Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
An ideal gas (1 mole, monatomic) is in the initial state P (see diagram) on an isothermal curve A at a temperature T0. It is brought under a constant volume 2V0 process to Q which lies on an adiabatic curve B intersecting the isothermal curve A at P0, V0, T0. The change in the internal energy of the gas (in terms of T0) during the process is  223=1.587

PhysicsThermodynamicsNEET
Options:
  • A 2.3T0
  • B -4.6T0
  • C -2.3T0
  • D 4.6T0
Solution:
2523 Upvotes Verified Answer
The correct answer is: -4.6T0
Temperature at state P=T0, since P lies on the isotherm of temperature T0. If T be the temperature at Q, then for the adiabatic process B, we have T0V0γ-1=T2V0γ-1

     T=T2γ-1=T022/3

Change in the internal energy of the gas is

      ΔU=CVT-T0=R-1T022/3-T0

             =3RT01-22/32×22/3=-4.6T0

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.