JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 5)
Consider a mixture of gas molecule of types A, B and C having masses mA < mB < mC. The ratio of their root mean square speeds at normal temperature and pressure is :
$${v_A} = {v_B} \ne {v_C}$$
$${1 \over {{v_A}}} > {1 \over {{v_B}}} > {1 \over {{v_C}}}$$
$${1 \over {{v_A}}} < {1 \over {{v_B}}} < {1 \over {{v_C}}}$$
$${v_A} = {v_B} = {v_C} = 0$$
Explanation
rms velocity of gas molecules is given as
$${v_{rms}} = \sqrt {{{3RT} \over m}} $$ ..... (i)
where, m = molar mass of the gas in kilograms per mole,
R = molar gas constant,
and T = temperature in kelvin.
According to question,
mA < mB < mC
From Eq. (i),
$${v_{rms}} \propto {1 \over {\sqrt m }}$$
$$\therefore$$ We can write,
vA > vB > vC or $${1 \over {{v_A}}} < {1 \over {{v_B}}} < {1 \over {{v_C}}}$$
$${v_{rms}} = \sqrt {{{3RT} \over m}} $$ ..... (i)
where, m = molar mass of the gas in kilograms per mole,
R = molar gas constant,
and T = temperature in kelvin.
According to question,
mA < mB < mC
From Eq. (i),
$${v_{rms}} \propto {1 \over {\sqrt m }}$$
$$\therefore$$ We can write,
vA > vB > vC or $${1 \over {{v_A}}} < {1 \over {{v_B}}} < {1 \over {{v_C}}}$$
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