

Write a Review

Share your experience with this question and solution.

Commentary

Send your comment, doubt, correction, or feedback to admin.

Student Reviews

What students say about this solution

No reviews yet. Be the first to write a review.

Similar Questions

Explore conceptually related problems

CG
CGP Question Assistant Question Bank + AI Help
Hi! Type your question or upload one screenshot. First I will search related questions from CGP Edu Question Bank. If none match, type YES and I will solve it with AI.
Upload only one screenshot at a time. Flow: Question Bank first → If not matched, type YES for AI solution.