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JEE Main 2018 · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2018

Q221JEE Main 2018Medium

Thermodynamic Processes

Two moles of an ideal monatomic gas occupies a volume V at 27oC. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

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Answer: (d)

  1. Adiabatic: TVγ−1=TV^{\gamma-1} = const, γ=53\gamma = \frac{5}{3}.
  2. T2=300(12)2/3=3001.587≈189T_2 = 300\left(\frac{1}{2}\right)^{2/3} = \frac{300}{1.587} \approx 189 K
  3. ΔU=n⋅32R ΔT=2×1.5×8.314×(189−300)≈−2.7\Delta U = n\cdot\frac{3}{2}R\,\Delta T = 2 \times 1.5 \times 8.314 \times (189 - 300) \approx -2.7 kJ
  4. So the answer is (d) 189 K, −2.7-2.7 kJ.
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