Kinetic Gas Equation

The kinetic theory of gases (KTG) tells us the method to determine the kinetic energy of a particular gas. It also helps us know the factors on which the kinetic energy of an ideal gas depends. This, in turn, will help us derive the kinetic gas equation. The kinetic theory of gases ascertains all the internal properties of an ideal gas, like the velocity and the factors on which it depends, its kinetic energy, and many more things at the microscopic level. Therefore, the KTG is 100% valid for ideal gas; however, partially valid for real gases.

Kinetic Gas Assumption

The kinetic gas assumption is one of the most crucial aspects of modern science which was established by British scientists, James Clerk Maxwell, and Ludwig Boltzmann, who was an Australian physicist in the nineteenth century. The simple kinetic model is based on some very important assumptions such as :

1. The gas is made up of a large number of the same molecules moving freely in an unspecific direction, which are separated by the distance that are compared according to their sizes.

2. The molecule undergoes collision with each other with no energy loss and with the container. Molecules release heat when there is the transfer of kinetic energy. With the help of these assumptions, the characteristics of gases come under the range of mathematical treatment.

This model completely describes a perfect gas or similar to a real gas but under a limit of extreme dilution and extreme temperature. However, such a description is not accurate enough to simulate the real characteristics and behaviour of gases at high density.

According to the kinetic theory, container pressure can be mathematically assigned to the aimless collision of molecules. Average energy totally depends upon the temperature of the gas. Therefore, pressure can be directly related to density and temperature. By this many other properties can be obtained like thermal and electrical conductivity, etc

Perfect Gas

Perfect gas is famously known as an ideal gas that obeys the physical behaviour to a specific relation between temperature, pressure and volume further described as the general gas law. The general gas law is basically derived from the KTG and is totally dependent on the assumption that there are a large number of molecules in a gas, which are in undefined motion and strictly follow the laws of motion established by Newton. The second assumption is that the size of the volume of the molecule should be small when compared to gas. No force takes place on the molecule except the time when the elastic collision is negligible. However, only perfect gas has these properties.

The behaviour that the real gases have are quite similar to the general law at high enough pressure and extreme temperature. Their high speed overcomes any sort of interaction. When gas is at its condensation point, also known as the point at which it liquefies, it does not follow the equation or even if any gas is in the mixture. The general law is the law that is applicable to almost every gas.

Now, we will go in-depth with the postulates of KTG followed by the kinetic gas equation derivation.

Important Postulates of KTG

• Gas is made up of small particles like atoms or molecules, where all the molecules of the same gas are identical, i.e., the same shape, size, and mass.

• Molecules are constantly in random motion along the straight line, i.e., along the x, y, or z-axis. So, the gas molecules have velocity and they are divided into three components along the three-axis, then:

v = vX i + vY j + vZ k

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The probability of molecules going along the x-axis is equal to the probability of going along the y-axis and then along the z-axis. Let’s say, if I ask you, in the next second, in which direction the gas molecules will go? Your answer will be, it can go along the x/y/z-axis, so the probability along each axis is the same.

So, numerically,  vX = vY = vZ. It means the probability of the average speed of the moving molecules along each axis is assumed the same.

The magnitude of u is: 

\[v = \sqrt{v_{x}^{2} + v_{y}^{2} + v_{z}^{2}}\]

Since vX  =  vY  = vZ. So, the resultant velocity becomes:

\[= v^{2} = 3vx^{2} = 3vy^{2} = 3vz^{2}\]

• All the collisions of gas molecules with themselves and with the walls of the container are elastic and there is no interatomic force of attraction among molecules.

• The kinetic energy of gases depends only on the absolute temperature of the gas.

• The volume occupied by gas molecules is negligible as compared to the volume of gas.

Now, let’s derive the kinetic gas equation:

Derivation of Kinetic Gas Equation

Let's consider an ideal gas contained in a cubical container having each side as ‘a’. The volume of the gas is a3.

