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MHT CET 2026 (13 May, Shift 1) · Pressure & Kinetic Energy of a Gas

Pressure & Kinetic Energy of a Gas question from MHT CET 2026 (13 May, Shift 1)

Q1MHT CET 2026 · 13 MAY · SHIFT 1Easy

Pressure & Kinetic Energy of a Gas

At what temperature, the kinetic energy of a gas molecule is half of the value at 27∘27^\circC?

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Answer: (b)

  1. Mean KE =32kBT= \frac{3}{2}k_BT, so E∝TE \propto T (in kelvin).
  2. T1=27+273=300T_1 = 27 + 273 = 300 K, so half the KE needs T2=150T_2 = 150 K.
  3. 150−273=−123∘150 - 273 = -123^\circC
  4. So the answer is (b) −123∘-123^\circC.
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