JEE Main 2023 · 8 April · evening · Single correct · +4 / −1

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at 27° C is

  1. (a) 627° C
  2. (b) 927° C
  3. (c) 327° C
  4. (d) 1227° C
Answer and solution

Answer (c) 327° C

Solution

Physics sir · 3 steps

Kinetic energy follows absolute temperature, so switch 27 °C to kelvin before you double anything.

  1. KE depends on T. By equipartition, the average kinetic energy of a gas molecule is proportional to its absolute temperature: K̅E̅ ∝ T. The gas is the same, so the degrees of freedom do not change and only T matters.

    KE₂/KE₁ = T₂/T₁

  2. Convert to kelvin. T₁ = 27 + 273 = 300 K. Doubling the kinetic energy means doubling the absolute temperature, so T₂ = 2 × 300 = 600 K.
  3. So. T₂ = 600 - 273 = 327°C, which is option (c).

The idea to keep. Any ratio built on KE ∝ T, or on pV = nRT, uses kelvin. Convert to kelvin first, scale, then convert back to Celsius only at the end.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Kinetic Theory of Gases

Asked about 130 times in JEE Main since 2020. This question is from 2023.