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The ideal gas law

Heat and matter · level 5, lesson 6 of 6

A question to start

NASA gives Mars's surface pressure as 636 Pa and its average temperature as 214 K. From just those two numbers, and what the air is made of, can you work out how much a cubic metre of Mars air weighs?

The idea

Boyle's law, Charles's law and the pressure law are all parts of one law:

\[pV = nRT.\]

Here \(p\) is the pressure in Pa, \(V\) the volume in m³, \(T\) the temperature in K, \(n\) the amount of gas in moles, and \(R = 8.31\) J per mole per K, the same for every gas. One mole of a gas has a mass \(M\), its molar mass: 0.029 kg for Earth's air, 0.0435 kg for Mars's, which is mostly heavy carbon dioxide.

The mass of gas in volume \(V\) is \(nM\), so the density is

\[\rho = \frac{nM}{V} = \frac{pM}{RT}.\]
Working out the density of the air Mars Earth
Pressure \(p\) 636 Pa 101,400 Pa
Temperature \(T\) 214 K 288 K
Molar mass \(M\) 0.0435 kg/mol 0.0290 kg/mol
Density, \(pM/RT\) 0.0156 kg/m³ 1.23 kg/m³
NASA's measured value about 0.016 kg/m³ 1.217 kg/m³
Earth: 1.2 kgMars: 0.016 kg

The law, from three numbers, gets very close to what NASA measures.

Worked examples

1. Mars air. \(\rho = \dfrac{636 \times 0.0435}{8.31 \times 214} = \dfrac{27.7}{1778} = 0.0156\) kg/m³.

2. Earth air. \(\rho = \dfrac{101{,}400 \times 0.029}{8.31 \times 288} = \dfrac{2941}{2393} = 1.23\) kg/m³.

3. Air in a habitat. A habitat of 200 m³ at 70 kPa and 293 K holds \(n = pV/RT = 70{,}000 \times 200/(8.31 \times 293) = 5{,}750\) moles of Earth-like air, a mass of \(5{,}750 \times 0.029 \approx 167\) kg.

Try it

  1. How many moles of gas are in 1 m³ at 100 kPa and 300 K?
  2. Find the density of Mars air in a deep basin where the pressure is 1,200 Pa and the temperature 230 K.
  3. Why is Mars air at the same pressure and temperature denser than Earth air would be?
  4. How many times denser is Earth's air than Mars's at their surfaces?
Answers
  1. \(n = 100{,}000 \times 1/(8.31 \times 300) \approx 40\) moles.
  2. \(1200 \times 0.0435/(8.31 \times 230) = 0.027\) kg/m³.
  3. Carbon dioxide molecules are heavier: 0.044 kg per mole against 0.029 for Earth's air.
  4. \(1.217/0.016 \approx 76\) times.

Common mistake

Mixing units. \(pV = nRT\) needs pressure in Pa (not kPa), volume in m³ (not litres) and temperature in K. Mixing in kPa or °C gives answers wrong by factors of a thousand or more.

Where it leads