Thermodynamics

JEE Main level, numerical answer, medium

134 of 190

Question 134 of 190

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 __________

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Answer: 150

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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