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JEE Advanced 2018 Paper 1 · Physics · Thermodynamics and Kinetic Theory of Gases · medium

One mole of a monatomic ideal gas, initially at pressure P0, volume V0 and temperature T0, is taken to volume 2V0 by a process in which its pressure…

One or more options correct

One mole of a monatomic ideal gas, initially at pressure P0P_0, volume V0V_0 and temperature T0T_0, is taken to volume 2V02V_0 by a process in which its pressure is directly proportional to its volume (P∝VP\propto V). Let RR be the gas constant. Which of the following is/are correct?
  1. A. The final temperature is 2T02T_0.
  2. B. The heat absorbed by the gas is 6RT06RT_0.
  3. C. The work done by the gas is 3RT03RT_0.
  4. D. The molar heat capacity of the gas in this process is 3R3R.
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Answer: B

Since P=P0V/V0P=P_0V/V_0, we have PV=P0V2/V0PV=P_0V^2/V_0, so T∝V2T\propto V^2 and Tf=4T0T_f=4T_0; option (A) is false. Work is the trapezoid area under the straight line: W=12(P0+2P0)V0=32P0V0=32RT0W=\tfrac12(P_0+2P_0)V_0=\tfrac32P_0V_0=\tfrac32RT_0, so (C) is false. The internal energy change is ΔU=32R(4T0−T0)=92RT0\Delta U=\tfrac32R(4T_0-T_0)=\tfrac92RT_0. The first law gives Q=ΔU+W=92RT0+32RT0=6RT0Q=\Delta U+W=\tfrac92RT_0+\tfrac32RT_0=6RT_0, so (B) is true. The molar heat capacity is C=Q/ΔT=6RT0/3T0=2RC=Q/\Delta T=6RT_0/3T_0=2R, so (D) is false. Only (B) is correct.
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