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JEE Main 2022 (26 Jul, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2022 (26 Jul, Shift 1)

Q163JEE Main 2022 · 26 JUL · SHIFT 1Medium

Thermodynamic Processes

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40 K40\text{ K} at constant pressure. The increase in the internal energy of the gas in this process is:

(Given R=8.3 J K−1 mol−1R = 8.3\text{ J K}^{-1}\text{ mol}^{-1})

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Answer: (b)

  1. ΔU=nCvΔT\Delta U = nC_v\Delta T for any process; monoatomic Cv=32RC_v = \frac{3}{2}R.
  2. ΔU=7×32×8.3×40=3486 J\Delta U = 7 \times \frac{3}{2} \times 8.3 \times 40 = 3486\text{ J}
  3. So the answer is (b) 3486 J.
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