Kinetic Gas Equation

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What is Kinetic Theory of Gases?

The kinetic theory of gases (KTG) tells us the method to determine the kinetic energy of a particular gas. It also helps us know that 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, viz:  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 Theory of Gases Definition

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 viz: 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 one 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:

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

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:

Derive Kinetic Gas Equation

Let 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 = 2a/x1

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

=  2mx1 * x1/2a = mx12/a

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

f1 = mx12/a

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

f2 = mx22/a,    fn = mxn2/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.

FAQ (Frequently Asked Questions)

Question 1: Let’s Suppose that the Mass of Each Molecule of the Gas is Halved and the Speed is Doubled, then the Ratio of Initial and the Final Pressure will be?

Answer: 

Given:

M1 = M, C1 = C

V1 = V


To find: P1

Solution:

P1 = M1/3V1C12

P2 = M2/3V2C22

P1/P2 = M1/M2 * V2/V1 * C12/C22

M/(M/2) * V/V * C2/4C2

P1/P2 = 1/2

Question 2: Derive the Average Translational K.E. of the Gas Molecule.

Answer: We know that the pressure of the gas molecule is:

P = M/3V C2

Or,

PV = M/3 C2

We know that PV = RT for one mole of gas

So,

M/3 C2 = RT

Or,

M/2 C2 = 3/2 RT

Since M = mN

mN/2 C2

Also, R/N = k

m/2 C2 = 3/2 RNT


Hence, the average translational K.E. is:

m/2 C2 = 3/2 kT

Question 3: What are the Limitations of the Kinetic Theory of Gases?

Answer: On increasing the temperature, the motion of the molecules in the liquid increases, and therefore, the interatomic force of attraction between these molecules reduces.

Question 4: Which is not an Assumption of the Kinetic Theory of Gases?

Answer: It is difficult to compress the gas particles (atoms/molecules) at high pressure.