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JEE Main 2016 · Physics · Thermodynamics and Kinetic Theory of Gases · easy

Two moles of a diatomic ideal gas (rigid molecules) are heated through 50 K in a process in which the pressure is directly proportional to the…

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Two moles of a diatomic ideal gas (rigid molecules) are heated through 50 K50\ \mathrm{K} in 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. 2.08 kJ2.08\ \mathrm{kJ}
  2. B. 2.49 kJ2.49\ \mathrm{kJ}
  3. C. 2.91 kJ2.91\ \mathrm{kJ}
  4. D. 3.33 kJ3.33\ \mathrm{kJ}
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Answer: (B) 2.49 kJ2.49\ \mathrm{kJ}

For P=kVP=kV the work is W=∫P dV=k2(V22−V12)=12(P2V2−P1V1)=12nR ΔTW=\int P\,dV=\dfrac{k}{2}(V_2^2-V_1^2)=\dfrac12(P_2V_2-P_1V_1)=\dfrac12nR\,\Delta T. The change in internal energy is ΔU=nCVΔT=n⋅52R ΔT\Delta U=nC_V\Delta T=n\cdot\dfrac52R\,\Delta T for a diatomic gas. Hence Q=ΔU+W=n(52R+12R)ΔT=3nR ΔTQ=\Delta U+W=n\left(\dfrac52R+\dfrac12R\right)\Delta T=3nR\,\Delta T, i.e. the molar heat capacity of this process is 3R3R. So Q=3×2×8.314×50=2494 J≈2.49 kJQ=3\times2\times8.314\times50=2494\ \mathrm{J}\approx2.49\ \mathrm{kJ}. Using CVC_V alone gives 2.08 kJ2.08\ \mathrm{kJ}, and CPC_P would give 2.91 kJ2.91\ \mathrm{kJ}; neither describes this process.
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