JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 28)
A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is kept at 300 K in a container. The ratio of their rms speeds $$\left[ {{{{V_{rms}}\,(helium)} \over {{V_{rms}}\,(\arg on)}}} \right],$$ is close to :
3.16
0.32
0.45
2.24
Explanation
We know,
Vrms = $$\sqrt {{{3RT} \over M}} $$
Where M = molar mass of the gas.
Here temperature is 300 K for both the gas. So temperature is constant. R is also a constant.
$$ \therefore $$ Vrms $$ \propto \,\,\sqrt {{1 \over M}} $$
$$ \therefore $$ $${{{V_{rms}}(helium|)} \over {{V_{rms}}\left( {\arg an)} \right)}} = \sqrt {{{{M_{Ar}}} \over {{M_{He}}}}} $$
= $$\sqrt {{{40} \over 4}} $$
= $$\sqrt {10} $$
= 3.16
Vrms = $$\sqrt {{{3RT} \over M}} $$
Where M = molar mass of the gas.
Here temperature is 300 K for both the gas. So temperature is constant. R is also a constant.
$$ \therefore $$ Vrms $$ \propto \,\,\sqrt {{1 \over M}} $$
$$ \therefore $$ $${{{V_{rms}}(helium|)} \over {{V_{rms}}\left( {\arg an)} \right)}} = \sqrt {{{{M_{Ar}}} \over {{M_{He}}}}} $$
= $$\sqrt {{{40} \over 4}} $$
= $$\sqrt {10} $$
= 3.16
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