A cyclist pumps her tyre at dawn in Jaipur, when the air is cool. She rides to college on a road that bakes in the sun all morning. By afternoon the tyre feels hard as stone, and she releases a little air. No air went in after dawn, and the tyre hardly grew. Something about the heat alone raised the pressure inside.
Four numbers describe a gas
A sample of gas in a closed vessel is described by four numbers. Its pressure \(p\) is the force it exerts on each square metre of wall, in pascal (Pa). Its volume \(V\) is the space it fills, in cubic metres. Its temperature \(T\) is measured in kelvin, and its amount is the number of moles \(n\). This chapter uses \(1\,\text{atm} = 1.013 \times 10^5\,\text{Pa}\) and \(1\,\text{L} = 10^{-3}\,\text{m}^3\).
Temperature in these laws is always the absolute temperature, \(T = t + 273\), where \(t\) is the Celsius reading. The chapter on temperature and thermal expansion introduced this scale. The reason for it will appear in a moment: the gas laws themselves point to the zero of the kelvin scale.
Boyle's law: squeeze a gas slowly
Hold the temperature fixed and squeeze a gas into half its volume. Its pressure doubles. Squeeze it into a third, and the pressure triples. The pressure and the volume change in opposite ratios, so their product stays the same. Robert Boyle found this in 1662.
For a fixed amount of gas at a constant temperature, the pressure is inversely proportional to the volume: \(pV = \text{constant}\), or \(p_1V_1 = p_2V_2\).
In Fig 1.1 the gas starts at \(1\,\text{atm}\) in \(3\,\text{L}\). At \(1.5\,\text{L}\) the pressure is \(2\,\text{atm}\), and at \(1\,\text{L}\) it is \(3\,\text{atm}\). The curve is a hyperbola, and it is called an isotherm, a line of one temperature. The squeeze must be slow. A fast squeeze warms the gas, and then the temperature is no longer fixed.
An air bubble of volume \(2\,\text{cm}^3\) is released at the bottom of a lake, where the pressure is \(3\,\text{atm}\). It rises to the surface, where the pressure is \(1\,\text{atm}\). The water temperature is the same everywhere. What is its volume at the surface, in \(\text{cm}^3\)?
Charles's law: heat a gas at one pressure
Now keep the pressure fixed and heat the gas. A balloon does this by itself: its rubber keeps the pressure inside close to the air outside. Warm a balloon from \(27\,^\circ\text{C}\) to \(52\,^\circ\text{C}\) and its volume grows from \(2.4\,\text{L}\) to \(2.6\,\text{L}\). The volume grows in the same ratio as the absolute temperature, \(325/300\). Jacques Charles found this in 1787.
For a fixed amount of gas at a constant pressure, the volume is proportional to the absolute temperature: \(V/T = \text{constant}\), or \(V_1/T_1 = V_2/T_2\).
A gas under a free, frictionless piston that carries a fixed load is also held at one pressure. The piston settles where the gas pressure balances the load and the air above it. So \(p = p_0 + mg/A\), where \(p_0\) is the air pressure, \(m\) the mass of the piston and its load, and \(A\) its area. Heating such a gas lifts the piston, and Charles's law gives the new volume.
Fig 1.2 holds the idea of absolute zero. Every gas, at every pressure, gives a straight line of \(V\) against \(t\). Every line points to zero volume at \(-273\,^\circ\text{C}\). No real gas reaches that point, because it turns liquid first. Still, the common point is the natural zero of temperature. Measured from it, \(V\) is simply proportional to \(T\). The line for \(2p_1\) lies lower, because by Boyle's law twice the pressure means half the volume.
A gas at constant pressure is warmed from \(20\,^\circ\text{C}\) to \(40\,^\circ\text{C}\). What happens to its volume?
The pressure law: heat a gas in a rigid vessel
Last, keep the volume fixed. A pressure cooker whose whistle stays closed is such a vessel. Heat the air inside and its pressure rises in the ratio of the absolute temperature. Air sealed at \(300\,\text{K}\) and heated to \(393\,\text{K}\), which is \(120\,^\circ\text{C}\), presses \(393/300 = 1.31\) times as hard. This is the pressure law of Gay-Lussac.
For a fixed amount of gas in a constant volume, the pressure is proportional to the absolute temperature: \(p/T = \text{constant}\), or \(p_1/T_1 = p_2/T_2\).
Now return to the cyclist. Her tyre barely changes volume, so the pressure law applies. Say it held \(3.0\,\text{atm}\) of air (the absolute pressure) at \(27\,^\circ\text{C}\) at dawn, and the hot road warmed it to \(57\,^\circ\text{C}\). Then \(p_2 = 3.0 \times 330/300 = 3.3\,\text{atm}\). The tyre's volume never entered the sum. Only the ratio of the temperatures in kelvin did.
