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NEET 2026 (Re-neet) · Thermodynamic Processes

Thermodynamic Processes question from NEET 2026 (Re-neet)

Q9NEET 2026 · RE-NEETEasy

Thermodynamic Processes

In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas ( γ=5/3\gamma = 5/3) decreases from 60 K to 50 K. The work done by the gas in the process is:

(Take the universal gas constant as R=8.3 J mol−1 K−1R = 8.3\text{ J mol}^{-1}\text{ K}^{-1})

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Answer: (d)

  1. Adiabatic work by gas: W=nR(T1−T2)γ−1W = \frac{nR(T_1 - T_2)}{\gamma - 1}.
  2. W=1×8.3×(60−50)5/3−1=832/3W = \frac{1 \times 8.3 \times (60 - 50)}{5/3 - 1} = \frac{83}{2/3}
  3. W=124.5W = 124.5 J
  4. So the answer is (d) 124.5 J.
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