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JEE Main 2018 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

One mole of a monatomic ideal gas is heated from 300 K to 600 K through a process in which the pressure is directly proportional to the volume.…

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One mole of a monatomic ideal gas is heated from 300 K300\ \mathrm{K} to 600 K600\ \mathrm{K} through a process in which the pressure is directly proportional to the volume. Taking R=8.314 J mol−1K−1R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}, the heat absorbed by the gas is nearly
  1. A. 3.74 kJ3.74\ \mathrm{kJ}
  2. B. 4.99 kJ4.99\ \mathrm{kJ}
  3. C. 1.25 kJ1.25\ \mathrm{kJ}
  4. D. 6.23 kJ6.23\ \mathrm{kJ}
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Answer: (B) 4.99 kJ4.99\ \mathrm{kJ}

For P=aVP=aV, the work is W=∫P dV=a2(V22−V12)=P2V2−P1V12=nRΔT2=150RW=\int P\,dV=\dfrac{a}{2}(V_2^2-V_1^2)=\dfrac{P_2V_2-P_1V_1}{2}=\dfrac{nR\Delta T}{2}=150R. The internal energy change is ΔU=32nRΔT=450R\Delta U=\tfrac32nR\Delta T=450R. By the first law Q=ΔU+W=600R=600×8.314=4988 J≈4.99 kJQ=\Delta U+W=600R=600\times8.314=4988\ \mathrm{J}\approx4.99\ \mathrm{kJ}. (Equivalently the molar heat capacity is C=Cv+R/2=2RC=C_v+R/2=2R.) The values 3.743.74 and 6.23 kJ6.23\ \mathrm{kJ} are the isochoric and isobaric heats.
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