Pressure In Gases – Boyle’s Law

A gas, made up of many particles called molecules that are in continuous random motion, colliding with one another and with the walls of the container. As the gas molecules strike and bounce off the walls, they exert a force on the walls. All these small forces add up so that a large number of collisions produces a total average force on the walls that is measurable. The force per unit area is the pressure. Hence, the pressure of a gas is due to the collisions of gas molecules with the walls of the container.

The pressure of a gas depends on the frequency of collision with the container walls and the size and number of molecules in the gas.

Boyle’s Law

Boyle’s Law states that the pressure of a fixed amount of gas is inversely proportional to the volume of the gas when the temperature is held constant.

$$pV = \text{constant}$$

$$P_{1}V_{1} = P_{2}V_{2}$$

, where

$P_{1}$, $P_{2}$ are the initial and final pressures respectively,

$V_{1}$, $V_{2}$ are the initial and final volumes respectively.


The animation above shows the compression of air within a cylinder and the resulting graph of volume against pressure. The graph shows that when the volume decreases, the pressure increases (inverse relationship). However, if a graph of volume against $\frac{1}{P}$ is plotted, it will be a straight line, whereby the gradient of the straight line is the constant of proportionality, k.

$$\begin{aligned} P &\propto \frac{1}{V} \\ P &= \frac{k}{V} \\ PV &= k \end{aligned}$$

boyle law graph

An intuitive explanation for the pressure being inversely proportional to the volume can be obtained via kinetic model of gases. We know that pressure is due to the collision of gas molecules with the walls of a container.


  • When the volume of the container is decreased to half, the number of gas molecules per unit volume will be doubled.
  • Since the number of gas molecules per unit volume is doubled, the number of collision of the gas molecules with the walls will also doubled.
  • Hence, the pressure will be doubled.

Gas pressure increases when:

  • number of molecules in the container increases
  • speed of molecules increases
  • molecules have larger mass

Relationship between Pressure, Volume & Temperature

The table below applies for a constant mass of gas.

IncreaseConstantIncrease$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

P is directly proportional to T
ConstantIncreaseIncrease$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

V is directly proportional to T
IncreaseDecreaseConstant$P_{1}V_{1} = P_{2}V_{2}$

P is inversely proportional to T

Explanation For The Relationship Between Pressure, Volume & Temperature

P is directly proportional to T (Constant V)

When you heat a container containing gas, the gas particles will gain kinetic energy and move faster. When the particles move faster, they will hit the walls of the container more often and more force will be exerted on the walls per unit area. This increases the pressure.

V is directly proportional to T (Constant P)

When a gas is heated, the temperature of the gas will rises which causes the molecules to move at higher speeds. When the particles move faster, they will hit the walls of the container more often. In order to keep the pressure constant, the gas will expand and occupy more volume.

P is inversely proportional to T (Constant V)

Explanation can be found in the above section.

Worked Examples

Example 1: Compressing Air With A Piston

Air at a pressure of $1.0 \times 10^{5} \text{ Pa}$ is confined in a cylinder that is fitted with a movable piston. The air is then compressed by pushing the piston, so that the same mass of air now occupies one-fifth of the original volume without any change in temperature. Calculate the new pressure of the air.

Click to show/hide answer

Let the initial volume be $V_{1}$ and the final volume be $V_{2}$, where $V_{2} = \frac{1}{5} V_{1}$.

Since $P_{1}V_{1} = P_{2}V_{2}$,

$$\begin{aligned} P_{2} &= \frac{P_{1}V_{1}}{V_{2}} \\ &= \frac{P_{1}V_{1}}{\frac{1}{5}V_{1}} \\ &= 5 P_{1} \\ &= 5 \times 1.0 \times 10^{5} \\ &= 5.0 \times 10^{5} \text{ Pa} \end{aligned}$$

Example 2

A specific amount of gas occupies a volume of 40 cm³ at a pressure of $1 \times 10^5 \text{ Pa}$. Determine its volume when the pressure is $2 \times 10^5 \text{ Pa}$, assuming constant temperature.

Click to show/hide answer

Using the equation $pV = \text{constant}$, we have

$$p_1V_1 = p_2V_2$$

Rearranging the equation gives:

$$\begin{aligned} V_2 &= \frac{p_1 \times V_1}{p_2} \\ &= \frac{1 \times 10^5 \, \text{Pa} \times 40 \, \text{cm}^3}{2 \times 10^5 \, \text{Pa}} \\ &= 20 \, \text{cm}^3 \end{aligned}$$

Example 3

How does the rise in temperature impact the kinetic energy of particles in a substance? Containers A and B contain an equal number and type of particles at an identical temperature. Despite the similar particle composition, container B has a greater volume than container A. Clarify whether the two containers exhibit the same gas pressure.

Click to show/hide answer

As temperature increases, the average kinetic energy of particles within the substance rises. However, the gas pressure in container B is not equivalent to that in A. Due to its larger volume, container B experiences a decrease in the number of particles per unit volume. Consequently, the collisions between gas particles and the inner container walls occur less frequently, resulting in a reduced average force exerted per unit area. Therefore, the gas pressure in container B is lower than in container A.

Back To Pressure (O Level Physics)

Back To O Level Physics Topic List

Mini Physics

As the Administrator of Mini Physics, I possess a BSc. (Hons) in Physics. I am committed to ensuring the accuracy and quality of the content on this site. If you encounter any inaccuracies or have suggestions for enhancements, I encourage you to contact us. Your support and feedback are invaluable to us. If you appreciate the resources available on this site, kindly consider recommending Mini Physics to your friends. Together, we can foster a community passionate about Physics and continuous learning.

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.