PrepRung
Sign inSign up free

JEE Main 2021 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

One mole of a monatomic ideal gas is taken from temperature T to 4T along a process in which pressure is directly proportional to volume (Ppropto V).…

Single correct option

One mole of a monatomic ideal gas is taken from temperature TT to 4T4T along a process in which pressure is directly proportional to volume (P∝VP\propto V). The heat absorbed by the gas is
  1. A. 1.5RT1.5RT
  2. B. 4.5RT4.5RT
  3. C. 6RT6RT
  4. D. 7.5RT7.5RT
Show answer and solution

Answer: (C) 6RT6RT

With P=kVP=kV and PV=RTPV=RT: kV2=RTkV^2=RT, so V∝TV\propto\sqrt T and volume doubles. ΔU=32R(3T)=4.5RT\Delta U=\dfrac32R(3T)=4.5RT. Work =∫P dV=k2(V22−V12)=12(4RT−RT)=1.5RT=\int P\,dV=\dfrac k2(V_2^2-V_1^2)=\dfrac12(4RT-RT)=1.5RT. Therefore Q=ΔU+W=6RTQ=\Delta U+W=6RT. (Equivalent: molar heat capacity C=Cv+R2=2RC=C_v+\dfrac{R}{2}=2R, Q=2R⋅3TQ=2R\cdot3T.)
Report a mistake

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake