A mixture of 2 moles of helium gas (atomic mass =4 amu) and 1 mole of argon gas (atomic mass =40 amu) is kept at 300 K in a container. The ratio of the rms speeds (( v rms ( text helium )/ v rms ( text argon ))) is

Question

A mixture of 2 moles of helium gas (atomic mass =4 amu) and 1 mole of argon gas (atomic mass =40 amu) is kept at 300 K in a container. The ratio of the rms speeds (( v rms ( text helium )/ v rms ( text argon ))) is (A) 0.32 (B) 0.45 (C) 2.24 (D) 3.16

Tardigrade Admin · Accepted Answer

v rms =√(3 RT / M ) Required ratio =√( M Ar / M He )=√(40/4)=√10 =3.16

Q. A mixture of 2 moles of helium gas (atomic mass $= 4$ amu) and 1 mole of argon gas (atomic mass $= 40$ amu) is kept at $300 K$ in a container. The ratio of the rms speeds $(\frac{v _{r m s} ( helium )}{v _{r m s} ( argon )})$ is

0.32

0.45

2.24

3.16

$v_{r m s} = \frac{3 RT}{M}$
Required ratio $= \frac{M _{A r}}{M _{He}} = \frac{40}{4} = 10$
$= 3.16$

Q. A mixture of 2 moles of helium gas (atomic mass =4 amu) and 1 mole of argon gas (atomic mass =40 amu) is kept at 300K in a container. The ratio of the rms speeds (vrms​( argon )vrms​( helium )​) is