## What is the kinetic theory of gases?

The Kinetic Theory of gases is a classical model of gases, according to which gases are composed of molecules/particles that are in random motion. While undergoing this random motion, kinetic energy in molecules can assume random velocity across all directions. It also says that the constituent particles/molecules undergo elastic collision, which means that the total kinetic energy remains constant before and after the collision. The average kinetic energy of the particles also determines the pressure of the gas.

## Assumptions of the kinetic theory of gases

The kinetic theory of gas makes some assumptions about the nature of an ideal gas; upon this assumption, it bases the arguments to explain the properties of gases. These assumptions are as follows,

- Smaller particles or molecules make up the gas. Their size is such that the sum of each gas particle is smaller as compared to the container volume. This statement is equivalent to saying that the mean distance between gas molecules is large as compared to their size. So, the time between the collision of particles and the wall of the container is negligible compared to the time between two successive collisions between particles themselves.
- The population of particles is very large, such that the statistical methods can be justifiably applied. This is also known as the thermodynamic limit. This limit is the ratio of the number of particles in gas to their volume, as both quantities approach infinity, their ratio remains constant.
- The molecules in kinetic theory are assumed to be perfectly hard spheres that undergo collision among themselves and the walls of the container elastically.
- The interaction between the molecules is negligible except when they are colliding, i.e. they move independently of each other, and feel each other's influence only when they are colliding.

**Work done on an ideal gas**

Consider a gas contained in a box fitted with a movable piston such that no gas leaks out. This gas exerts some pressure on the piston, which is given by

$P=\frac{F}{A}$ ......(1)

where F is the force on the piston and A is the cross-sectional area of the piston.

Now, as we compress the piston, we have to work against the pressure of the gas to decrease its volume. The differential work done in compressing the gas by moving the piston by a small amount dx along its axis (say x-direction) is given by

$dW=-Fdx$

From equation (1),

$dW=-PAdx$

Now since $Adx=dV$,

$dW=-PdV$ ......(2)

The negative sign indicates that work done on the gas is positive when it is compressed (i.e., when its volume decreases).

**The pressure of ideal gas**

Building upon the assumptions above, we derive the pressure due to an ideal gas in a container. As the particles of gas collide with the walls of the container, they exert force on them which we measure as the pressure of the gas. To analyze this situation quantitatively, consider a volume of gas in a box, and the box has a movable piston on one side. If the piston was left free, each time a molecule collided with the piston, it would gain some momentum, and as time passed, more and more molecules collided with the piston, hence it would pick up some finite momentum and fly off the box. To stop the piston from flying away requires some external force, this force per unit area of the face of the piston.

To calculate the force on a piston per collision, we require the rate of change of momentum as the particles collide elastically with the piston wall and reverse their momentum. Assume that the piston's axis lies along the x-axis, so that when the molecule. Mathematically speaking the change in momentum when one molecule of gas reverses its x component of velocity after bouncing off the piston wall is,

$\u2206P=m{v}_{x}-(-m{v}_{x})$

$\u2206P=2m{v}_{x}$

Now the number of such collisions depends upon the number of molecules hitting the piston per second. Suppose that there are N molecules in volume V, so the number density of molecules is,

$n=\frac{N}{V}$

To determine the number of molecules hitting the piston in time t, note that the particle has to have a certain velocity and be close enough to the piston, if it happens to be far away it would not make it to the piston in time t. This distance is

$d={v}_{x}t$ ......(3)

So the range of volume molecules lying in which collide with the piston in time t is given by,

$V=AdV$

$V=A{v}_{x}t$

Where A is the cross-sectional area of the piston.

So, the number of molecules hitting piston in time 't' is the product of the number density of the molecules and the range of volume 'V', i.e.

${n}_{hitting}=nV\phantom{\rule{0ex}{0ex}}{n}_{hitting}=nA{v}_{x}t$

So, the number of molecules hitting per unit time is,

$n\text{'}=nA{v}_{x}$

Now, the total force on the piston per unit time is,

$F=n\text{'}\u2206P\phantom{\rule{0ex}{0ex}}F=nA{v}_{x}\left(2m{v}_{x}\right)\phantom{\rule{0ex}{0ex}}F=2mnA{{v}_{x}}^{2}$

Since the pressure is force per unit area, the pressure exerted by the gas on the piston is given by,

$P=FA\phantom{\rule{0ex}{0ex}}P=2mn{v}_{x}{}^{2}$ .....(4)

The logic used to arrive at the pressure in equation (4) is somewhat incomplete, the missing point is that all molecules do not have the same velocity and do not move in the x-direction, so the x component of velocity is different for all the molecules, so what we want is average of squared velocity overall molecules, which we write as,

${{v}^{2}}_{x,avg}=<{{v}_{x}}^{2}>$

Now, on average, half of the molecules moving along x-direction have velocities pointing along the positive x-axis, and half of them have velocities along the negative x-direction, the expression for pressure reduces to,

$P=\frac{2mn{{v}_{x}}^{2}}{2}$

$P=mn{{v}_{x}}^{2}$

The last piece to put together is the fact from the point of view of a gas molecule. There is nothing special about any particular direction, all the directions are equivalent, so the average motion of molecules in all directions are all equal, i.e.

