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JEE Main 2014 (11 Apr, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2014 (11 Apr, Shift 1)

Q230JEE Main 2014 · 11 APR · SHIFT 1Medium

Thermodynamic Processes

During an adiabatic compression, 830 J of work is done on 2 moles of a diatomic ideal gas to reduce its volume by 50%. The change in its temperature is nearly: (R = 8.3 JK−1{}^{-1} mol−1{}^{-1})

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Answer: (c)

  1. Adiabatic: Q=0Q = 0, so work done on gas =ΔU=nCvΔT= \Delta U = nC_v\Delta T.
  2. Diatomic: Cv=52RC_v = \frac{5}{2}R. 830=2×52×8.3×ΔT=41.5 ΔT830 = 2 \times \frac{5}{2} \times 8.3 \times \Delta T = 41.5\,\Delta T
  3. ΔT=20\Delta T = 20 K
  4. So the answer is (c) 20 K.
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