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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2023
  • Easy
  • Numerical

The temperature at which the kinetic energy of oxygen molecules becomes double than its value at \(27^{\circ}\,C\) is \(............^{\circ}\,C\)

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Answer & Solution

Correct Answer

(C) \(327\)

Step-by-step Solution

Detailed explanation

Kinetic energy \(=\frac{f}{2} kT , T\) is absolute temperature.

If \(K _1\) is kinetic energy at \(27^{\circ} C\).

\(K _2\) is kinetic energy at new temperature \(T\).

\(\frac{ K _1}{ K _2}=\frac{ T _1}{ T _2} \Rightarrow \frac{1}{2}=\frac{300}{ T }\)

\(T =600\,K\)

\(T =327^{\circ}\,C\)

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