${v}^{2}={{v}_{x}}^{2}={{v}_{y}}^{2}={{v}_{z}}^{2}$

So, we can write it as,

${v}^{2}={\frac{1}{2}{v}_{x}}^{2}+\frac{1}{3}{{v}_{y}}^{2}+\frac{1}{3}{{v}_{z}}^{2}$

Finally

$P=\frac{mn{v}^{2}}{3}$ ......(5)

**Ideal gas equation**

Now the result (5) can be stated in terms of average kinetic energy by noting the fact that the expression for kinetic energy is,

$<KE>=<\frac{1}{2}m{v}^{2}>$ ......(6)

Using this, equation (5) can be rewritten in the form,

$P=\frac{2}{3}n<KE>$

The equation gives the pressure of a gas in terms of its number density and average kinetic energy of the particles.

One of the assumptions of kinetic theory was that the particle of gas does not interact except at the time of the collision. This means that there is no potential energy associated with the system. The total energy of the system comprises kinetic energy alone. So, the total internal energy 'U' of gas can be written as,

$U=n<KE>$

Then equation (6) in terms of internal energy U can be written as,

$PV=\frac{2}{3}U$ ......(7)

## Context and application

The kinetic theory of gas is studied under following courses

- Bachelors in Technology (Mechanical Engineering)
- Masters in Technology (Mechanical Engineering)
- Bachelors in Science (Chemistry/ Physics)
- Masters in Science (Chemistry/Physics)

## Practice problems

- A cylinder and piston system has an ideal gas in it. Suppose that the diameter of the cylinder is 2m, find out the work done by the piston if it moves 2.5 cm inwards compressing the gas at constant pressure 1 Pascal.

- 0.078J
- 0.35J
- 0.049J
- 2J

**Answer**: a) 0.078J

**Explanation**: The work done by the piston is pressure P times the change in volume of the gas dV, which is given by the product of the area of the piston's face times the distance moved. So the expression is W=PAdx. Upon substituting the given values in this expression, we get the answer as 0.078J.

**2.** Determine the pressure of an ideal gas whose particles have average kinetic energy KE=6 mJ confined to a unit volume box.

- 4 mJ
- 4 J
- 6.7 mJ
- 3J

**Answer**: a) 4mJ

**Explanation**: The pressure is related to the kinetic energy as $P=\frac{2}{3}KE$. Upon substituting the given values in the formula, we get the answer as 4mJ.

**3.** Find the average internal energy of a gas confined to 2 cubic meters volume with a pressure P=5 Pascals.

- 12J
- 10J
- 11J
- 15J

**Answer**: d) 15J

**Explanation**: The pressure P, volume V and internal energy U of an ideal gas are related as $PV=\frac{2}{3}U$. Putting the values in the expression gives the value of U as 15J.

**4. ** Which of the following factor determines the average kinetic energy of the ideal gas?

- The temperature of the gas
- The volume of the gas
- The mass of the gas molecules
- The atomic number of the gas molecules

**Answer**: a) The temperature of the gas

**Explanation**: The temperature quantifies the internal energy of a gas, in the case of an ideal gas, kinetic energy is the only type of energy it can possess, so the temperature, in turn, determines the kinetic energy of the ideal gas.

**5.** A cylinder containing an ideal gas is moving on a lorry with some uniform speed. How would the temperature of the gas be affected?

- The temperature will increase.
- The temperature will decrease.
- The temperature will remain the same.
- The information is insufficient.

**Answer**: c) The temperature will remain the same

**Explanation**: The cylinder and particles of an ideal gas are moving at the same speed, i.e. at the speed of a lorry. There is no relative motion between the walls of the cylinder and gas particles due to the motion of the lorry. Since the speed of the lorry does not contribute to the collision process in the cylinder, the temperature remains the same.

## Formulas

Work done on ideal gas,

$dW=-PdV$

The pressure of ideal gas,

$PV=\frac{2}{3}U$

The internal energy of ideal gas,

$U=n<KE>\phantom{\rule{0ex}{0ex}}U=\frac{n}{2}m<{v}^{2}>$

## Common Mistakes

In the case of an ideal gas, the particles are ideally assumed to have zero volume and compressible to zero volume too. Since this is not strictly true for any real gas, the postulate of zero sizes must be relaxed and the size must be taken into account. The ideal gas particles were postulated to have zero interaction amongst themselves except when colliding, this postulate is also approximately true in the case of real gases in only some situations.

**Related Concepts**

- Boltzmann Equation
- Gas Laws
- Heat
- Thermodynamics
- Interatomic potential

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