Question

Consider an ideal gas at pressure P, volume V and temperature T. The mean free path for molecules of the gas is L. If the radius of gas molecules, as well as pressure, volume and temperature of the gas are doubled, then the mean free path will be: 

विकल्प

• ${\displaystyle \frac{5\text{L}}{2}}$

• ${\displaystyle \frac{\text{L}}{4}}$

• "L"/8

• 2L

Answer 1

${\displaystyle \frac{\text{L}}{4}}$

Explanation:

The mean free path of gas molecules is,

${\displaystyle \overline{\lambda }=\frac{1}{\sqrt{2}}\frac{{\text{k}}_{\text{B}}\text{T}}{\pi {\text{d}}^2\text{P}}}$

∴ ${\displaystyle \text{L}=\frac{1}{\sqrt{2}}\frac{{\text{k}}_{\text{B}}\text{T}}{4\pi {\text{r}}^2\text{P}}}$

According to given conditions,

${\displaystyle \text{L'}=\frac{1}{\sqrt{2}}\frac{{\text{k}}_{\text{B}}\text{T'}}{4\pi \left(\text{r'}\right)\text{P'}}}$

${\displaystyle \frac{\text{L'}}{\text{L}}=\frac{\text{T'}}{{\left(\text{r'}\right)}^2\text{P'}}\times \frac{{\text{r}}^2\text{P}}{\text{T}}}$

= ${\displaystyle \frac{2\text{T}}{{\left(2\text{r}\right)}^2\left(2\text{P}\right)}\times \frac{{\text{r}}^2\text{P}}{\text{T}}}$

= ${\displaystyle \frac{1}{4}}$

∴ L' = ${\displaystyle \frac{\text{L}}{4}}$