Ideal Gases and the Relationship Between Pressure and Volume

The Particle Model of an Ideal Gas

According to the particle model, a gas is composed of identical particles in continuous random motion. In this model, particles do not exert any forces on each other except during collisions. Several specific assumptions define an ideal gas: the particles occupy a negligible volume compared to the container, they undergo elastic collisions, and they obey Newton's laws of motion. While theoretical, many real gases behave like ideal gases when they are at conditions near room temperature and atmospheric pressure.

Explaining Temperature and Pressure Through Particles

Temperature is a macroscopic property that describes the 'hotness' of a gas, which can be measured with a thermometer. On a microscopic level, temperature is directly related to the movement of the gas particles. The higher the temperature of the gas, the higher the average speed of its particles. It is important to note that temperature is a property of the gas as a whole; it is incorrect to describe an individual particle as being hot or cold. Instead, we describe the behaviour of individual particles in terms of their average speed.

Pressure is also a macroscopic property. A gas exerts a force on any surface it touches because of the random motion of its particles. When these particles collide with the walls of a container, each collision exerts a tiny force. The sum of these individual microscopic collisions results in a steady average force per unit area, which we define as pressure. Like temperature, pressure is not a property of a single particle but of the entire gas sample.

Example: Increasing Temperature in a Sealed Container

Consider a sample of gas held in a sealed container. If the temperature of the gas is increased, the pressure it exerts on the container walls also increases. This happens for two reasons:

1. The particles move faster on average, meaning they collide with the container walls more frequently.
2. Because they are moving faster, the collisions with the walls are more forceful on average.

Both effects combine to increase the average force per unit area, resulting in higher pressure.

The Relationship Between Pressure and Volume

When the volume of a sealed container of gas is decreased (for example, by pushing in a plunger) while keeping the temperature constant, the gas pressure increases. This is because the average speed of the particles remains the same due to the constant temperature, but the particles have less distance to travel between the walls. Consequently, the frequency of collisions between the gas particles and the container walls increases. Since the average force of each collision is unchanged but the collisions happen more often, the pressure rises.

Conversely, if the volume is increased at a constant temperature, the gas pressure decreases. For a fixed amount of ideal gas at a constant temperature, the relationship is:

$PV = \text{constant}$

In practical scenarios, compressing a gas often causes its temperature to rise. However, if the gas is allowed to return to its original temperature: such as by reaching thermal equilibrium with its surroundings: the relationship $PV = \text{constant}$ will hold true for the initial and final states.

Applying Boyle's Law

For a fixed mass of gas at a constant temperature, if the initial pressure and volume are $P_1$ and $V_1$, and the final pressure and volume are $P_2$ and $V_2$, we can state:

$P_1 V_1 = P_2 V_2$

This shows that pressure is inversely proportional to volume, which can be written as:

$P = \frac{\text{constant}}{V}$ or $P \propto \frac{1}{V}$

When plotted on a graph, the relationship between $P$ and $V$ for a fixed mass of gas at a constant temperature forms an inverse proportion curve.

Worked Example: Calculating Changes in Pressure

Question: A sample of gas in a sealed container has a pressure of $1.0 \times 10^5$ Pa and a volume of 60 cm³. The volume is reduced to 40 cm³ without any gas escaping. What is the new pressure of the gas?

Step 1: Calculate the constant using initial values. Initial $PV = 1.0 \times 10^5 \text{ Pa} \times 60 \text{ cm}^3 = 6.0 \times 10^6 \text{ Pa cm}^3$.

Step 2: Use the constant to find the new pressure. Since $PV$ remains constant: $P \times 40 \text{ cm}^3 = 6.0 \times 10^6 \text{ Pa cm}^3$. $P = \frac{6.0 \times 10^6 \text{ Pa cm}^3}{40 \text{ cm}^3} = 1.5 \times 10^5 \text{ Pa}$.

Alternative Method: Using $P_1 V_1 = P_2 V_2$ directly: $P_2 = \frac{P_1 V_1}{V_2}$ $P_2 = \frac{1.0 \times 10^5 \text{ Pa} \times 60 \text{ cm}^3}{40 \text{ cm}^3}$ $P_2 = 1.5 \times 10^5 \text{ Pa}$.