Gas Laws: Pressure, Volume and Temperature

For a fixed mass of ideal gas, pressure, volume and temperature are linked. The three gas laws each hold one variable constant while the other two change.

Boyle’s law (constant temperature)

$p_1V_1 = p_2V_2$

At constant temperature, pressure is inversely proportional to volume. Compressing a gas (reducing volume) increases the pressure.

Particle explanation: Reducing volume means particles hit the walls more often per second, increasing pressure.

Pressure-temperature law (constant volume)

$\frac{p_1}{T_1} = \frac{p_2}{T_2}$

At constant volume, pressure is directly proportional to absolute temperature. Heating a gas in a sealed container increases the pressure.

Particle explanation: Higher temperature means particles move faster and hit the walls harder and more often, increasing pressure.

Charles’s law (constant pressure)

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

At constant pressure, volume is directly proportional to absolute temperature. Heating a gas at constant pressure causes it to expand.

Important: use kelvin

All gas law equations require temperature in kelvin. Using Celsius gives incorrect results.

Worked example: Boyle’s law

A gas occupies 80 cm$^3$ at 120 kPa. The temperature stays constant while the pressure increases to 200 kPa. Find the new volume.

$V_2 = \frac{p_1V_1}{p_2} = \frac{120 \times 80}{200} = 48\text{ cm}^3$

No unit conversion was needed because both pressures are in kPa and volumes in cm$^3$.

Worked example: pressure-temperature

A sealed container of gas is at 300 K and 100 kPa. It is heated to 450 K. Find the new pressure.

$p_2 = \frac{p_1 T_2}{T_1} = \frac{100 \times 450}{300} = 150\text{ kPa}$

Common errors and how to correct them

Using Celsius in the equation. Always convert to kelvin first. A common error: 0 $^\circ$C treated as “zero temperature” makes the entire calculation invalid.

Applying Boyle’s law when temperature changes. Boyle’s law requires constant temperature. If temperature also changes, use the combined gas equation or the appropriate single law.

How to apply this in an exam

State which quantity is constant, write the correct equation, convert temperature to kelvin, substitute, and solve. When paired units match (both in kPa, both in cm$^3$), you do not need to convert to SI.