Gases and the absolute scale of temperature

Cambridge·Physics·2.1.3·Extended

Gas pressure and temperature at constant volume

Gas particles move randomly and collide with the walls of their container. Each collision exerts a force on the wall. Gas pressure is the force exerted by these collisions per unit area of the wall.

For a fixed mass of gas at constant volume, increasing the temperature increases the particles’ average kinetic energy. The particles move faster, so they collide with the walls more frequently and exert a greater force when they collide. The force per unit area therefore increases, so the pressure rises.

Decreasing the temperature has the opposite effect. The particles’ average kinetic energy decreases, so they move more slowly. They collide with the walls less frequently and exert less force during collisions. The pressure therefore falls.

Gas pressure and volume at constant temperature

Gas particles move continually and collide with the walls of their container. Each collision exerts a force on the wall. The combined force per unit area of the wall is the gas pressure.

For a fixed mass of gas, decreasing the volume confines the same particles to a smaller space. The particles collide with each unit area of the walls more frequently, so the force per unit area increases. The pressure therefore rises.

Increasing the volume gives the same particles more space. They collide with each unit area of the walls less frequently, so the pressure falls.

When the temperature remains constant, the particles’ average kinetic energy remains unchanged. The change in pressure is therefore explained by the change in collision frequency per unit area, not by a change in the particles’ average kinetic energy.

Kelvin and Celsius temperature conversion

The kelvin is the unit of the absolute temperature scale. Its zero point is absolute zero, the lowest possible temperature. Absolute zero is approximately $$-273,{}^{\circ}\text{C}$$.

A temperature interval of $$1,\text{K}$$ is the same size as an interval of $$1,{}^{\circ}\text{C}$$. The scales give different numbers for the same temperature because their zero points are different.

Converting between kelvin and Celsius

To convert a temperature from degrees Celsius to kelvin, add 273:

T (in K)=θ (in ∘C)+273T\text{ (in K)} = \theta\text{ (in }{}^{\circ}\text{C)} + 273

Here, $$T$$ is the temperature in kelvin and $$\theta$$ is the temperature in degrees Celsius. For example, $$20,{}^{\circ}\text{C}$$ is approximately $$293,\text{K}$$.

To convert from kelvin to degrees Celsius, subtract 273:

θ (in ∘C)=T (in K)−273\theta\text{ (in }{}^{\circ}\text{C)} = T\text{ (in K)} - 273

For example, $$250,\text{K}$$ is approximately $$-23,{}^{\circ}\text{C}$$. The conversions are approximate because the difference between the scale zero points is rounded to 273.

Boyle's law

When a fixed mass of gas is compressed while its temperature stays constant, its pressure increases. If the gas expands under the same conditions, its pressure decreases.

Boyle’s law states that pressure is inversely proportional to volume. This means that multiplying the volume by a given factor divides the pressure by the same factor. For example, doubling the volume halves the pressure.

The relationship can be written as:

pV=constantpV = \text{constant}

Here, $$p$$ is the absolute pressure of the gas and $$V$$ is its volume. The product stays constant only when the mass of gas and its temperature remain constant.

Boyle’s law: Comparing two states

If the same gas changes from pressure $$p_1$$ and volume $$V_1$$ to pressure $$p_2$$ and volume $$V_2$$, the two products are equal:

p1V1=p2V2p_1V_1 = p_2V_2

This relationship can be rearranged to find an unknown pressure or volume. Use the same pressure unit for both pressures and the same volume unit for both volumes.

For example, a gas at $$200\ \text{kPa}$$ occupies $$3.0\ \text{L}$$. It expands to $$5.0\ \text{L}$$ at constant temperature. Its final pressure is:

p2=p1V1V2=200 kPa×3.0 L5.0 L=120 kPap_2 = \frac{p_1V_1}{V_2} = \frac{200\ \text{kPa} \times 3.0\ \text{L}}{5.0\ \text{L}} = 120\ \text{kPa}

The pressure decreases as the volume increases, as Boyle’s law predicts.

Boyle’s law: Pressure–volume graph

A graph of pressure against volume is a decreasing curve, not a straight line. As volume increases, pressure falls: for example, at twice a given volume, the pressure is half its original value.

Gases and the absolute scale of temperature | Cambridge IGCSE (9–1) Physics 0972 (Extended) (2026–2028) Revision Notes | KnowledgeSet