JEE Main 2022PhysicsThermodynamicsMediumMCQ

JEE Main 2022 — Thermodynamics Question with Solution

JEE Main 2022 (28 Jun Shift 1)

Question

Statement - I : When μ amount of an ideal gas undergoes adiabatic change from state P1,V1,T1 to state P2,V2,T2, then work done is W=μRT2-T11-γ, where γ=CPCV and R= universal gas constant.

Statement - II : In the above case, when work is done on the gas, the temperature of the gas would rise.

Choose an option

Show full solutionCorrect option: A
Correct answer
ABoth statement-I and statement-II are true.

Step-by-step explanation

The work done in an adiabatic process is given by,

Wadiabatic =nRTf-Ti1-γ→ statement I is true.

For adiabatic process ∆Q=0. From the first law of thermodynamics,

∆Q=W+ΔU

0=W+ΔU

ΔU=-W

If work is done on the gas, i.e. work is negative

∴ΔU is positive.

∴ The temperature will increase. Hence, statement II is also true.

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About this question

This is a previous-year question from JEE Main 2022, 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.