Question 172: Thermodynamics & KTG

Question

Consider two boxes containing ideal gases $A$ and $B$ such that their temperatures, pressures and number densities are same. The molecular size of $A$ is half of that of $B$ and mass of molecule $A$ is four times that of $B$. If the collision frequency in gas $B$ is $32 	imes 10^{18} / \mathrm{s}$ then collision frequency in gas $A$ is $\_\_\_\_$ /s.

Accepted Answer

For an ideal gas, the collision frequency of a molecule is given by the kinetic theory of gases :
$$ \mathrm{Z}=\sqrt{2} \pi \mathrm{~d}^2 \mathrm{v}_{	ext {avg }} \mathrm{n} $$
Where :
d is the molecular diameter (or size).
$\mathrm{V}_{	ext {avg }}$ is the average speed of the gas molecules.
n is the number density (number of molecules per unit volume).
The average speed $\left(\mathrm{v}_{	ext {avg }}ight)$ of gas molecules depends on the absolute temperature ( T ) and the mass of a single molecule (m) according to the formula :
$$ \mathrm{v}_{\mathrm{avg}}=\sqrt{\frac{8 \mathrm{kT}}{\pi \mathrm{~m}}} $$
Where k is the Boltzmann constant.
Substituting this into the collision frequency formula gives :
$$ \mathrm{Z}=\sqrt{2} \pi \mathrm{~d}^2\left(\sqrt{\frac{8 \mathrm{kT}}{\pi \mathrm{~m}}}ight) \mathrm{n} $$
It is given that the temperatures (T), pressures, and number densities (n) are exactly the same for both gases A and B.
$$ \mathrm{Z} \propto \frac{\mathrm{~d}^2}{\sqrt{\mathrm{~m}}} $$
$$ \Rightarrow \frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\left(\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{~d}_{\mathrm{B}}}ight)^2 	imes \sqrt{\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{~m}_{\mathrm{A}}}} $$
Molecular size of A is half that of $\mathrm{B} \Rightarrow \mathrm{d}_{\mathrm{A}}=\frac{1}{2} \mathrm{~d}_{\mathrm{B}} \Rightarrow \frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}=\frac{1}{2}$
Mass of molecule A is four times that of $\mathrm{B} \Rightarrow \mathrm{m}_{\mathrm{A}}=4 \mathrm{~m}_{\mathrm{B}} \Rightarrow \frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}}}=\frac{1}{4}$.
Collision frequency in gas B is $\mathrm{Z}_{\mathrm{B}}=32 	imes 10^{18} / \mathrm{s}$.
$$ \frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\left(\frac{1}{2}ight)^2 	imes \sqrt{\frac{1}{4}} $$
$$\Rightarrow $$ $\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{1}{4} 	imes \frac{1}{2}$
$$\Rightarrow $$ $\frac{\mathrm{Z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{1}{8} \Rightarrow \mathrm{Z}_{\mathrm{A}}=\frac{\mathrm{Z}_{\mathrm{B}}}{8}=\frac{32 	imes 10^{18}}{8}=4 	imes 10^{18} / \mathrm{s}$
So, the collision frequency in gas A is $4 	imes 10^{18} / \mathrm{s}$.

Answer

$8 	imes 10^{18}$

Answer

$2 	imes 10^{18}$

Answer

$32 	imes 10^{18}$

Answer

$4 	imes 10^{18}$

Thermodynamics & KTG

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