Gas mixtures, partial pressures and diffusion

Each gas in a mixture

Air is about $78\%$ nitrogen and $21\%$ oxygen by volume, with small amounts of argon and carbon dioxide. When you pump up a tyre, all of these gases push on the walls. Each gas pushes as if the others were not there. Gas molecules are so far apart that, for an ideal gas, a molecule does not notice another kind.

John Dalton stated this in 1801 as the law of partial pressures. The total pressure of a mixture of gases that do not react is the sum of the partial pressures:

The partial pressure of a gas is the pressure that it would have if it alone filled the vessel. For gas $i$, $P_i = \dfrac{n_iRT}{V}$. Add up all the gases, and $P_{\text{total}} = \dfrac{n_{\text{total}}RT}{V}$.

Divide one by the other. $\dfrac{P_i}{P_{\text{total}}} = \dfrac{n_i}{n_{\text{total}}} = x_i$. The mole fraction $x_i$ is the amount of gas $i$ divided by the total amount. So

The mole fractions add up to $1$. At the same $T$ and $P$, the volume of each gas is also proportional to $n$, so the mole fraction equals the volume fraction. That is why the composition of air is given as percentages by volume.

A worked example

A $10.0\ \text{L}$ tank at $300\ \text{K}$ holds $1.20\ \text{mol}$ of $\text{N}_2$, $0.30\ \text{mol}$ of $\text{O}_2$ and $0.10\ \text{mol}$ of Ar.

1. Total amount. $1.20 + 0.30 + 0.10 = 1.60\ \text{mol}$.
2. Total pressure. $P = \dfrac{nRT}{V} = \dfrac{1.60 \times 0.08314 \times 300}{10.0} = 3.99\ \text{bar}$.
3. Mole fractions. $x_{\text{N}_2} = 0.7500$, $x_{\text{O}_2} = 0.1875$, $x_{\text{Ar}} = 0.0625$.
4. Partial pressures. $P_{\text{N}_2} = 2.99\ \text{bar}$, $P_{\text{O}_2} = 0.748\ \text{bar}$ and $P_{\text{Ar}} = 0.249\ \text{bar}$. They add up to $3.99\ \text{bar}$.

Notice that nitrogen has three quarters of the molecules, and so it has three quarters of the pressure.

A gas collected over water

A gas that does not dissolve much in water can be collected by pushing the water out of an upside-down jar. The gas in the jar is wet. Water evaporates into the jar until its vapour pressure is reached. So the gas in the jar is a mixture of the gas and water vapour:

We raise the jar until the water level inside matches the level outside. Then the pressure in the jar equals the atmospheric pressure. The vapour pressure of water, sometimes called the aqueous tension, depends only on the temperature:

Example. Zinc reacts with acid: $\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$. The hydrogen is collected over water at $25\ ^\circ\text{C}$. The jar holds $150\ \text{mL}$ at a total pressure of $100.0\ \text{kPa}$.

1. Hydrogen alone. $P_{\text{H}_2} = 100.0 - 3.17 = 96.83\ \text{kPa}$.
2. Amount of hydrogen. We want $n$, so $n = \dfrac{PV}{RT}$. With $R = 8.314\ \text{kPa L K}^{-1}\ \text{mol}^{-1}$ and $T = 298.15\ \text{K}$, we get $n = 5.859 \times 10^{-3}\ \text{mol}$.
3. Zinc used. One mole of zinc gives one mole of hydrogen. The mass is $5.859 \times 10^{-3}\ \text{mol} \times 65.38\ \text{g/mol} = 0.383\ \text{g}$.

If you forget the water vapour and use $100.0\ \text{kPa}$, you get a mass $3\%$ too large.

Diffusion and effusion

Diffusion is the mixing of gases by random motion. A drop of perfume in one corner of a room is smelled in the other corner after a while. Effusion is the escape of a gas through a very small hole into a vacuum, for example air leaking out of a balloon. In effusion the molecules pass through one at a time, without bumping into each other.

Thomas Graham measured the rates of both in the early nineteenth century. At the same temperature and pressure, the rate is inversely proportional to the square root of the molar mass:

This is Graham's law. Hydrogen, $M = 2.016\ \text{g/mol}$, escapes $\sqrt{32.00/2.016} = 3.98$ times faster than oxygen, $M = 32.00\ \text{g/mol}$. The reason is that all gases have the same average kinetic energy at the same temperature. A light molecule must move faster to have the same energy. We prove this in the next course.

