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JEE Main 2019 (9 Jan, Shift 2) · Molecular Speeds & Maxwell Distribution

Molecular Speeds & Maxwell Distribution question from JEE Main 2019 (9 Jan, Shift 2)

Q81JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Molecular Speeds & Maxwell Distribution

A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27oC. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about: [Take R = 8.3 J/K mole]

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Answer: (c)

  1. n=1528=0.536n = \dfrac{15}{28} = 0.536 mol, T1=300T_1 = 300 K.
  2. Doubling vrmsv_{rms} needs T2=4T1=1200T_2 = 4T_1 = 1200 K, so ΔT=900\Delta T = 900 K.
  3. Closed vessel: Q=nCVΔT=0.536×52(8.3)×900Q = nC_V\Delta T = 0.536 \times \dfrac{5}{2}(8.3) \times 900
  4. Q≈1.0×104Q \approx 1.0\times10^{4} J =10= 10 kJ
  5. So the answer is (c) 10 kJ.
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