Kinetic Theory of Gases

Kinetic Theory of Gases

Edited By Shivani Poonia | Updated on Oct 10, 2024 01:49 PM IST

The Kinetic Theory of Gases is a fundamental model that explains the macroscopic properties of gases in terms of the motion and interactions of their molecules. According to the kinetic theory, the temperature of a gas is directly related to the average kinetic energy of its molecules, which is proportional to the temperature in Kelvin. Gas molecules are in constant, random motion.

This Story also Contains
  1. Kinetic Energy and Molecular Speeds
  2. Some Solved Examples
  3. Summary
Kinetic Theory of Gases
Kinetic Theory of Gases

Kinetic Energy and Molecular Speeds

As you have studied in the previous section the molecules of a gas are always in motion and are colliding with each other and with the walls of the container. Due to these collisions, the speeds and the kinetic energies of the individual molecules keep on changing. However, at a given temperature, the average kinetic energy Of the gas molecules remains constant.

If at a given temperature, n1, molecules have speed u1, n2, molecules have speed u2, n3 molecules have speed u3, and so on. Then, the total kinetic energy (EK) of the gas at this temperature is given by:
$E_K=\frac{1}{2} m\left(n_1 v_1^2+n_2 v_2^2+n_3 v_3^2+\ldots \ldots \ldots\right)$
where m is the mass of the molecule. The corresponding average kinetic energy $\overline{E_k}$ of the gas will be:

$
\overline{\mathrm{E}_{\mathrm{K}}}=\frac{1}{2} \frac{\mathrm{m}\left(\mathrm{n}_1 \mathrm{v}_1^2+\mathrm{n}_2 \mathrm{v}_2^2+\mathrm{n}_3 \mathrm{v}_3^2+\ldots \ldots \ldots\right)}{\left(\mathrm{n}_1+\mathrm{n}_2+\mathrm{n}_3+\ldots \ldots \ldots\right)}
$

If the term $\frac{\left(n_1 v_1^2+n_2 v_2^2+n_3 v_3^2+\ldots \ldots \ldots\right)}{\left(n_1+n_2+n_3+\ldots \ldots\right)}=\bar{v}^2$ then the average kinetic energy is given by :
$
\overline{\mathrm{E}_\kappa}=\frac{1}{2} \mathrm{mv}^2
$
where $v$ is given by
$
v=\sqrt{\frac{\left(n_1 v_1^2+n_2 v_2^2+n_3 v_3^2+\ldots \ldots \ldots\right)}{\left(n_1+n_2+n_3+\ldots \ldots \ldots\right)}}
$

This 'v' is known as root-mean-square speed urms

Maxwell-Boltzmann Distribution of speeds

According to it

  • Molecules have different speeds due to frequent molecular collisions with the walls and among themselves.
  • Rare molecules have either very high or very low speed.
  • Maximum number of molecules of the gas have maximum velocity which is called most probable velocity and after Vmp velocity decreases.
  • Zero velocity is not possible.
  • All these velocities increase with the increase in temperature but fraction of molecules having these velocities decreases.
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Average Speed, uav

It is the average of different velocities possessed by the molecules

$\begin{aligned} & u_{a v}=\frac{u_1+u_2+u_3}{n} \\ & u_{a v}=\frac{n_1 u_1+n_2 u_2+n_3 u_3}{n_1+n_2+n_3}\end{aligned}$

Here n1, n2, n3 are the number of molecules having u1, u2, u3 velocities respectively.
Relation between Uav, temperature and molar mass is given as

$\mathrm{u}_{\mathrm{av}}=\sqrt{8 \mathrm{RT} / \pi \mathrm{M}}=\sqrt{8 \mathrm{PV} / \pi \mathrm{M}}$

Most Probable Speed, ump

The most probable speed(ump) of a gas at a given temperature is the speed possessed by the maximum number of molecules at that temperature. Unlike average speed and root mean square speed, the most probable speed cannot be expressed in terms of the individual molecular speeds.

The most probable speed(ump) is related to absolute temperature (T) by the expression:

$u_{m p}=\sqrt{\frac{2 R T}{M}}=\sqrt{\frac{2 P V}{M}}$

Root Mean Square Speed urms

It is the square root of the mean of the square of the velocities of different molecules.$\begin{aligned} & u_{r m s}=\frac{\sqrt{u_1^2+u_2^2+\ldots}}{n} \\ & =\frac{\sqrt{n_1 u_1^2+n_2 u_2^2+n_3 u_3^2}}{n_1+n_2+n_3} \\ & u_{r m s}=\sqrt{3 R T / M} \\ & u_{r m s}=\sqrt{3 P V / M}=\sqrt{3 P / d}\end{aligned}$

