Real gases, critical constants and liquefaction

When the ideal gas equation fails

The ideal gas equation treats molecules as points with no forces between them. Neither is true. At a high pressure, molecules are close together. Then their size matters, and so do their attractions. At a low temperature, molecules move slowly and attract each other more effectively. Eventually the gas turns into a liquid, which an ideal gas can never do.

The compressibility factor

To measure how far a gas is from ideal, we use the compressibility factor of one mole:

For an ideal gas, $Z = 1$ at every pressure and temperature. We can also read $Z$ as the ratio $\dfrac{V_{\text{real}}}{V_{\text{ideal}}}$, the real molar volume divided by the ideal one at the same $P$ and $T$.

• $Z < 1$: the gas is more compressible than an ideal gas. Attractions are winning. They pull the molecules closer, so the volume is smaller.
• $Z > 1$: the gas is less compressible than an ideal gas. The size of the molecules is winning. They cannot be pushed into the space that the other molecules fill.

The picture shows $Z$ against pressure for five gases at $400\ \text{K}$. All of them start at $Z = 1$ at zero pressure, because at low pressure every gas is ideal. Carbon dioxide has strong attractions, so its $Z$ falls far below $1$ and comes back up only at several hundred bar. Hydrogen and helium have very weak attractions, so their $Z$ goes above $1$ at once. At a high pressure, every gas has $Z > 1$. The molecules are almost touching, and repulsion dominates. The curves of the picture are computed from the van der Waals equation. Measured curves have the same shapes.

The van der Waals equation

Johannes van der Waals made two corrections to $PV = nRT$ in 1873.

1. The volume. The molecules have a size, so a molecule can move only in the space that the others leave free. We subtract a volume $nb$ from $V$. The constant $b$ is the excluded volume of one mole of molecules. It is about four times the real volume of the molecules.

2. The pressure. A molecule near the wall is pulled back by the molecules behind it. It hits the wall less hard, so the pressure that we measure is less than the pressure that the gas would have without attractions. We add a correction to the measured pressure. Both the number of molecules that hit the wall and the pull on each one are proportional to $\dfrac{n}{V}$. So the correction is proportional to $\dfrac{n^2}{V^2}$, and we write $\dfrac{an^2}{V^2}$.

The result is the van der Waals equation:

The constant $a$ measures the attraction, and $b$ measures the size. They depend on the gas:

A large $a$ goes with strong attractions, and so with a gas that is easy to liquefy. Ammonia, which forms hydrogen bonds, has the largest $a$ in the table.

Other books give $a$ in $\text{L}^2\ \text{atm mol}^{-2}$ or in $\text{Pa m}^6\ \text{mol}^{-2}$. The conversion is $1\ \text{L}^2\ \text{bar mol}^{-2} = 0.1\ \text{Pa m}^6\ \text{mol}^{-2}$. Values in textbooks differ in the last digit, so use the numbers that a question gives you.

Example. One mole of carbon dioxide is in a $1.00\ \text{L}$ vessel at $300\ \text{K}$. For an ideal gas, $P = \dfrac{RT}{V} = 24.94\ \text{bar}$. The van der Waals equation gives

The attraction lowers the pressure by $3.64\ \text{bar}$, and the size of the molecules raises it by $1.11\ \text{bar}$. The result is $Z = \dfrac{22.41 \times 1.00}{0.08314 \times 300} = 0.899$.

At low pressure, $V$ is large. Then $V - b$ is almost $V$ and $\dfrac{a}{V^2}$ is small, and the van der Waals equation becomes the ideal gas equation. A good approximation at low pressure is $Z \approx 1 + \left(b - \dfrac{a}{RT}\right)\dfrac{P}{RT}$.

Set $a = 3.6$, $b = 0.043$ and $T = 300\ \text{K}$, and move the volume down to $0.2\ \text{L}$. The pressure of the real gas goes below the ideal pressure and then crosses it. Now set $a = 0$. What does the blue curve do?

The Boyle temperature

At low pressure, $Z \approx 1 + \left(b - \dfrac{a}{RT}\right)\dfrac{P}{RT}$. The bracket is zero at one temperature, the Boyle temperature:

At $T_B$, the gas follows Boyle's law, $PV$ = constant, over a wide range of low pressures. Below $T_B$ the attraction wins at low pressure and $Z$ dips below $1$. Above $T_B$ the size wins and $Z > 1$ at all pressures. For nitrogen $T_B = 426\ \text{K}$ and for carbon dioxide $T_B = 1026\ \text{K}$. For hydrogen $T_B = 112\ \text{K}$ and for helium $T_B = 17\ \text{K}$. At room temperature both are far above their Boyle temperatures, so $Z > 1$ for both.

Liquefaction and the critical temperature

In 1869 Thomas Andrews compressed carbon dioxide at different temperatures and measured the pressure. The picture shows the same experiment from the van der Waals equation.