A weather balloon holds \(2.0\,\text{m}^3\) of helium at \(1.0\,\text{atm}\) and \(300\,\text{K}\) on the ground. It rises to a height where the pressure is \(0.50\,\text{atm}\) and the temperature is \(250\,\text{K}\). Its thin skin keeps the pressure inside equal to the air outside. Find its volume there.
State 1: \(p_1 = 1.0\,\text{atm}\), \(V_1 = 2.0\,\text{m}^3\), \(T_1 = 300\,\text{K}\). State 2: \(p_2 = 0.50\,\text{atm}\), \(T_2 = 250\,\text{K}\), \(V_2\) unknown.
\(V_2 = V_1 \times \dfrac{p_1}{p_2} \times \dfrac{T_2}{T_1}\).
\(V_2 = 2.0 \times \dfrac{1.0}{0.50} \times \dfrac{250}{300}\), which is \(2.0 \times 2 \times 0.833 = 3.3\,\text{m}^3\).
The balloon swells to about \(3.3\,\text{m}^3\). Higher still it swells until the skin bursts, which is why such balloons are only partly filled at launch.
Convert every temperature to kelvin before it enters a gas law. Using Celsius gives nonsense: heating from \(27\,^\circ\text{C}\) to \(127\,^\circ\text{C}\) would seem to multiply the volume by nearly five. In fact it multiplies it by \(400/300 = 1.33\).
Avogadro's law: counting molecules
The three laws hold for a fixed amount of gas. What happens when the amount changes? Blow into a balloon and it grows, at the same pressure and temperature. In 1811 Amedeo Avogadro made a bold guess about why. Equal volumes of different gases, at the same temperature and pressure, hold the same number of molecules. A litre of hydrogen and a litre of carbon dioxide side by side contain equal counts, although the carbon dioxide is 22 times heavier.
At the same temperature and pressure, equal volumes of all gases contain equal numbers of molecules. So at fixed \(p\) and \(T\), the volume of a gas is proportional to the number of molecules in it.
Chemists count molecules in moles. One mole is \(N_A = 6.02 \times 10^{23}\) particles, the Avogadro constant. The mass of one mole is the molar mass \(M\). It is \(2\,\text{g/mol}\) for hydrogen, \(4\) for helium, \(28\) for nitrogen, \(32\) for oxygen and about \(29\) for air. A sample of mass \(m\) holds \(n = m/M\) moles.
By Avogadro's law, one mole of any gas fills the same volume at the same \(p\) and \(T\). At STP, which is \(0\,^\circ\text{C}\) and \(1\,\text{atm}\), that volume is \(22.4\,\text{L}\). So \(11.2\,\text{L}\) of nitrogen at STP is half a mole: \(3.01 \times 10^{23}\) molecules with a mass of \(14\,\text{g}\).
What is the density of oxygen (\(M = 32\,\text{g/mol}\)) at STP, in g/L?
One equation for every gas
Put the laws together. Volume is proportional to \(1/p\) (Boyle), to \(T\) (Charles) and to \(n\) (Avogadro). So \(V\) is proportional to \(nT/p\), and \(pV/nT\) is a constant. Avogadro's law makes this constant the same for every gas. It is the gas constant \(R\).
A gas that obeys \(pV = nRT\) exactly is an ideal gas. The value \(R = 8.31\,\text{J/(mol K)}\) follows from STP. Put \(p = 1.013 \times 10^5\,\text{Pa}\), \(V = 0.0224\,\text{m}^3\), \(n = 1\) and \(T = 273\,\text{K}\) into \(R = pV/nT\). Two other forms are useful. Counting molecules, \(n = N/N_A\). So \(pV = N(R/N_A)T = NkT\). The gas constant per molecule is the Boltzmann constant, \(k = R/N_A = 1.38 \times 10^{-23}\,\text{J/K}\).
The third form uses density. A mass \(m\) of gas holds \(n = m/M\) moles, so \(pV = (m/M)RT\). Dividing by \(V\) gives \(p = \rho RT/M\). For air at \(300\,\text{K}\) and \(1\,\text{atm}\), \(\rho = pM/RT\), which is \(1.013 \times 10^5 \times 0.029/(8.31 \times 300)\), or \(1.18\,\text{kg/m}^3\). The same form gives the number density, \(N/V = p/kT = 2.45 \times 10^{25}\) molecules in each cubic metre.
The ideal gas equation also explains the straight-line graphs of the gas laws. In a rigid vessel, \(p = (nR/V)\,T\). A graph of \(p\) against \(T\) is then a straight line through the origin with slope \(nR/V\). The same gas in a vessel twice as large gives a line half as steep. In the same way, \(V = (nR/p)\,T\) at a fixed pressure.
