Think of the tires of a bicycle you have to inflate with air- when you blow air into it, the tire swells up and becomes hard. This occurs as the air molecules within acquire energy, move at high speed and in turn create pressure through impacting the inner walls. This is the common event which explains the basic principle of Kinetic theory of Gases, which is the subject of Class 11 physics chapter 12. The Kinetic Theory explains the behavior of gases at a molecular level, which assists students in grasping the relationship between temperature, pressure, and volume. This chapter holds special importance to students who will encounter major board exams of CBSE, as well as JEE and NEET, as it establishes the basis of thermodynamics and how real gases behave which are taught again later in their more advanced study and in board exams too.
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The NCERT Class 11 Physics Notes of Chapter 12 explain these complicated concepts by means of simple definitions and short explanations. Students will have an opportunity to study such major ideas as molecular nature of matter, gaseous behaviour, ideal gas approximations, law of equipartition of energy, gas specific heat capacities and collision theory and mean free path. The NCERT Class 11 Kinetic theory of gases notes are designed in a way that will improve the ability to learn in the short term and do final-minute revisions. Although they give an excellent overview, they pay more attention to the clarity of concepts rather than step-by-step derivations which are usually carried out independently. They contain some of the most important concepts, formulas, diagrams, and most importantly, exam practice questions, so they are a sure way to prepare yourself whether you are a school or competitive exam student.
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The Kinetic Theory NCERT Notes are crucial for exam preparation and help students gain a better understanding. These notes contain the important derivations and concepts required for exams. Students can use the notes to prepare as well as to review their knowledge.
Some properties of gases
Gases can be compressed easily.
Gases have neither a definite volume nor a definite shape.
When a substance is in the gaseous state, its density is very small in comparison to the density in its solid or liquid state.
$\begin{aligned} & V \propto \frac{1}{P} \\ & P V=\text { constant }\end{aligned}$
$P_1 V_1=P_2 V_2$
$V_t=V_0\left(1+\frac{t}{273}\right)$
$\begin{aligned} & V_1=V_0\left(1+\frac{t_1}{273}\right)=V_0\left(\frac{273+t_1}{273}\right) \\ & V_2=V_0\left(1+\frac{t_2}{273}\right)=V_0\left(\frac{273+t_2}{273}\right) \\ & \therefore \frac{V_1}{V_2}=\frac{273+t_1}{273+t_2}=\frac{T_1}{T_2} \\ & \frac{V_1}{T_1}=\frac{V_2}{T_2} \\ & \frac{V}{T}=a \text { cons } \tan t \\ & V \propto T\end{aligned}$
Therefore at constant pressure, the volume of a given mass of a gas is directly proportional to its absolute temperature. This is Charles’ Law in terms of absolute temperature.
It strictly obeys Boyle's Law, Charles’ Law and the law of pressure under all conditions of temperature and pressure.
Its pressure coefficient and volume coefficient are exactly equal to each other.
Its molecules are infinitesimally small.
There is no force attraction between its molecules.
Equation of State for an Ideal Gas
i. Let the temperature of the gas be kept constant at T₁ and the pressure is changed from P₁ to P₂. If in this process the volume changes from V₁ to V2 then by Boyle’s law,
$$
\begin{aligned}
& P_2 V_2=P_1 V_1 \\
& V_2=\frac{P_1 V_1}{P_2}
\end{aligned}
$$
ii. Let the pressure of the gas is kept constant at P₂ and its absolute temperature is changed from T₁ to T₂. If in this process the volume of the gas changes from V1 to V₂ then by Charles’ law,
$$
\begin{aligned}
& \frac{V_1}{T_1}=\frac{V_2}{T_2} \\
& V_2=\frac{T_1 V_1}{T_2}
\end{aligned}
$$
From eqn (i) and (ii), we have
$\begin{aligned} & \frac{P_1 V_1}{P_2}=\frac{T_1 V_2}{T_2} \\ & \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\end{aligned}$
Hence for a gas, the volume of PV/T remains constant. Let this constant be equal to K, then
$\begin{aligned} & \frac{P V}{T}=K \\ & P V=K T\end{aligned}$
We can write K = nk, k is the same for all gases. It is called Boltzmann constant.
