Heat and Thermodynamics 4 Question 14
16. 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 $\left[\frac{v _{\text {rms (helium) }}}{v _{\text {rms (argon) }}}\right]$ is close to
(2019 Main, 9 Jan I)
(a) 0.32
(b) 2.24
(c) 3.16
(d) 0.45
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Solution:
- Root mean square (rms) velocity of the molecules of a gas is given as
$$ v _{rms}=\sqrt{\frac{3 R T}{M}} $$
where, $M$ is the atomic mass of the gas.
$$ \begin{array}{llrl} & \Rightarrow & v _{\text {rms }} & \propto \sqrt{\frac{1}{M}} \\ \therefore & & \frac{v _{\text {rms (helium) }}}{v _{\text {rms (argon) }}} & =\sqrt{\frac{M _{\text {argon }}}{M _{\text {helium }}}} \end{array} $$
Given, $\quad M _{\text {argon }}=40 u$ and $M _{\text {helium }}=4 u$
Substituting these values in Eq. (i), we get
$$ \frac{v _{\text {rms (helium) }}}{v _{\text {rms (argon) }}}=\sqrt{\frac{40}{4}}=\sqrt{10}=3.16 $$