If the same amount of each gas passes through the hole, the time is inversely proportional to the rate, so $\dfrac{t_1}{t_2} = \sqrt{\dfrac{M_1}{M_2}}$. A heavier gas takes longer.

At different pressures, the rate of effusion is also proportional to the pressure: $r \propto \dfrac{P}{\sqrt{M}}$. A higher pressure puts more molecules at the hole.

Graham's law is exact for effusion into a vacuum. For diffusion in air it is only a guide, because the molecules collide with air molecules all the time and the spreading is much slower.

The ammonia and hydrogen chloride tube. Put cotton wool with ammonia at one end of a glass tube $100\ \text{cm}$ long and cotton wool with hydrochloric acid at the other end. Ammonia, $M = 17.03\ \text{g/mol}$, and hydrogen chloride, $M = 36.46\ \text{g/mol}$, travel towards each other. When they meet, they form white ammonium chloride, $\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl}$. The rate ratio is $\sqrt{36.46/17.03} = 1.463$. So the distances travelled have the same ratio. Ammonia covers $\dfrac{1.463}{1 + 1.463} = 59.4\%$ of the length. The ring forms $59.4\ \text{cm}$ from the ammonia end, nearer the hydrochloric acid.

Separating uranium isotopes. Uranium hexafluoride gas contains $^{235}\text{UF}_6$ with $M = 349.03\ \text{g/mol}$ and $^{238}\text{UF}_6$ with $M = 352.04\ \text{g/mol}$. The lighter one effuses faster by a factor of $\sqrt{352.04/349.03} = 1.0043$. That is only $0.43\%$ more in one step. So thousands of steps in a row are needed to enrich the uranium.

Put gas A at $2\ \text{g/mol}$ and gas B at $32\ \text{g/mol}$. The lighter gas covers most of the tube.

Practise with new numbers

Try it in Python

The program finds the total pressure and the partial pressure of each gas in the tank of the worked example. It ends with the Graham's law ratio for hydrogen and oxygen.

Common mistakes

• Giving every gas the total pressure. Each gas has its own partial pressure, and they add up to the total.
• Forgetting the water vapour. A gas collected over water is wet. Subtract the vapour pressure of water from the total.
• Turning Graham's law upside down. The lighter gas is the faster gas.
• Using the ratio of masses without the square root. The rate depends on the square root of the molar mass.
• Using Graham's law at different temperatures. The law is for the same temperature and pressure. A hotter gas moves faster.
• Adding partial pressures of gases that react. Dalton's law is for gases that do not react.

Remember these

• $P_{\text{total}} = P_1 + P_2 + \cdots$, and $P_i = x_i\,P_{\text{total}}$, where $x_i = n_i/n_{\text{total}}$.
• Gas over water: $P_{\text{gas}} = P_{\text{atm}} - P_{\text{water}}$.
• Graham's law: $\dfrac{r_1}{r_2} = \sqrt{\dfrac{M_2}{M_1}}$ at the same $T$ and $P$.
• For the same amount of gas, the time goes as $\sqrt{M}$.
• Effusion rate goes as $P/\sqrt{M}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A cylinder holds $2.0\ \text{mol}$ of helium and $3.0\ \text{mol}$ of neon at a total pressure of $4.5\ \text{bar}$. What is the partial pressure of helium?

2. Hydrogen is collected over water at $20\ ^\circ\text{C}$ when the air pressure is $99.0\ \text{kPa}$. What is the pressure of the dry hydrogen?

3. Methane, $\text{CH}_4$ with $M = 16.04\ \text{g/mol}$, and sulfur dioxide, $\text{SO}_2$ with $M = 64.07\ \text{g/mol}$, escape through a pinhole. Which is faster, and by what factor?

4. Why does the ring of ammonium chloride form nearer to the hydrochloric acid end of the tube?

5. Is the total pressure of a mixture of an ideal gas A and an ideal gas B larger than the pressure of A alone in the same vessel? Why?

References

• OpenStax, Chemistry 2e, Chapter 9, sections 9.3 (partial pressures, gas collected over water) and 9.4 (effusion and diffusion).
• The vapour pressures of water are standard values from the CRC Handbook of Chemistry and Physics.