$P V=n k T \quad \ldots \ldots(i i i)$
If P, V, and T are the same, then n is also the same for all gases.
If we take 1 gram-molecule (1 mole) of a gas-constant it will be the same for all gases. Then it is called the universal gas constant and it is represented by R. Then,
$P V=R T$
Where R = Nk, where N = Avogadro’s number and k = Boltzmann constant
It is called the ideal gas equation or perfect gas equation.
$P V=\mu R T \quad \ldots \ldots(i v)$
The following are the fundamental assumptions of the kinetic theory of gases:
Let us consider a perfect gas contained in a cubical vessel of side (l) and perfectly elastic walls. Let n be the total total number of molecules of a gas in the vessel and m is the mass of the molecule. Let the sides of the cube represent mutually perpendicular X, Y, and Z-axes.
Consider a molecule molecule P moving in a random direction with velocity v1. Let v1x, v1y, and v1z be the components of the velocity v1 along three mutually perpendicular X, Y, and Z axes, respectively.
Then
$v_1^2=v_{1 x}^2+v_{1 y}^2+v_{1 z}^2$
The component of velocity with which the molecule P will strike the face DEFA is v1x and the momentum of the molecule X- direction before collisions
$=m v_{1 x}$
As the molecule and the wall are perfectly elastic, during collision speed of the molecule does not change and only the direction of velocity is changed.
Therefore, The momentum along X- axis after collisions
$=m\left(-v_{1 x}\right)=-m v_{1 x}$
The change in the momentum during one collision along X-axis
$=m v_{1 x}-\left(-m v_{1 x}\right)=2 m v_{1 x}$
The molecule suffering reflection at face DEFA traverses a distance l and strikes the face CGBO. It is equal to the force exerted on it. Hence, the force exerted by the molecule on the wall A will be mvx2/l Since there are n molecules, the total force exerted on the wall A is given by
$\frac{m}{l}\left(v_{x_1}^2+v_{x_2}^2+\ldots \ldots v_{x_n}^2\right)$
Where vxn is the x-component of velocity of the molecule 1, vx2 of molecule 2,.... And so on. Since the area of the wall is l2, the pressure on the wall A is given by
$\begin{aligned} & P=\frac{m}{l^3}\left(v_{x_1}^2+v_{x_2}^2+\ldots \ldots v_{x_n}^2\right) \\ & =\frac{m n}{l^3}\left(\frac{\left(v_{x_1}^2+v_{x_2}^2+\ldots \ldots v_{x_n}^2\right)}{n}\right) \\ & \text { But } l^3=\vee \\ & \therefore P=\frac{m n}{V}\left(\frac{\left(v_{x_1}^2+v_{x_2}^2+\ldots \ldots v_{x_n}^2\right)}{n}\right)\end{aligned}$
Now, $\left(\frac{\left(v_{x_1}^2+v_{x_2}^2+\ldots \ldots v_{x_n}^2\right)}{n}\right)$ is the average value of vx2 for all the n molecules. This can be written as
$\overline{v_x^2}$
Therefore,
$P=\frac{m n}{V} \bar{v}_x^2 \quad \ldots \ldots(i)$
We know that for any molecule, we have
$$
v^2=\overline{v_x^2}+\overline{v_y^2}+\overline{v_z^2} \quad \ldots \ldots
$$
Because this equation is true for every molecule, this will also be true for the mean value of all the molecules.
