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JEE Main 2017 · Physics · Thermodynamics and Kinetic Theory of Gases · medium

One mole of a monoatomic ideal gas (γ = 5/3) at 300 K is compressed adiabatically and reversibly to 1/8 of its initial volume. Taking R = 8.314 J…

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One mole of a monoatomic ideal gas (γ=5/3\gamma = 5/3) at 300 K300\ \mathrm{K} is compressed adiabatically and reversibly to 18\tfrac18 of its initial volume. Taking R=8.314 J mol−1 K−1R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}, the work done on the gas is nearly
  1. A. 1.12×104 J1.12\times10^{4}\ \mathrm{J}
  2. B. 1.87×104 J1.87\times10^{4}\ \mathrm{J}
  3. C. 1.50×104 J1.50\times10^{4}\ \mathrm{J}
  4. D. 4.85×103 J4.85\times10^{3}\ \mathrm{J}
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Answer: (A) 1.12×104 J1.12\times10^{4}\ \mathrm{J}

For an adiabatic process TVγ−1TV^{\gamma-1} is constant, so T2=300×82/3=300×4=1200 KT_2 = 300\times 8^{2/3} = 300\times4 = 1200\ \mathrm{K}. No heat flows, so the work done on the gas equals the rise in internal energy: W=nCVΔT=1×32(8.314)(900)≈1.12×104 JW = nC_V\Delta T = 1\times\tfrac32(8.314)(900) \approx 1.12\times10^{4}\ \mathrm{J}. Using CP=52RC_P = \tfrac52R would give 1.87×104 J1.87\times10^4\ \mathrm{J}, and using 82/58^{2/5} in place of 82/38^{2/3} gives the smallest option.
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