• At $330\ \text{K}$ the pressure rises smoothly as the volume falls. The gas never liquefies.
• At $280\ \text{K}$, the curve of the van der Waals equation has an S-shaped loop. The real gas does not follow the loop. As the volume falls, the pressure rises until a liquid starts to form. Then the pressure stays constant, along the dashed line, while more and more gas turns into liquid. When all of it is a liquid, the pressure rises steeply again, because a liquid is hard to compress. The height of the dashed line is the vapour pressure at that temperature. It is placed so that the area of the loop above it equals the area below it. This is the equal area construction of Maxwell.
• At one temperature the flat part shrinks to a point. This is the critical point. The temperature is the critical temperature $T_c$, the pressure the critical pressure $P_c$ and the molar volume the critical volume $V_c$.

Above $T_c$ a gas cannot be liquefied by pressure alone, however high. Below $T_c$ it can be. A substance above both $T_c$ and $P_c$ is a supercritical fluid. It is not a gas and not a liquid. Supercritical carbon dioxide is used to remove caffeine from coffee.

For a van der Waals gas the critical point is where the isotherm has a flat point of inflection. Setting the first and second slopes to zero gives

From these, $Z_c = \dfrac{P_cV_c}{RT_c} = \dfrac{3}{8} = 0.375$ for every van der Waals gas. For carbon dioxide, the constants of the table give $T_c = 304.0\ \text{K}$, $P_c = 74.0\ \text{bar}$ and $V_c = 0.128\ \text{L/mol}$. The measured values are $304.2\ \text{K}$, $73.8\ \text{bar}$ and $0.094\ \text{L/mol}$. The equation is good for $T_c$ and $P_c$, and it overstates $V_c$. The measured $Z_c$ is about $0.27$ and not $0.375$.

We can go the other way, and find $a$ and $b$ from measured critical constants. From $T_c$ and $P_c$, $b = \dfrac{RT_c}{8P_c}$ and $a = \dfrac{27R^2T_c^2}{64P_c}$.

The critical temperatures decide how a gas is stored:

At room temperature, $298\ \text{K}$, ammonia and carbon dioxide are below $T_c$. Pressure alone liquefies them, and that is why ammonia and carbon dioxide are sold in steel cylinders as liquids. Nitrogen, oxygen, hydrogen and helium are above $T_c$. They must first be cooled below $T_c$, and in practice far below it. One cooling method uses the Joule-Thomson effect. A real gas that expands through a small hole cools, because the molecules do work against their attraction. Hydrogen and helium must first be cooled by other means, because at room temperature they warm up on expanding.

Practise with new numbers

Try it in Python

The program solves the van der Waals equation for the molar volume of carbon dioxide. The equation is a cubic in $V$, and numpy.roots finds the roots. It also prints the critical constants and the Boyle temperature.

Common mistakes

• Using the volume $V$ in place of $V - nb$. The molecules cannot use the volume that they occupy themselves.
• A wrong sign for the pressure correction. The measured pressure is lower than the ideal one because of attraction, so we add $\dfrac{an^2}{V^2}$ to the measured pressure.
• Reading $Z < 1$ as a sign of repulsion. $Z < 1$ means that attraction wins. $Z > 1$ means that the size of the molecules wins.
• Mixing the Boyle temperature and the critical temperature. $T_B = \dfrac{a}{Rb}$ and $T_c = \dfrac{8}{27}T_B$.
• Thinking that pressure liquefies every gas at any temperature. Above $T_c$ it does not.

Remember these

• $Z = \dfrac{PV_m}{RT}$. $Z < 1$: attraction wins. $Z > 1$: size wins. Every gas has $Z \to 1$ as $P \to 0$.
• $\left(P + \dfrac{an^2}{V^2}\right)(V - nb) = nRT$. $a$ is the attraction, $b$ is the excluded volume.
• $T_c = \dfrac{8a}{27Rb}$, $P_c = \dfrac{a}{27b^2}$, $V_c = 3b$, $Z_c = \tfrac{3}{8}$. Boyle temperature $T_B = \dfrac{a}{Rb}$.
• Below $T_c$ a gas can be liquefied by pressure. Above $T_c$ it cannot.
• A gas with a large $a$ has a high $T_c$ and is easy to liquefy.

Check yourself

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

1. A gas has $Z = 0.85$ at a certain pressure. What does this say about the forces in the gas?

2. Which gas is closer to ideal at $300\ \text{K}$ and $50\ \text{bar}$, helium or methane? Why?

3. The critical temperature of nitrogen is $126.2\ \text{K}$. Can nitrogen be liquefied at $25\ ^\circ\text{C}$ by raising the pressure?

4. Hydrogen has $a = 0.2476$ and $b = 0.02661$. What is its critical temperature?

5. Why does the pressure correction in the van der Waals equation have $n^2/V^2$ and not $n/V$?

References

• OpenStax, Chemistry 2e, Chapter 9, section 9.6 (non-ideal gas behaviour).
• The van der Waals constants and critical constants are standard values from the CRC Handbook of Chemistry and Physics.