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.

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

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

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. Hence,

- 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.

#### Worked 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.

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

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

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

James Odhiambo Odundo scribbled

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