WBJEE · Chemistry · States of Matter
The average speed of \(\mathrm{H}_2\) at \(T_1 \mathrm{~K}\) is equal to that of \(\mathrm{O}_2\) at \(T_2 \mathrm{~K}\). The ratio \(T_1: T_2\) is
- A \(1: 6\)
- B \(16: 1\)
- C \(1: 4\)
- D None of these
Answer & Solution
Correct Answer
(D) None of these
Step-by-step Solution
Detailed explanation
\(\left(\mathrm{C}_{\mathrm{av}}\right)_{\mathrm{H}_2}=\left(\mathrm{C}_{\mathrm{av}}\right)_{\mathrm{O}_2}\) \(\sqrt{\frac{8 \mathrm{RT}_1}{\pi \mathrm{M}_{\mathrm{H}_2}}}=\sqrt{\frac{8 \mathrm{RT}_2}{\pi \mathrm{M}_{\mathrm{O}_2}}}\)…
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