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JEE Main 20204 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2020 · 4 September · Evening Shift · Physics · Thermodynamics

The change in the magnitude of the volume of an ideal gas when a...

The change in the magnitude of the volume of an ideal gas when a small additional pressure ∆P∆ P is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity ∆T∆ T at constant pressure. The initial temperature and pressure of the gas were 300K300Kand 2atm2atm respectively. If ∣ΔT∣=C∣ΔP∣| \Delta T | = C | \Delta P | then value of CC in (K/atm)( K / atm ) is __________

Correct answer

0

Step-by-step solution

PV=nRTPV = nRT

PΔV+VΔP=0P \Delta V + V \Delta P = 0

ΔV=−ΔPPV....(i)\Delta V = - \dfrac{\Delta P}{P}V . . . . ( i )

In second case

PΔV=−nRΔTP \Delta V = - nR \Delta T

ΔV=−nRΔTP...(ii)\Delta V = - \dfrac{nR \Delta T}{P} . . . ( ii )

equating (i)( i ) and (ii)( ii )

nRΔTP=−ΔPPV\dfrac{nR \Delta T}{P} = - \dfrac{\Delta P}{P}V

ΔT=ΔPVnR\Delta T = \Delta P\dfrac{V}{nR}

c=vnRc = \dfrac{v}{nR}

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

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