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JEE Main 2021 (1 Sep, Shift 2) · Heat, Work & First Law of Thermodynamics

Heat, Work & First Law of Thermodynamics question from JEE Main 2021 (1 Sep, Shift 2)

Q16JEE Main 2021 · 1 SEP · SHIFT 2Medium

Heat, Work & First Law of Thermodynamics

The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0∘^\circC without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is x10\frac{x}{10}. Then the value of x (round off to the nearest integer) is ___________. (Given R = 8.31 J mol−1{}^{-1} K−1{}^{-1})

Numerical answer type. Work it out, then open the answer.

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Answer: 25

  1. At constant pressure: ΔU=nCvΔT\Delta U = nC_v\Delta T, W=nRΔTW = nR\Delta T
  2. Diatomic with rotation, no vibration: Cv=52RC_v = \frac{5}{2}R
  3. ΔUW=52=2510\frac{\Delta U}{W} = \frac{5}{2} = \frac{25}{10}
  4. So x=25x = 25.
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