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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Answer:

Correct Answer: 16. (c)

Solution:

  1. 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.

$$ \begin{array}{llrl} & \Rightarrow & v _{\text {rms }} & \propto \sqrt{\frac{1}{M}} \\ \therefore & & \frac{v _{\text {rms (helium) }}}{v _{\text {rms (argon) }}} & =\sqrt{\frac{M _{\text {argon }}}{M _{\text {helium }}}} \cdots(i) \end{array} $$

Given, $\quad M _{\text {argon }}=40 u$ and $M _{\text {helium }}=4 u$

Substituting these values in Eq. (i), we get

$$ \frac{v _{\text {rms (helium) }}}{v _{\text {rms (argon) }}}=\sqrt{\frac{40}{4}}=\sqrt{10}=3.16 $$



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