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 mm ( text helium )/ v rma ( 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 mm ( text helium )/ v rma ( 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 A e/ M He ) =√(40/4)=√10 =3.16 .

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

$0.32$

$0.45$

$2.24$

$3.16$

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

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

0.32

0.45

2.24

3.16