Question

A mixture of 2 moles of heìium 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

$\left(\frac{v_{rms}\left(helium\right)}{v_{rms}\left(argon\right)}\right)$ is

• A. $0.32$
• B. $0.45$
• C. $2.24$
• D. $3.16$

Answer 1

The correct option is D $3.16$

Given that,

Temperature $T=300K$

Molecular mass of Helium $M_{He}=4$

Molecular mass of Argon $M_{Ar}=40$

We know that,

$v_{rms}=\sqrt{\frac{3kT}{2NM}}$

Where,

K = Boltzmann constant

T= Absolute temperature

N = Number of moles

M = molecular mass

So, $v_{rms}\propto \sqrt{\frac{1}{M}}$

Here, only mass is varying and all factors and number of moles are constant

Now, the ratio of rms speed is

$\frac{{{\left(v_{rms}\right)}_H}_e}{{\left(v_{rms}\right)}_{Ar}}=\sqrt{\frac{M_{Ar}}{M_{He}}}$

$\frac{{{\left(v_{rms}\right)}_H}_e}{{\left(v_{rms}\right)}_{Ar}}=\sqrt{\frac{40}{4}}$

$\frac{{{\left(v_{rms}\right)}_H}_e}{{\left(v_{rms}\right)}_{Ar}}=\sqrt{10}$

$\frac{{{\left(v_{rms}\right)}_H}_e}{{\left(v_{rms}\right)}_{Ar}}=3.16$

Hence, the ratio of rms speed is $3.16$