The temperature at which the velocity of oxygen will be half that of hydrogen at NTP is

1092°C

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1492°C

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273 K

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819°C

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Solution

The correct option is D
819°C

$G i v e n v_{o_{2}} = \frac{1}{2} v_{H_{2}} ∴ \sqrt{\frac{3 R T}{32}} = \frac{1}{2} \sqrt{\frac{3 R \times 273}{2}}; \frac{T}{32} = \frac{273}{8} ∴ T = 4 \times 273 = 1092 K = 1092 - 273 \Rightarrow t = 819^{\circ} C$

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