JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 24)
The root mean square speed of molecules of a given mass of a gas at 27$$^\circ$$C and 1 atmosphere pressure is 200 ms$$-$$1. The root mean square speed of molecules of the gas at 127$$^\circ$$C and 2 atmosphere pressure is $${{x \over {\sqrt 3 }}}$$ ms$$-$$1. The value of x will be _________.
Answer
400
Explanation
Given, T1 = 27$$^\circ$$C = 27 + 273 = 300K, p1 = 1 atm, v1 = 200 ms$$-$$1, T2 = 127$$^\circ$$C = 400 K, p2 = 12 atm, v2 = ?
As we know that,
Root mean square speed, $${v_{rms}} = \sqrt {{{3RT} \over m}} $$
$$\therefore$$ $${{{v_1}} \over {{v_2}}} = \sqrt {{{{T_1}} \over {{T_2}}}} = \sqrt {{{300} \over {400}}} = \sqrt {{3 \over 4}} $$
$$ \Rightarrow {v_2} = \sqrt {{4 \over 3}} {v_1} = {2 \over {\sqrt 3 }} \times 200 = {{400} \over {\sqrt 3 }}$$ ms$$-$$1
$$ \Rightarrow {x \over {\sqrt 3 }} = {{400} \over {\sqrt 3 }} \Rightarrow x = 400$$
As we know that,
Root mean square speed, $${v_{rms}} = \sqrt {{{3RT} \over m}} $$
$$\therefore$$ $${{{v_1}} \over {{v_2}}} = \sqrt {{{{T_1}} \over {{T_2}}}} = \sqrt {{{300} \over {400}}} = \sqrt {{3 \over 4}} $$
$$ \Rightarrow {v_2} = \sqrt {{4 \over 3}} {v_1} = {2 \over {\sqrt 3 }} \times 200 = {{400} \over {\sqrt 3 }}$$ ms$$-$$1
$$ \Rightarrow {x \over {\sqrt 3 }} = {{400} \over {\sqrt 3 }} \Rightarrow x = 400$$
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