JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 10)
Consider a sample of oxygen behaving like an ideal gas. At 300 K, the ratio of root mean square (rms) velocity to the average velocity of gas molecule would be :
(Molecular weight of oxygen is 32g/mol; R = 8.3 J K$$-$$1 mol$$-$$1)
(Molecular weight of oxygen is 32g/mol; R = 8.3 J K$$-$$1 mol$$-$$1)
$$\sqrt {{{3\pi } \over 8}} $$
$$\sqrt {{3 \over 3}} $$
$$\sqrt {{8 \over 3}} $$
$$\sqrt {{{8\pi } \over 3}} $$
Explanation
$${V_{rms}} = \sqrt {{{3RT} \over M}} $$
$${V_{avg}} = \sqrt {{8 \over \pi }{{RT} \over M}} $$
$${{{V_{rms}}} \over {{V_{avg}}}} = \sqrt {{{3\pi } \over 8}} $$
$${V_{avg}} = \sqrt {{8 \over \pi }{{RT} \over M}} $$
$${{{V_{rms}}} \over {{V_{avg}}}} = \sqrt {{{3\pi } \over 8}} $$
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