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

The NCERT text explains, 'The top of a cloud of smoke holds together for hours. This happens because molecules in a gas have a finite though small size, so they are bound to undergo collisions. As a result, they cannot move straight unhindered; their paths keep getting incessantly deflected.' These incessant collisions, characterized by the mean free path, prevent rapid diffusion.

The rate of collisions is given by $n\pi d^2 $ according to the NCERT text. The time between two successive collisions (collision time) $\tau$ is the inverse of the collision rate, i.e., $\tau = 1 / (n\pi d^2 )$. The mean free path $l$ is then $ \tau$, showing a direct relationship between these quantities. The question asks what is DIRECTLY related to the rate of collisions; the time between collisions is inversely related to the rate, so shorter time means higher rate.

The NCERT text states, 'The average distance between two successive collisions, called the mean free path $l$, is : $l = \tau$'. Rearranging this gives $\tau = l / $.

The NCERT text states, 'In this derivation, we imagined the other molecules to be at rest. But actually all molecules are moving and the collision rate is determined by the average relative velocity of the molecules. Thus we need to replace $ $ by $ $ in Eq. (12.38). A more exact treatment gives $l = 1 / (\sqrt{2} n \pi d^2)$ (12.40).' The $\sqrt{2}$ factor arises from considering the relative motion of all molecules.

The NCERT text mentions, 'Using, the kinetic theory of gases, the bulk measurable properties like viscosity, heat conductivity and diffusion can be related to the microscopic parameters like molecular size. It is through such relations that the molecular sizes were first estimated.' This highlights the role of kinetic theory, and particularly concepts like mean free path, in connecting the macroscopic and microscopic worlds.

$v_{rms}\propto\sqrt{T}$. "Increased by 3 times" $\Rightarrow v\to 4v$, so $T\to 16T$. $T_1 = 223\ \text{K}$, $T_2 = 16 \times 223 = 3568\ \text{K} = 3295^\circ\text{C}$.

Slope of $T$-$V$ at fixed $P$ $\propto P/nR$ → steeper curve = higher pressure → $P_1 > P_2 > P_3$.

Absolute final pressure $= 11+1 = 12$ atm. $n_2 = \dfrac{PV}{RT} = \dfrac{(12\times1.01\times10^5)(0.03)}{(100/12)(300)} \approx 14.54$ mol. $\Delta n = 18.20-14.54 = 3.66$ mol; mass $= 3.66\times32 \approx 117\ \text{g} \approx 0.116\ \text{kg}$.

Internal energy conserved $\Rightarrow P_f = \dfrac{p_1V_1 + p_2V_2}{V_1+V_2} = \dfrac{1(2)+2(3)}{5} = \dfrac{8}{5} = 1.6$ atm.