JEE Main 2020PhysicsThermodynamicsMediumNumerical

JEE Main 2020 — Thermodynamics Question with Solution

JEE Main 2020 (09 Jan Shift 2)

Question

Starting at temperature 300K, one mole of an ideal diatomic gas γ=1.4 is first compressed adiabatically from volume V1 to V2=V116. It is then allowed to expand isobarically to volume 2V2 . If all the processes are the quasi-static then the final temperature of the gas (in oK ) is (to the nearest integer) ___________.

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Show full solutionCorrect answer: 1819
Correct answer
1819

Step-by-step explanation

PVy=constant
TV-1=constant
300V11.4-1=TBV1162/5
TB=300×28/5
Now for BC process
VBTB=VcTc
Tc=VcTBvB=2×300×28/5
Tc=1818.859
Tc=1819K

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About this question

This is a previous-year question from JEE Main 2020, covering the Thermodynamics chapter of Physics. 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.