$\therefore v^{\overline{2}}=\bar{v}_x^2+\bar{v}_y^2+\bar{v}_z^2 \quad \ldots \ldots(i i i)$
The directions of the molecules are quite at random, that is, all directions are equally possible. Hence, the values, of $\overline{v_x^2}, \overline{v_y^2}, \overline{v_z^2}$ will be equal, ie. $\overline{v_x^2}=\overline{v_y^2}=\overline{v_z^2}$
Therefore from eqn (ii), we have
$$
\begin{aligned}
\overline{v^2} & =3 \overline{v_x^2} \\
\overline{v_x^2} & =\frac{1}{3} \overline{v^2}
\end{aligned}
$$
or
$$
P=\frac{1}{3} \frac{m n}{V} \overline{v^2} \ldots \ldots(i i i)
$$
This is the equation for the pressure of the gas. $\overline{v^2}$
is called the mean-square velocity of the molecules. Since, mn is the mass of the whole gas and V is the volume so, mn/V is the density ⍴ of the gas therefore, the above expression can be written as
$P=\frac{1}{3} \rho \bar{v}^2 \quad \ldots .(i v)$
Again,
$P=\frac{1}{3} \rho v^2=\frac{2}{3}\left(\frac{1}{2} \rho \overline{v^2}\right)=\frac{2}{3} E$
Where $E=\left(\frac{1}{2} \rho \overline{v^2}\right)$ is the transnational kinetic energy of the gas per unit volume. Thus, the pressure is equal to two-thirds of its transnational kinetic energy per unit volume.
$v_{r m s}=\sqrt{\overline{v^2}} \sqrt{\frac{v_1^2+v_2^2+v_3^2}{n}}$
$\begin{aligned} & P=\frac{1}{3} \rho \overline{v^2} \\ & v_{r m s}=\sqrt{\overline{v^2}}=\sqrt{\frac{3 P}{\rho}}\end{aligned}$
Let us consider 1 gram molecule of a gas at absolute temperature, T, occupying a volume V. According to Kinetic theory, the pressure of the gas
$\begin{aligned} & P=\frac{1}{3} \cdot \frac{m N}{V} \overline{v^2} \ldots \ldots(i) \\ & P V=\frac{1}{3} m N \overline{v^2} \because \ldots(i i) \\ & \because P V=R T \ldots \ldots(i i i) \end{aligned} $
And mN = M
Comparing eqn (ii) and (iii)
$\begin{aligned} & \frac{1}{3} M \overline{v^2}=R T \\ & \overline{v^2}=\frac{3 R}{M} T \\ & \sqrt{\overline{v^2}}=\sqrt{\frac{3 R T}{M}} \\ & v_{r m s}=\sqrt{\frac{3 R T}{M}} \\ & v_{r m s} \propto \sqrt{T}\end{aligned}$
Thus, the root mean square speed of the molecules of a gas is directly proportional to the square root of the absolute temperature of the gas.
Relation between temperature and molecular kinetic energy:-
$$
\frac{1}{2} M \overline{v^2}=\frac{1}{2} M\left(\frac{3 R T}{M}\right)
$$
$\frac{(3 / 2) R T}{N}=\frac{3}{2}\left(\frac{R}{N}\right) T$
R/N = k (Boltzmann constant)
$$
\text { Average kinetic energy of } 1 \text { molecule }=\frac{3}{2} K T
$$
It means different gases at the same temperature have the same average kinetic energy per molecule and this energy is directly proportional to the absolute temperature of the gas.
Thus, the average kinetic energy of a molecule per degree of freedom is 0.5KT which is the law of equipartition of energy.
We know
$P V=\frac{1}{3} m N v^{\overline{2}} \quad \ldots .(i)$
According to kinetic theory, the average kinetic energy of translation of molecules of gas is directly proportional to its absolute temperature,
$\frac{1}{2} m \overline{v^2}=\alpha T$
i.e., at a given temperature, average kinetic energy of a gas is constant and N is also constant, so eqn (i)
$P V=\frac{2}{3} N \cdot \frac{1}{2} m \overline{v^2}=\frac{2}{3} N \alpha T$
= constant at a given temperature
This is Boyle's Law.
From eqn(i)
$P V=\frac{1}{3} m N \overline{v^2}$
Also the average K.E. of a molecule,
$\begin{aligned} & \frac{1}{2} m \overline{v^2}=\alpha T \\ & P V=\frac{2}{3} N \cdot \alpha T \\ & V \propto T\end{aligned}$
for constant value P.
Thus at a given pressure for a given mass of a gas, the volume is directly proportional to the absolute temperature. This is Charles' law.
