Kinetic Theory

JEE Main level, single correct, medium

134 of 168

Question 134 of 168

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at 27∘C27^{\circ}C is

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Answer: (B) 327∘C327^{\circ}C

The kinetic energy is given by KE=f2kTKE=\dfrac{f}{2}kT.
Let the kinetic energy be K1{K}_{1} at 27∘C27^{\circ}C and the kinetic energy be K′K' at T′T' temperature. As given in the question, K′=2K1K'=2{K}_{1}. Hence,
K′K1=T′T⇒2=T′300⇒T′=600K\dfrac{K'}{{K}_{1}}=\dfrac{T'}{T} \Rightarrow 2=\dfrac{T'}{300} \Rightarrow T'=600K
In Celsius the temperature is T′−273=327∘CT'-273=327^{\circ}C.

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