The Kinetic Theory of Gases. Ideal Gases | iCalculator™

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Thermodynamics Learning Material
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13.6The Kinetic Theory of Gases. Ideal Gases

In this Physics tutorial, you will learn:

  • The definition of mole
  • What is the Avogadro's number and what does it represent?
  • How can we calculate the number of particles in a substance?
  • What is an ideal gas?
  • What is the formula of the Ideal Gas Law?
  • What is the Boltzmann constant and how can we express the ideal gas law in terms of it?
  • How to calculate work done by the gas in a thermodynamic process?
  • How to find the work from a P - V graph?
  • What are state variables and path-dependent variables?


How do the molecules of a gas move? Do they have a determined direction of motion or they move randomly in the space?

Is it possible to calculate the kinetic energy of each individual particle of a gas? Why? What method do we use to calculate the average kinetic energy of all particles in a gas? Which is the closest physical quantity to the average kinetic energy of particles?

What do you understand with the term "ideal"? What is an ideal gas according to your opinion?

In this tutorial, we will discuss about the behavior of gas particles in various circumstances. There are a number of factors, quantities and parameters to be considered in this regard. Therefore, the study of gases behavior is considered as an entire "theory", i.e. the kinetic theory of gases.

The Meaning of Mole. Avogadro's Number

When you go to buy rice, you take pockets of 1 kg from the shelves. Everybody would laugh if hears you saying to the seller: "Please, give me 7432 grains of rice."

Similarly, it is meaningless to consider the gas molecules individually; they always are taken as "pockets" of particles. In tutorial 1.1 "Units. Systems of Units. Fundamental and Derived SI Units", it was explained that the standard (SI) unit of amount of matter is not kilogram but mole (mol) instead. It represents a determined number of particles used as a standard of measurement for all materials.

By definition, one mole (in short mol) of a substance represents the mass of 6.02 × 1023 particles of that substance.

The number 6.02 × 1023 is known as Avogadro's number (NA) and it represents the number of particles contained in 12 g of Carbon - 12.

In other words, 1 mol of carbon-12 has a mass of 12 grams and contains 6.022140857 x 1023 of carbon atoms (written in 10 significant figures). We say molar mass M of carbon-12 is 12 g/mol = 0.012 kg/mol.

Mole is one of the seven fundamental quantities in science. It differs from mass as for the same number of particles (for example for one mole) different materials have different masses. Therefore, mole represents better than mass the number of particles contained in a sample of material.

We can calculate the number of moles n in a certain amount of a known substance of mass m and molar mass M through the equation

n = m/M

Example 1

The molar mass of oxygen O2 is 32 g/mol. How many molecules are there in 160 g O2?

Solution 1

First, we must calculate the number of moles n contained in 160 g oxygen O2. We have M = 32 g/mol and m = 160 g. Thus,

n = m/M
= 160 g/32 g/mol
= 5 moles

Hence, the number N of O2 molecules contained in 160 g oxygen O2 is:

N = n × Na
= 5 moles × 6.02 × 1023 molecules/mol
= 30.10 × 1023 molecules
= 3.01 × 1024 molecules of O2

Ideal Gases

An ideal gas is a model (not a reality) used to describe easier the behaviour of gases. However, real gases in certain conditions can be assumed as ideal gases with a good approximation.

A gas must meet the following condition to be considered as ideal:

  1. Particles do not collide with each other but only with the walls of the container. For this, they must have negligible dimensions (negligible volume).
  2. Particles are equally sized and do not interact through intermolecular forces (attraction or repulsion) with other gas particles.
  3. Particles move in random directions in accordance with the Newton's Laws of Motion.
  4. Particles experience perfectly elastic collisions with the container walls, i.e. they don't lose energy during such collisions.

In reality, there are no ideal gases. All gas particles possess a volume within the system (very small, but not zero anyway), a condition which violates the first assumption. Also, gas particles can have different sizes; especially when there is a mixture of gases such as the air. Yet, gases in a thermodynamic system have intermolecular forces with neighboring gas particles, especially at low temperatures where the particles move slowly and interact with each other. Finally, even though gas particles can move randomly, they do not experience perfect elastic collisions with the container walls based on the law of conservation of energy and momentum within the system.

Thermodynamic systems at low pressure and high temperature have a similar behavior to ideal gases.

The Ideal Gas Law

The ideal gas law is the equation of state for an ideal gas, which establishes the relation between the four parameters of a gas sample. These four parameters are pressure, volume, temperature and number of moles of the gas sample. Given that all gases behave quite ideally at low pressures, we can calculate the unknown fourth parameter when three of them are known.

The Ideal Gas Law is an empirical law of physics. It was derived from experiment and observation rather than from theory. The experiments show that the ratio of the products

Pressure × Volume/Number of moles × Absolute temperature

is always constant. This constant is always the same for all gas samples. It is known as the ideal gas constant and is denoted by R. For ideal gases it has a value of R = 8.31 J/mol × K.

Therefore, we can write:

R = P × V/n × T

Rearranging in order to remove the fraction, we obtain the Ideal Gas Law:

P × V = n × R × T

In chemistry, there is another unit used for the ideal constant R. It stems from the fact that in chemistry pressure is usually measured in atmospheres (atm), volume in litres (L) and temperature in Kelvin degree (K).

Example 2

Find the value of ideal gas constant R in atm × L/mol × K given that its value is R = 8.31 J/mol × K.

Solution 2

From previous sections, we known the following conversion factors:

1 atm = 101 325 Pa (the exact value, not rounded)
1 L = 1 dm3 = 0.001 m3

Given that

R = P × V/n × T

It is obvious that we have to convert only the units in the numerator, as those in the denominator are the same in both cases. Thus,

1 atm × 1L = 101 325 Pa × 0.001 m3
= 101 325 N/m2 × 0.001 m3
= 101.325 N × m
= 101.325 J


1 atm × L = 101.325 J


8.31 J = 8.31/101.325 atm × L
=0.082 atm × L

Hence, we obtain the value of R in atm × L/mol × K equal to 0.082 atm × L/mol × K.

There is a very important quantity known as the Boltzmann Constant, k, which is used to write the Ideal Gas Law in an alternative form. It is defined by the equation:

k = R/Na

Substituting the known values, we obtain for the value of Boltzmann Constant k:

k = 8.31 J/mol × K/6.02 × 1023 particles/mol
= 1.38 × 10-23 J/K

The equation of Boltzmann Constant k, helps us write the Ideal gas Law in an alternative form as stated earlier. Rearranging the last equation, we have:

Na = R/k

Also, we have explained earlier that

Na = N/n

where N is the total number of particles in a gas and n is the number of moles. Thus, combining the last two equations, we obtain:

R/k = N/n


n × R = N × k

Therefore, substituting the right part of the above expression in the equation of Ideal Gas Law P × V = n × R × T, we obtain:

P × V = N × k × T

Note the difference between the two versions of the Ideal gas law. The original equation is written in terms of moles, while this last form involves the number of gas molecules.

Example 3

A gas sample exerts 2 kPa pressure on the walls of a 40 L container at 270. What is the number of gas molecules in the sample?

Solution 3


P = 2 kPa = 2000 Pa
V = 40 L = 0.04 m3
T = 27°C = 300 K

Also, we known that R = 8.31 J/mol × K.

Let's calculate the number of moles first. From the Ideal Gas Law, we have

P × V = n × R × T
n = P × V/R × T
= 2000 × 0.04/8.31 × 300
= 0.032 moles

Now, let's calculate the number of molecules N in the gas sample. We have:

n = N/na

Rearranging the last equation, we obtain:

N = n × Na
= 0.032 moles × 6.02 × 1023 molecules/mol
= 0.19264 × 1023 molecules
= 1.9264 × 1022 molecules

Work in a Thermodynamic Process

Let's find an expression for calculation of the work done by a gas during a slow thermal process. Consider a cylinder filled with gas with a freely moveable piston above it as shown in the figure below.

Physics Tutorials: This image provides visual information for the physics tutorial The Kinetic Theory of Gases. Ideal Gases

Since the piston moves freely, the gas pressure remains constant because it is balanced from the atmospheric pressure, i.e. the pressure exerted by the air from above the piston. The work done by the gas to raise the piston by Δy as shown in the figure below, is

Physics Tutorials: This image provides visual information for the physics tutorial The Kinetic Theory of Gases. Ideal Gases

Given that

Force = Pressure × Area

we have

Wby gas=Pgas × Apiston × ∆y
=Pgas × (Apiston × ∆y)
=Pgas × ∆V

This expression gives the left part of the ideal gas law formula.

The formula indicates that:

Volume increases ⇒ ΔV > 0 ⇒ Wby gas > 0
Volume decreases ⇒ ΔV < 0 ⇒ Wby gas < 0
Volume constant ⇒ ΔV = 0 ⇒ Wby gas = 0

as we discussed earlier.

Example 4

12 g of helium gas is heated from 20 ºC to 100 ºC, at a constant pressure. What is the work done by helium during the heating process? (Take MHe = 4 g/mol)

Solution 4

First, let's write the clues to create a clearer idea about the situation. We have:

m = 12 g
M = 4 g/mol
t1 = 20°C
t2 = 100°C
P1 = P2 = P
Wby gas = ?

From the first two clues we can work out the number of moles, n. We have:

n = m/M
= 12 g/4 g/mol
= 3 moles

Also, from the next two clues, we obtain the change in temperature:

∆T = t2 - t1
= 100°C - 20°C
= 80°C
= 80 K

Applying the ideal gas law for the two given instants (states) 1 and 2, we obtain

P × V1 = n × R × T1


P × V2 = n × R × T2

Subtracting the first equation from the second, we obtain

P × V2 - P × V1 = n × R × T2 - n × R × T1


P × ∆V = n × R × ∆T

The first part of the last equation gives the work done by the gas. Therefore, the left part will also give the work done by the gas. Hence, we can write:

Wby gas = n × R × ∆T
= 3 moles × 8.31 J/mol × K × 80 K
= 1994.4 J

This result means helium does 1994.4 J of work against the environment during thermal expansion.

Finding Work from P - V Diagram

It is clear that in a P - V diagram, work is mathematically represented through the area under the graph. The simplest case is when pressure is constant as shown in the graph below.

Physics Tutorials: This image provides visual information for the physics tutorial The Kinetic Theory of Gases. Ideal Gases

This reasoning can be extended in the processes with non-constant pressure as well. Thus, even if the pressure is changing, we can divide the curve over P - V diagram into small, constant pressure steps. The total area under the curve is therefore the sum of small rectangular areas. As we increase the number of rectangles by taking thinner rectangles, the error in calculation becomes smaller.

Physics Tutorials: This image provides visual information for the physics tutorial The Kinetic Theory of Gases. Ideal Gases

If we consider a certain rectangle (the i-th rectangle), the change in volume is ΔVi and the actual pressure is pi. Therefore, the work done to change the volume for this interval is

∆Wi = pi × ∆Vi

Thus, the sum of all such intervals gives the total work. If we divide the graph in n squares, we can write for the total work:

W = ni=1 pi × ∆Vi

Example 5

A gas sample expands from 3 L to 11 L against the standard atmospheric pressure. What is the work done by the gas during this expansion?

Solution 5

Since the pressure is constant, we can use the P∆V formula to find the work done by the gas.

To find the result in joules, we must use the units Pa and m3. We know that the standard atmospheric pressure in pascals is Patm = 100 000 Pa. Also, we have V1 = 3 L = 0.003 m3 and V2 = 11 L = 0.011 m3. Therefore,

ΔV = V2 - V1
= 0.011 m3 - 0.003 m3
= 0.008 m3

Now, we can apply the work formula for constant volume:

Wby gas = Pgas × ∆V
= 100 000 Pa × 0,008 m3
= 800 J

State Variables and Path-Dependent Variables

a) State variables

For a fixed amount of gas, we can find the temperature if we know the pressure and volume of the gas. Consider the example below:

Example 6

A P - V diagram for 0.2 moles of an ideal gas is shown below. Determine the initial and final temperatures of the gas sample.

Physics Tutorials: This image provides visual information for the physics tutorial The Kinetic Theory of Gases. Ideal Gases

Solution 6

In the graph, "i" stands for initial and "f" for final state of the gas. Let's apply the ideal gas law for the points i and f given that P1 = P2 = P = 4 atm = 400 000 Pa, Vi = 3 L = 0.003 m3 and Vf = 5 L = 0.005 m3. We have for the initial state i:

P × Vi = n × R × Ti


Ti = P × Vi/n × R
= 100 000 × 0.003/0.2 × 8.31
= 180.5 K

For the final state, we have:

P × Vf = n × R × Tf
Tf = P × Vf/n × R
= 100 000 × 0.005/0.2 × 8.31
= 300.8 K

We draw two conclusions from the above example:

  1. Every point on a P - V diagram has a definite temperature for a fixed amount of gas.
  2. The points on a P - V diagram are called "states". A gas can undergo an infinite number of intermediate states when it moves from the initial to the final state (from i to f).

As explained earlier, pressure, volume and temperature are known as "state variables". This means all these three quantities have an infinite number of values from the initial to the final state. These values either change slowly or remain constant during a thermodynamic process.

Internal energy is also a state variable. Internal energy of a gas sample is a function of temperature only. For a definite amount of gas, every specific temperature value corresponds to a specific amount of internal energy. Therefore, for a given state, the internal energy of gas has a fixed value. When the gas sample returns to the previous state, the internal energy takes the previous value as well.

State variables are macroscopic parameters. For a given state, a gas sample has fixed values of P, V, T and U. Although the speed or energy of a certain molecule are in continuous change, the values of state variables are determined by an average of effects produced by billions of molecules.

b) Path-dependent variables

Heat and work are not state variables. It is meaningless if we talk about heat or work contained in a system. Internal energy is a property of a system. Heat and work are not properties of system in a given state, but characteristics of the process between two states. Remember, both heat and work represent energies transferred between systems.

Consider two different processes between two fixed states as shown in the graph below:

Physics Tutorials: This image provides visual information for the physics tutorial The Kinetic Theory of Gases. Ideal Gases

In the process 1 (path 1) the gas expands at constant volume.

When the gas undergoes the process 2 (path 2), it expands from the same initial state to the same final state as in the process 1, but during the process 2 pressure is first increased then decreased to the original value, probably by placing and removing weights on the piston.

Let's compare the change in the internal energy ΔU, work W and heat Q during the two above processes.

Since the temperatures Ti and Tf have fixed values in the states "i" and "f" regardless of the intermediate stages, we have T1 = T2. Since for a gas sample equal changes in temperatures imply equal changes in internal energy, we can write:

∆U1 = ∆U2

Comparing the areas under the P - V graphs (which correspond to the works in each process), we conclude that

W1 < W2

Applying the First Law of Thermodynamics Q = ΔU + W for both states 1 and 2, we obtain

Q1 < Q2

Hence, we conclude that although the initial and final states are the same, the gas sample absorbs more heat when it moves along the path 2, because it does more work along this path. This is like walking from one village to another along two different paths: one requires climbing over a hill, the other walking along a horizontal path. Both paths bring us to the same destination, but one of them requires much more energy compared to the other.

It is clear from the above example that heat and work depend not only on the initial and final states, but also on the intermediate states the gas undergoes during a thermodynamic process. Therefore, heat and work are not state variables, because they depend on the path followed between the initial and final states. Such variables are called "path-dependent variables".

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