JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2026
- Medium
- MCQ
The r.m.s speed of oxygen molecules at \(47^{\circ} C\) is equal to that of the hydrogen molecules kept at _________ \({ }^{\circ} C\). (Mass of oxygen molecule/mass of hydrogen molecule \(=32 / 2\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(-253\)
Step-by-step Solution
Detailed explanation
\( V_{rma}=\sqrt{\frac{3RT}{M}} \)
\( V_{rmsO_{2}}=V_{rmsH_{2}} \)
\( T_{O_{2}}=273+47=320K \)
\( \sqrt{\frac{3RT_{O_{2}}}{M_{O_{2}}}}=\sqrt{\frac{3R~T_{H_{2}}}{M_{H_{2}}}} \)
\( \frac{T_{2}}{M_{O_{2}}}=\frac{T_{H_{2}}}{M_{H_{2}}} \)
\( \frac{320}{32}=\frac{T_{H_{2}}}{2} \)
\( T_{H_{2}}=20K \)
\( T_{H_{2}}=-253^{\circ}C \)
\( V_{rmsO_{2}}=V_{rmsH_{2}} \)
\( T_{O_{2}}=273+47=320K \)
\( \sqrt{\frac{3RT_{O_{2}}}{M_{O_{2}}}}=\sqrt{\frac{3R~T_{H_{2}}}{M_{H_{2}}}} \)
\( \frac{T_{2}}{M_{O_{2}}}=\frac{T_{H_{2}}}{M_{H_{2}}} \)
\( \frac{320}{32}=\frac{T_{H_{2}}}{2} \)
\( T_{H_{2}}=20K \)
\( T_{H_{2}}=-253^{\circ}C \)
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