A classroom is \(10\,\text{m}\) long, \(6\,\text{m}\) wide and \(3\,\text{m}\) high. The air in it is at \(300\,\text{K}\) and \(1\,\text{atm}\). Take \(M = 29\,\text{g/mol}\) for air. Find the number of moles of air, the number of molecules and the mass of the air.
The volume is \(10 \times 6 \times 3 = 180\,\text{m}^3\). So \(n = pV/RT\), which is \(1.013 \times 10^5 \times 180/(8.31 \times 300)\), or \(7.31 \times 10^3\,\text{mol}\).
The number of molecules is \(N = nN_A = 7.31 \times 10^3 \times 6.02 \times 10^{23}\), which is \(4.40 \times 10^{27}\).
The mass is \(nM = 7.31 \times 10^3 \times 0.029 = 212\,\text{kg}\), about the mass of three students.
Estimate the mass of air in a bedroom \(4\,\text{m}\) by \(3\,\text{m}\) by \(3\,\text{m}\), in kilograms. Take the density of air as about \(1.2\,\text{kg/m}^3\).
A gas cylinder of volume \(16.62\,\text{L}\) holds gas at \(300\,\text{K}\) and \(3.0 \times 10^5\,\text{Pa}\). Take \(R = 8.31\,\text{J/(mol K)}\). How many moles of gas are in it?
A rigid gas cylinder at \(27\,^\circ\text{C}\) reads \(20\,\text{atm}\). Some gas leaks out, and the pressure falls to \(15\,\text{atm}\) at the same temperature. What fraction of the gas escaped?
Dalton's law: each gas presses as if alone
Air is a mixture: about \(78\,\%\) nitrogen, \(21\,\%\) oxygen and \(1\,\%\) argon by number of molecules. The molecules of an ideal gas do not pull on one another. So each gas in the mixture strikes the walls as if the others were not there. The pressure a gas would exert alone in the whole vessel is its partial pressure.
The total pressure of a mixture of gases that do not react with each other is the sum of their partial pressures: \(p = p_1 + p_2 + \dots\), where \(p_i = n_iRT/V\).
The law follows from the ideal gas equation. The total amount is \(n_1 + n_2\), so \(pV = (n_1 + n_2)RT = n_1RT + n_2RT\). Dividing by \(V\) gives \(p = p_1 + p_2\). For air at \(1\,\text{atm}\), the oxygen supplies \(0.21 \times 101.3 = 21\,\text{kPa}\) of the pressure. This partial pressure decides how much oxygen each breath carries, a point that matters in Ladakh.
A vessel of volume \(8.31\,\text{L}\) at \(300\,\text{K}\) holds \(0.40\,\text{mol}\) of nitrogen and \(0.10\,\text{mol}\) of oxygen. Find the partial pressure of each gas and the total pressure.
For each mole, \(RT/V = 8.31 \times 300/(8.31 \times 10^{-3})\), which is \(3.0 \times 10^5\,\text{Pa}\).
Nitrogen: \(p_1 = 0.40 \times 3.0 \times 10^5 = 1.2 \times 10^5\,\text{Pa}\). Oxygen: \(p_2 = 0.10 \times 3.0 \times 10^5 = 0.30 \times 10^5\,\text{Pa}\).
Total: \(p = 1.5 \times 10^5\,\text{Pa}\), the pressure of \(0.50\,\text{mol}\) alone in the vessel.
Vessel A (\(2\,\text{L}\)) holds a gas at \(3\,\text{atm}\). Vessel B (\(6\,\text{L}\)) holds the same gas at \(1\,\text{atm}\) and the same temperature. A tap between them is opened. What is the final pressure, in atm?
Robert Boyle trapped air in the short, closed arm of a J-shaped glass tube with mercury. Each time he poured more mercury into the long arm, the trapped air shrank. Twice the pressure of mercury gave half the volume. In 1662 he published the table and called the property "the spring of the air". His assistant Robert Hooke built the pumps, and Hooke later found the law of springs as well.
A pressure cooker traps steam and air under a heavy weight on its vent. As the cooker heats up, the gas inside is nearly a fixed volume, so its pressure rises with its temperature. When the pressure lifts the weight, steam escapes in a whistle and the pressure drops a little. The trapped pressure lets water boil well above \(100\,^\circ\text{C}\), so dal cooks in far less time.
- Using the Celsius temperature in a gas law.Always convert: \(T = t + 273\). Only the kelvin scale starts at absolute zero.
- Using gauge pressure, the reading above the atmosphere, in \(pV = nRT\).The gas laws need the absolute pressure: the gauge reading plus \(1\,\text{atm}\).
- Putting litres and grams into \(pV = nRT\) with \(R = 8.31\).With \(R = 8.31\,\text{J/(mol K)}\), use Pa, m³ and kg/mol. Convert \(1\,\text{L} = 10^{-3}\,\text{m}^3\) and \(32\,\text{g/mol} = 0.032\,\text{kg/mol}\).