Thus, if λ₁, λ₂, λ₃, ….λₙ are the successive free paths traversed in the total time t, then,
$\lambda_1+\lambda_2+\ldots . \lambda_n=\bar{v} t$
$\begin{aligned} \lambda & =\frac{\lambda_1+\lambda_2+\lambda_3+\ldots . \lambda_n}{N} \\ & =\frac{\bar{v} t}{N}\end{aligned}$
We know that,
$\begin{aligned} \lambda & =\frac{\text { total dis } \tan \text { ce covered in time } t}{\text { number of collision in time } t} \\ & =\frac{\bar{v} t}{n \times \pi d^2 \bar{v} t}=\frac{1}{\pi n d^2}\end{aligned}$
But by experiment, Maxwell found that
$\lambda=\frac{1}{\sqrt{2}\left(\pi \sigma^2 n\right)}$
The mean free path is inversely proportional to the number of molecules per unit volume (n) which is proportional to the density of the gas.
For N molecule of an ideal gas, we have
$\begin{aligned} & P V=N k T \\ & \therefore n=\frac{N}{V}=\frac{P}{k T} \\ & \lambda=\frac{k T}{\sqrt{2} \pi d^2 P}\end{aligned}$
This is the mean free path of the molecules of a gas is directly proportional to its absolute temperature and inversely proportional to the pressure of the gas.
Avagadro’s Number: The Avagadro’s number is the number of molecules in 1 mole of a substance. It is denoted by N and its value is 6.0221 X 10 ²³.
Q1: Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory PV = $\frac{2}{3}E$ , E is
(a) the total energy per unit volume
(b) only the translational part of energy because rotational energy is very small compared to the translational energy
(c) only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum
(d) the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero
Answer:
As per the kinetic theory of gases, the perpendicular forces exerted by the molecules on the walls while in motion are only responsible for the pressure exerted due to the gas molecules. So, for molecules striking at angles other than 90 degrees, no pressure will be exerted. Hence, in this case, only translational motion change leads to pressure on the wall. So, PV = $\frac{2}{3}E$, which represents the translational motion part and hence (c) is the correct option.
Q2: When an ideal gas is compressed adiabatically, its temperature rises the molecules on average have more kinetic energy than before. The kinetic energy increases,
(a) because of collisions with moving parts of the wall only
(b) because of collisions with the entire wall
(c) because the molecules get accelerated in their motion inside the volume
(d) because of the redistribution of energy amongst the molecules
Answer:
The number of collisions per second between the molecules and walls increases and the mean free path becomes smaller as an ideal gas is compressed. This, in turn, increases the temperature of the gas, which increases the overall kinetic energy of the gas molecules as KE depends on the temperature. Hence, (a) is the correct option.
Q3: In a diatomic molecule, the rotational energy at a given temperature
(a) obeys Maxwell's distribution
(b) have the same value for all molecules
(c) equals the translational kinetic energy for each molecule
(d) is $\left(\frac{2}{3}\right) ^{rd}$ the translational kinetic energy for each molecule
Answer:
If we assume a diatomic molecule along the $z$-axis, its energy along that axis will be zero. The total energy of a diatomic molecule can be expressed as: $E=\frac{1}{2} m v x^2+\frac{1}{2} m v y^2+\frac{1}{2} m v z^2+\frac{1}{2} I x W x^2+\frac{1}{2} I x W y^2$.
The number of independent terms in the expression $=5$. The above expression obeys Maxwell's distribution as their velocities can be predicted with Maxwell's findings. In this case, for each molecule 3 translational and 2 rotational energies are associated. So, at any temperature, rotational energy $=\left(\frac{2}{3}\right) ^{rd}$ translational KE
Hence, (a) and (d) are the correct options.
Frequently Asked Questions (FAQs)
While notes may focus on key concepts, understanding and practicing important derivations like pressure formula and rms speed is crucial for board exams and JEE/NEET preparation.
It is a theory that describes how gases behave regarding the notions that the particles of gas are always in random motion and that the pressure of the gas results due to the collisions betw
It aids the macroscopic nature of gases, including pressure, temperature, and gas volume, by studying the microscopic behavior of gas molecules.
It is mainly applicable to ideal gases but can be helpful in comprehending real ones to lesser extent. The deviations happen in the high pressure and low temperatures.
Numerical and conceptual questions about ideal gas behavior, specific heat and molecular motion are used in JEE, NEET and other engineering/medical entrance examinations in
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