JEE Main 2016PhysicsThermodynamicsMediumMCQ

JEE Main 2016 — Thermodynamics Question with Solution

JEE Main 2016 (03 Apr)

Question

n moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:

Choose an option

Show full solutionCorrect option: C
Correct answer
C9 P0V04nR

Step-by-step explanation

Temperature T will be maximum where product of PV is max equation of line

P=−P0V0V+3P0

Multiplying both sides by V,

PV=−P0V0V2+3P0V

For maxima,            d(PV) dV =0

⇒               −2 P 0 V 0 V+3 P 0 =0     ⇒  V= 3 V 0 2

and P=3P02

⇒   T= PV nR = 9 P 0 V 0 4 nR

Alternate method :

Since initial and final temperature are equal.

Hence maximum temperature is at middle of line.

PV=nR T max

⇒Tmax=3P023V02nR=9P0V04nR

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About this question

This is a previous-year question from JEE Main 2016, 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.