An ideal gas in a closed container is slowly | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The mean free path $\lambda$ of molecules of an ideal gas is given by the formula:$\lambda = \frac{V}{\sqrt{2} \pi d^2 N}$where $V$ is the volume

Vedclass · Accepted Answer

$(B)$ and $(C)$

Vedclass · Answer

(D) The mean free path $\lambda$ of molecules of an ideal gas is given by the formula:$\lambda = \frac{V}{\sqrt{2} \pi d^2 N}$where $V$ is the volume of the container,$d$ is the molecular diameter,and $N$ is the number of molecules.Since the gas is in a closed container,the volume $V$ and the number of molecules $N$ remain constant as the temperature increases. Therefore,the mean free path $\lambda$ remains unchanged.The mean collision time $	au$ is defined as the ratio of the mean free path to the average speed of the molecules $(v_{av})$:$	au = \frac{\lambda}{v_{av}}$Since the average speed $v_{av}$ of gas molecules is proportional to $\sqrt{T}$,we have:$	au \propto \frac{1}{\sqrt{T}}$As the temperature $T$ increases,the average speed of the molecules increases,which causes the mean collision time $	au$ to decrease.Thus,statements $(B)$ and $(C)$ are correct.

Similar Questions

If the radius of an $O_{2}$ molecule is $40 \; \mathring{A}$,$T = 27^{\circ}C$,and $P = 1 \; \text{atm}$,find the relaxation time.

In Van der Waals equation $\left( P + \frac{a}{V^2} \right)(V - b) = RT$,what do $a$ and $b$ represent?

What is mean free path? Derive the equation for the mean free path.

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters $1\,\mathring{A}$ and $2\,\mathring{A}$. The gases may be considered under identical conditions of temperature,pressure,and volume.

Vedclass Test Series

Exam Paper Generator

Online Exam Module