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If there are ‘n’ molecules, each having mass ‘m’, the total mass (M) of the container will be:

M = m * n….(1)

Let A1, A2,...., An be the molecules with their respective velocities as c1, c2,...., cn.

Let (x1, y1, z1), (x2, y2, z2),...., (xn, yn, zn) be the rectangular components of the velocities c1, c,..., cn along with the three mutually perpendicular directions viz: OX, OY, and OZ.

Since it is an ideal gas, so the mass of each molecule is ‘m’.

If x1 is the velocity component of A1 along OX, then the momentum of A1 is:

= mx1

After the collision, the momentum of A1 becomes:

= - mx1

So, the change in momentum is:

= - mx1 - (- mx1) = - 2mx1

By the law of conservation of momentum, the momentum transferred by the molecule A1

to the wall will be:

= + 2 mx1

The time between the two successive collisions is ‘2a’ and the distance covered by the molecule is ‘x1’ is given by:

\[t = \frac{2a}{x_{1}}\]

In every second, the momentum transferred to the wall = momentum transferred in one wall * no. of collisions

\[= \frac{2mx_{1} * x_{1}}{2a}\]

\[= \frac{mx{_{1}}^{2}}{a}\]

By Newton’s second law, the rate of change of momentum equals the force exerted by the molecule A on the wall, so:

\[= f_{1} = \frac{mx{_{1}}^{2}}{a}\]

Similarly, the force exerted by each molecule on the wall will be:

\[= f_{2} = \frac{mx{_{2}}^{2}}{a}\]

\[= f_{n} = \frac{mx{_{n}}^{2}}{a}\]

So, the total force exerted on the walls along the x-axis:

FX = f1 + f2 +....+ fn

\[= \frac{m}{a} (x_{1}^{2} + x_{2}^{2} + . . . . + x_{n}^{2})\]

The pressure exerted on the walls will be:

\[P_{x} = \frac{m}{a^{3}}(x_{1}^{2} + x_{2}^{2} + . . . + x_{n}^{2})\]

Similarly, the pressure along the y and z-axis will be:

\[P_{y} = \frac{m}{a^{3}}(y_{1}^{2} + y_{2}^{2} + . . . + y_{n}^{2})\], and

\[P_{z} = \frac{m}{a^{3}}(z_{1}^{2} + z_{2}^{2} + . . . + x_{n}^{2})\]

Since the molecular density is uniform throughout the gas; therefore, the pressure of the gas molecules is the same in all directions.

So,:

\[P = \frac{p_{x} + p_{y} + p_{z}}{3}\]

\[P = \frac{m}{3a^{3}} [(x_{1}^{2} + y_{1}^{2} + z_{1}^{2}) + (x_{2}^{2} + y_{2}^{2} + z_{2}^{2} + (x_{n}^{2} + y_{n}^{2} + z_{n}^{2})]\]

\[= \frac{m}{3V} [c_{1}^{2} + c_{2}^{2} + . . . + c_{n}^{2})]\] (Since \[V = a^{3}\])

\[= \frac{mn}{3V} [(c_{1}^{2} + c_{2}^{2} + . . . + c_{n}^{2}){n}]\]…..(2)

From eq (1) in eq (2):

\[= \frac{M}{3V} [\frac{(c_{1}^{2} + c_{2}^{2} + . . . + c_{n}^{2})}{n}]\]

\[P = \frac{M}{3V} C^{2} (C^{2} = \frac{(c_{1}^{2} + c_{2}^{2} + . . . + c_{n}^{2})}{n} \text{ is the R.M.S speed)}\]

\[P = \frac{1}{3} \rho C^{2}\] ....(3)

Or,

\[C = \sqrt{\frac{3P}{\rho}}\]

⇒ Knowing the value of P & \[\rho\], we can determine the R.M.S velocity of the gas at a given temperature.

Now, let’s proceed with FAQs on KTG.