The kinetic energy is given by KE=2fkT.
Let the kinetic energy be K1 at 27∘C and the kinetic energy be K′ at T′ temperature. As given in the question, K′=2K1. Hence,
K1K′=TT′⇒2=300T′⇒T′=600K
In Celsius the temperature is T′−273=327∘C.
JEE Main 2023 — Physics Thermodynamics
The temperature at which the kinetic energy of oxygen molecules becomes double than its value at 27∘C is
Held on 8 Apr 2023 · Verified 6 Jul 2026.
927∘C
327∘C
1227∘C
627∘C
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