Relation between uav, ump and urms

The three types of molecular speeds, namely, most probable speed(vmp), average speed (vav) and root mean square speed(vrms) of a gas at a given temperature are related to each other as follows:
$\begin{aligned} & v_{m p}: v_{a v}: v_{m s}=\sqrt{\frac{2 R T}{M}}: \sqrt{\frac{8 R T}{\pi M}}: \sqrt{\frac{3 R T}{M}} \\ & v_{m p}: v_{a v}: v_{r m s}=1.414: 1.596: 1.732 \\ & v_{m p}: v_{a v}: v_{r m s}=1: 1.128: 1.224\end{aligned}$

For a particular gas, at a particular temperature:
vmp < vav < vrms

Recommended topic video on (Kinetic Theory of Gases)

Some Solved Examples

Example 1: The ratio among most probable velocity, mean velocity, and root mean square velocity is given by

1)$1: 2: 3$

2)$1: \sqrt{2}: \sqrt{3}$

3)$\sqrt{2}: \sqrt{3}: \sqrt{\frac{8}{\pi}}$

4) $\sqrt{2}: \sqrt{\frac{8}{\pi}}: \sqrt{3}$

Solution

Most probable speed of gas molecules -

$V_{m p}=\sqrt{2 R T / M}$

M- Molecular Mass, R- Gas Constant, T- Temperature
most probable velocity: mean velocity : $V_{r m s}$

$=\sqrt{\frac{2 R T}{M}}: \sqrt{\frac{8 R T}{\pi M}}: \sqrt{\frac{3 R T}{M}}=\sqrt{2}: \sqrt{\frac{8}{\pi}}: \sqrt{3}$

Hence, the answer is the option (4).

Example 2: Calculate the $u_{r m s}$ (in m/sec) of $\mathrm{O}_2$ if its density at 1 atm pressure and $0^{\circ} \mathrm{O}$ is 1.429g/l.

Answer upto one decimal places

1) 462.2

2)46.1

3)56.3

4)44.2

Solution

It is the square root of the mean of the square of the velocities of different molecules.$\begin{aligned} & u_{r m s}=\frac{\sqrt{u_1^2+u_2^2+\ldots \ldots}}{n} \\ & =\frac{\sqrt{n_1 u_1^2+n_2 u_2^2+n_3 u_3^2}}{n_1+n_2+n_3} \\ & u_{r m s}=\sqrt{3 R T / M} \\ & u_{r m s}=\sqrt{3 P V / M}=\sqrt{3 P / d}\end{aligned}$

We know that rms velocity is given as:

$u_{r m s}=\frac{3 P}{d}$

Now, $P=1 \mathrm{~atm}=101.3 \times 10^3 \mathrm{~Pa}$

And d = 1.42g/litre = 1.42kg/m^3

$u_{r m s}=\sqrt{\frac{3 \times 101.32 \times 10^3}{1.429}}=462.21 \mathrm{~m} / \mathrm{sec}$

Hence, the answer is the option (1).

Identify the correct labels A. B and C in the following graph from the options given below:

Example 3: Root mean square speed $\left(V_{r m s}\right)$ ; most probable speed $\left(V_{m p}\right)$, Average speed $\left(V_{a v}\right)$

1) A - $\left(V_{m p}\right)$, B- $\left(V_{a v}\right)$, C- $\left(V_{r m s}\right)$

2)A - $\left(V_{m p}\right)$, B- $\left(V_{r m s}\right)$ , C- $\left(V_{r m s}\right)$

3)A - $\left(V_{a v}\right)$, B- $\left(V_{r m s}\right)$ ,C- $\left(V_{a v}\right)$

4)A - $\left(V_{r m s}\right)$ ,B- $\left(V_{m p}\right)$, C- $\left(V_{a v}\right)$

Solution
The three types of molecular speeds, namely, most probable speed(vmp), average speed (vav), and root mean square speed(vrms) of a gas at a given temperature are related to each other as follows:
$\begin{aligned} & v_{m p}: v_{a v}: v_{m s}=\sqrt{\frac{2 R T}{M}}: \sqrt{\frac{8 R T}{\pi M}}: \sqrt{\frac{3 R T}{M}} \\ & v_{m p}: v_{a v}: v_{r m s}=1.414: 1.596: 1.732 \\ & v_{m p}: v_{a v}: v_{r m s}=1: 1.128: 1.224\end{aligned}$

For a particular gas, at a particular temperature:
vmp < vav < vrms

It follows from the above relationship that:
Average speed(vav) =0.921 x Root mean square speed(vrms)
Most probable speed(vmp) = 0.817 x Root mean square speed(vrms)

CRMS > CAverage > CMPS.

A - VMPS, B - VAverage , C - VRMS

Therefore, Option(1) is correct.

Summary

The kinetic theory of gases explains the macroscopic properties with respect to the motion and interaction of individual molecules in the gas. According to this theory, these gas molecules are subject to continuous, random motion, where the impacts with each other and the walls of the container are perfectly elastic. The theory is founded on several key postulates: gas molecules occupy negligible volume compared with the container; they interact with each other only through elastic collisions

Articles

Get answers from students and experts
Back to top