JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
The temperature at which the kinetic energy of oxygen molecules becomes double than its value at \(27^{\circ}\,C\) is \(............^{\circ}\,C\)
- A \(1227\)
- B \(927\)
- C \(327\)
- D \(627\)
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 \(T\). \(\frac{ K _1}{ K _2}=\frac{ T _1}{ T _2} \Rightarrow \frac{1}{2}=\frac{300}{ T }\) \(T =600\,K\)…
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