The Gas Law for Air

In chemistry you used the combined gas law, $P_1V_1/T_1 = P_2V_2/T_2$, for a fixed amount of gas in a container. The atmosphere has no container, so meteorologists write the same law with density (mass per volume) in place of volume. It ties together pressure, temperature and density, three things a weather balloon measures, and it explains why warm air rises.

Key idea

For air, $$P = \rho R T$$ with $P$ in pascals, density $\rho$ in kg/m³, temperature $T$ in kelvins and $R = 287$ J/(kg·K) for dry air. At the same pressure, warmer air is less dense than colder air.

Why it works

Start from the ideal gas law, $PV = nR^*T$, where $R^* = 8.314$ J/(mol·K). The number of moles is the mass divided by the mass of one mole: $n = m/M$, and a mole of dry air has a mass of $M = 0.02897$ kg. So $P = \dfrac{m}{V}\cdot\dfrac{R^*}{M}\,T = \rho\,RT$, with $R = 8.314 \div 0.02897 \approx 287$ J/(kg·K). The temperature must be absolute because pressure comes from molecules hitting surfaces, and their energy of motion is proportional to the kelvin temperature: at 0 K they would not move at all.

Worked example: Air at sea level

Find the density of air at 101,325 Pa and 15 °C.

Convert to kelvins (temperature scales): $T = 15 + 273.15 = 288.15$ K.
Solve $P = \rho R T$ for density: $\rho = \dfrac{P}{RT} = \dfrac{101{,}325}{287 \times 288.15}$.
$\rho \approx 1.225$ kg/m³: a cubic meter of air has a mass of about 1.2 kg.

Answer: About 1.23 kg/m³

Worked example: Warm air and cold air

At 1000 hPa (100,000 Pa), compare the density of air at 35 °C with air at −5 °C.

Warm air: $T = 308.15$ K, so $\rho = 100{,}000 \div (287 \times 308.15) \approx 1.131$ kg/m³.
Cold air: $T = 268.15$ K, so $\rho = 100{,}000 \div (287 \times 268.15) \approx 1.299$ kg/m³.
At the same pressure, $\rho$ is proportional to $1/T$, so the ratio is $308.15/268.15 \approx 1.15$: the cold air is about 15% denser.

Answer: Warm air 1.13 kg/m³, cold air 1.30 kg/m³

Worked example: Finding the pressure

A balloon measures air at −20 °C with a density of 0.70 kg/m³. What is the pressure?

$T = -20 + 273.15 = 253.15$ K.
$P = \rho R T = 0.70 \times 287 \times 253.15 \approx 50{,}860$ Pa.
That is about 509 hPa, so the balloon is near 5.5 km, where pressure has halved.

Answer: About 509 hPa

Common mistake

Putting Celsius into the gas law. With $T = -5$ in the formula, a cold day would give air a negative density, and 0 °C would divide by zero. Always add 273.15 first. Check: for air near the ground, $T$ should be roughly 230 to 320 K, and the density about 1 to 1.4 kg/m³.

Move the probe and read the air density tile. At 5.5 km the density is about 0.70 kg/m³, a bit more than half of the ground's, because the air there is also colder.

This is why warm air rises. A bubble of air warmer than its surroundings, at the same pressure, is less dense, so the denser air around it pushes it up, the way water pushes up a cork. Heat a pocket of air by 3 °C and it becomes about 1% less dense: enough to float upward. Try it in Gas in a Box: heat the box and watch the pressure climb, then let it expand and watch the gas spread out.

The Gas Law for Air

Practice

Builds on

Leads to

Sources and further reading