JEE Main 2020PhysicsThermodynamicsMediumNumerical

JEE Main 2020Thermodynamics Question with Solution

JEE Main 2020 (04 Sep Shift 2)

Question

The change in the magnitude of the volume of an ideal gas when a small additional pressure P is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity T at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm respectively. If  |ΔT|=C|ΔP| then value of C in (K/atm)  is __________

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Show full solutionCorrect answer: 150
Correct answer
150

Step-by-step explanation

PV=nRT

PΔV+VΔP=0

ΔV=-ΔPPV  ....(i)

  In second case

PΔV=-nRΔT

ΔV=-nRΔTP    ...(ii)

equating (i) and (ii)

nRΔTP=-ΔPPV

ΔT=ΔP VnR

c=vnR

Putting the value of V, n and  R , C = 150 

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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.