PrepRung
Sign inSign up free

JEE Main 2022 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

Two moles of a diatomic ideal gas (rigid rotor, Cv=5/2R) are heated from 300 K to 400 K in a process in which the pressure is directly proportional…

Single correct option

Two moles of a diatomic ideal gas (rigid rotor, Cv=52RC_v=\tfrac52R) are heated from 300 K300\ \mathrm{K} to 400 K400\ \mathrm{K} in a process in which the pressure is directly proportional to the volume. The heat absorbed by the gas is
  1. A. 400R400R
  2. B. 500R500R
  3. C. 600R600R
  4. D. 700R700R
Show answer and solution

Answer: (C) 600R600R

For P=aVP=aV, PV=aV2=nRTPV=aV^{2}=nRT, so work in a small step is PdV=nR dT2PdV=\dfrac{nR\,dT}{2} (since 2aV dV=nR dT2aV\,dV=nR\,dT). Hence W=nRΔT2=2R(100)2=100RW=\frac{nR\Delta T}{2}=\frac{2R(100)}{2}=100R. Change in internal energy =nCvΔT=2⋅52R⋅100=500R=nC_v\Delta T=2\cdot\frac52R\cdot100=500R. By the first law Q=500R+100R=600RQ=500R+100R=600R. Equivalent molar heat capacity is Cv+R2=3RC_v+\frac R2=3R. Using CvC_v alone gives 500R500R and using CpC_p gives 700R700R.
Report a mistake

Original question written in the JEE Main 2022 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake