Kinetic Theory

Behaviour of Gases

In gases, molecules are far from each other as compared to solids and liquids. Also, mutual interaction between molecules is negligible except when they collide. Hence, understanding the behaviour and properties of gases is much easier than solids or liquids.

Suggested Videos

previous arrow
next arrow
previous arrownext arrow


Behaviour of Gases

Gases which are at low pressures or high temperatures (higher than their liquefaction of solidification range) approximately satisfy a simple relation between their pressure, volume, and temperature given by –

PV = KT … (1)

where T is the temperature in Kelvin, K is a constant for the given sample (which can vary with the volume of the gas), V is the volume, and P is the pressure of the gas. Let’s look at this now with the perspective of atoms and molecules. Since the gas constant K varies with the volume of the gas, it is proportional to the number of molecules in the sample. Hence, we have

K = Nk

where N is the number of molecules in the sample and k is Boltzmann’s constant which is also denoted as kB. From equation (1), we have


⇒ K = \( \frac {PV}{T} \)

⇒ NkB = \( \frac {PV}{T} \)

⇒ kB = \( \frac {PV}{NT} \)

Hence, for two samples 1 and 2, we have

\( \frac {P_1V_1}{N_1T_1} \) = \( \frac {P_2V_2}{N_2T_2} \) = constant = kB … (2)

Form the above equation, we can conclude that if P, V, and T are same then N is also the same for all gases. This is Avagadro’s Hypothesis which states: ‘The number of molecules per unit volume is the same for all gases at a fixed temperature and pressure’.

Avogadro’s Number

The number of molecules in 22.4 liters of any gas is 6.02 x 1023. This is Avogadro’s number and is denoted by NA. The mass of 22.4 liters of any gas is equal to its molecular weight at STP (Standard Temperature and Pressure; Temperature = 273K and Pressure = 1 atm). This amount of substance is called a mole.

By studying chemical reactions, Avogadro had guessed the equality in numbers in equal volumes of gas at fixed temperature and pressure. This hypothesis is justified by Kinetic theory. Therefore, the perfect gas equation is written as,

PV = μ RT … (3)

where μ is the number of moles and R = NA. kB is a universal constant. The temperature T is absolute temperature. By choosing the Kelvin scale for absolute temperature, we get, R = 8.314 J mol–1K–1. Hence,

μ = \( \frac {M}{M_0} \) = \( \frac {N}{N_A} \) … (4)

where M is the mass of the gas containing N molecules, M0 is the molar mass, and NA is Avogadro’s number. From equations (3) and (4), we have

PV = kB NT or P = kB nT

where n is the number of molecules per unit volume. kB is the Boltzmann’s constant having value 1.38 × 10–23 J K–1 (in SI units). Equation (3) can also be written as

P = \( \frac {pRT}{M_0} \) … (5)

where p is the mass density of the gas.

Ideal Gas

Any gas that satisfies equation (3) at all pressures and temperatures is called an ideal gas. However, this is for theoretical purposes only since no gas can be truly ideal. The figure below shows the behaviour of a real gas at three different temperatures.


You can observe that the real gas departs from the ideal gas behaviour at three different temperatures. Also, all curves approach the ideal gas line for low pressures and high temperatures. It simply means that without molecular interactions, all gases behave like an ideal gas.

Boyle’s Law

If we fix μ and T in equation (3), we get

PV = constant … (6)

So, by keeping the temperature constant, the pressure of a given mass of gas is inversely proportional to the volume of the gas. This is Boyle’s law. The following figure shows a comparison of the experimental and theoretical P-V curves as predicted by Boyle’s law.


From the figure, it is evident that the agreement is good at high temperatures and low pressures.

Charles’ Law

Now, if you fix P, from equation (1), we get V ∝ T. Hence, at a fixed pressure, the volume of a gas is inversely proportional to its absolute temperature (Charles’ Law). This is represented in the figure below:


Dalton’s Law of Partial Pressures

Now consider a mixture of ideal gases which don’t interact with each other in a vessel of volume V at temperature T and pressure P. Let the mixture contain μ1 moles of gas 1, μ2 moles of gas2, etc. In this mixture, we find that

PV = (μ1 + μ2 +… ) RT … (7)

i.e. P = (μ1 \( \frac {RT}{V} \) + μ2 \( \frac {RT}{V} \) + …) … (8)

Or P = (P1 + P2 + …) … (9)

Where P1 = μ1\( \frac {RT}{V} \) is the pressure gas 1 would exert at the same conditions of volume and temperature if no other gases were present. This is called the partial pressure of the gas. Hence, the total pressure of a mixture of ideal gases is simply the sum of the partial pressures. This is Dalton’s law of partial pressures.

Solved Examples For You

Q1. The figure below shows a plot of \( \frac {PV}{T} \) versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.


  1. What does the dotted plot signify?
  2. Which is true: T1 > T2 or T1 < T2?
  3. What is the value of \( \frac {PV}{T} \) where the curves meet on the y-axis?
  4. If we obtained similar plots for 1.00×10–3 kg of hydrogen, would we get the same value of \( \frac {PV}{T} \) at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of \( \frac {PV}{T} \) (for the low-pressure high-temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.)


  • The dotted line is parallel to P. Hence, it remains unchanged even when P is changed. Therefore, it corresponds to the behaviour of an ideal gas.
  • The dotted line represents an ideal gas. From the figure, we can see that T1 is closer to the dotted line than T2. Now, a real gas approaches the behaviour of an ideal gas at high temperatures. Hence, T1>T2.
  • When the two curves meet, the value of \( \frac {PV}{T} \) is μR. This is because the ideal gas equation for μ moles is: PV = μRT.

The molecular mass of oxygen = 32 g. Therefore,

μ = \( \frac {1 × 10^{-3}}{32 × 10^{-3}} \) kg = \( \frac {1}{32} \)

We also know that the value of the universal constant R is 8.31 Jmole-1K-1. Therefore,

\( \frac {PV}{T} \) = \( \frac {1}{32} \) × 8.31 = 0.26 JK-1

  • If we obtained similar plots for 1 x 10-3 kg of hydrogen, then we will not get the same value of \( \frac {PV}{T} \) at the point where the curve meets the axis. The reason is that the molecular mass of hydrogen is different from that of oxygen.

Let’s calculate the mass of hydrogen needed to obtain the same value of \( \frac {PV}{T} \). We have,

\( \frac {PV}{T} \) = 0.26 JK-1

Molecular mass of hydrogen = M = 2.02 u.

We know that μ = \( \frac {m}{M} \) … where m is the mass of hydrogen. Also, we know that \( \frac {PV}{T} \) = μR at constant temperature. Therefore,

\( \frac {PV}{T} \) = \( \frac {m}{M} \)R

m = \( \frac {PV}{T} \) × \( \frac {M}{R} \) = \( \frac {0.26 × 2.02}{8.31} \) = 6.32 x 10-2 gram or 6.32 x 10-5 kg.

Share with friends

Customize your course in 30 seconds

Which class are you in?
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Dr. Nazma Shaik
Gaurav Tiwari
Get Started

Leave a Reply

Your email address will not be published. Required fields are marked *

Download the App

Watch lectures, practise questions and take tests on the go.

Customize your course in 30 seconds

No thanks.