Question 284: Thermodynamics & KTG

Question

The number of air molecules per cm$$^3$$ increased from $$3	imes10^{19}$$ to $$12	imes10^{19}$$. The ratio of collision frequency of air molecules before and after the increase in number respectively is:

Accepted Answer

1. The collision frequency (f) is given by the formula :
$$ f = \sqrt{2} \pi d^2 v n_v $$
Where:
- $d$ is the diameter of the molecule,
- $v$ is the average velocity of the molecules, and
- $n_v$ is the number density (number of molecules per unit volume).
2. From this equation, we can see that the collision frequency (f) is directly proportional to the number density ($n_v$), because all the other variables ($d$ and $v$) are constant :
$$ f \propto n_v $$
3. Therefore, the ratio of two different collision frequencies ($f_1$ and $f_2$) is equal to the ratio of their corresponding number densities ($n_{v1}$ and $n_{v2}$):
$$ \frac{f_1}{f_2} = \frac{n_{v1}}{n_{v2}} $$
4. Given that the number density increased from $3 	imes 10^{19}$ to $12 	imes 10^{19}$, we can substitute these values into the equation to find the ratio of the collision frequencies:
$$ \frac{f_1}{f_2} = \frac{3 	imes 10^{19}}{12 	imes 10^{19}} $$
5. Simplifying this equation gives:
$$ \frac{f_1}{f_2} = 0.25 $$
So, the ratio of the collision frequency of air molecules before and after the increase in number is 0.25.

Answer

1.25

Answer

0.25

Answer

0.50

Answer

0.75

Thermodynamics & KTG

The number of air molecules per cm$^3 increased from 3\times10^{19} to 12\times10^{19}$. The ratio of collision frequency of air molecules before and after the increase in number respectively is:

Choose the correct answer: