Kinetic Theory of Gases MCQs for NEET — Physics Questions with Answers

$$\begin{gathered} \mathrm{The} \mathrm{van} \mathrm{der} \mathrm{Waal}\text{'}\mathrm{s} \mathrm{equation} \mathrm{for} 1 \mathrm{mole} \mathrm{of} \mathrm{gas} \mathrm{is} \\ \left(\mathrm{P}+\frac{\mathrm{a}}{{\mathrm{V}}^2}\right) \left(\mathrm{V}-\mathrm{b}\right)=\mathrm{RT} \\ \mathrm{if} \mathrm{we} \mathrm{put}, {\mathrm{P}}_{\mathrm{r}}=\frac{\mathrm{P}}{{\mathrm{P}}_{\mathrm{C}}}, {\mathrm{V}}_{\mathrm{r}}=\frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{c}}} \mathrm{and} {\mathrm{T}}_{\mathrm{r}}=\frac{\mathrm{T}}{{\mathrm{T}}_{\mathrm{C}}}; \\ \left(\because {\mathrm{P}}_{\mathrm{C}}=\frac{\mathrm{a}}{27 {\mathrm{b}}^2}, {\mathrm{V}}_{\mathrm{C}}=3\mathrm{b} \mathrm{and} {\mathrm{T}}_{\mathrm{C}}=\frac{8\mathrm{a}}{27 \mathrm{Rb}}\right) \\ \left({\mathrm{P}}_{\mathrm{r}}\left(\frac{{\mathrm{a}}^2}{27 {\mathrm{b}}^2}\right)+\frac{\mathrm{a}}{{\mathrm{V}}_{\mathrm{r}}^2{\left(3\mathrm{b}\right)}^2}\right) \left({\mathrm{V}}_{\mathrm{r}}\left(3\mathrm{b}\right)-\mathrm{b}\right)={\mathrm{RT}}_{\mathrm{r}} \left(\frac{8\mathrm{a}}{27\mathrm{Rb}}\right) \\ \mathrm{This} \mathrm{equation} \mathrm{is} \mathrm{called} \mathrm{van} \mathrm{der} \mathrm{Waal}\text{'}\mathrm{s} \mathrm{equation} \mathrm{of} \mathrm{law} \mathrm{of} \mathrm{corresponding} \mathrm{states}. \end{gathered}$$

(C)  $\mathrm{V} \mathrm{\alpha } \mathrm{n}$ at same temperature and Pressure.

According to the NCERT text, 'The mean free path $l$ is the average distance covered by a molecule between two successive collisions'.

The formula for mean free path is $l = 1 / (\sqrt{2} \pi n d^2)$. This equation clearly shows that $l$ is inversely proportional to both the number density ($n$) and the square of the diameter of the molecule ($d^2$). The NCERT text also states, 'As expected, the mean free path given by Eq. (12.40) depends inversely on the number density and the size of the molecules'.

The formula for mean free path is $l = 1 / (\sqrt{2} \pi n d^2)$. If $n$ is doubled, the denominator becomes $2n$, making the mean free path half of its original value ($l' = 1 / (\sqrt{2} \pi (2n) d^2) = l/2$).

At constant pressure, according to the ideal gas law ($PV = N k_B T$), increasing temperature ($T$) leads to a decrease in number density ($n = N/V$). Since $l$ is inversely proportional to $n$, an increase in temperature (and thus a decrease in $n$) will lead to an increase in the mean free path.

The NCERT text states, 'In a highly evacuated tube n [number density] is rather small and the mean free path can be as large as the length of the tube'.

The NCERT 'POINTS TO PONDER' section notes, 'What is different is the mean free path which in a gas is 100 times the interatomic distance and 1000 times the size of the molecule.' A large mean free path implies that molecules travel significant distances before colliding, suggesting ample space and facilitating motion that can be described as approximately unhindered for these distances.

The mean free path $l$ is inversely proportional to $d^2$ ($l \propto 1/d^2$). If $d$ is halved, then $d^2$ becomes $(d/2)^2 = d^2/4$. Therefore, $l$ would be proportional to $1/(d^2/4) = 4/d^2$, meaning the mean free path would be quadrupled.

In Example 12.9, the calculation for air molecules gives $l = 2.9 \times 10^{-7} m$ and $d = 2 \times 10^{-10} m$. The ratio $l/d \approx (2.9 \times 10^{-7} m) / (2 \times 10^{-10} m) \approx 1450$, which is approximately 1500. The NCERT explicitly states '$l \approx 1500 d$' after the calculation.