When Is the Ideal-Gas Model a Good Approximation?

Compress 1.00 mol of gas at 300 K from 100 kPa to 1000 kPa. The ideal-gas equation predicts that its volume falls from 24.9 L to 2.49 L: exactly one-tenth of its original volume.

Would a real gas shrink by exactly that factor? Not necessarily. At high pressure, gas particles are crowded together, so their own volume and the forces between them can no longer be ignored.

Before going further, predict: would lowering the pressure or raising the temperature make the ideal-gas model more accurate? Usually, both. Lower pressure helps particles stay further apart, while higher temperature makes intermolecular attractions less significant compared with the particles’ kinetic energy.

01Quick decision rule

Use the ideal-gas model when a gas is at relatively low pressure and high temperature, particularly when it is far from condensing into a liquid.

Be more cautious when the gas is at high pressure, low temperature, or has particularly strong intermolecular attractions.

These are general trends, not guarantees. The identity of the gas matters, and different non-ideal effects can compete.

For HSC calculations: if the question tells you to assume ideal behaviour, use the ideal-gas equation as instructed. If the question asks you to evaluate whether the model is realistic, explain how the conditions affect its assumptions.

02What does the ideal-gas model actually assume?

Imagine a gas as a collection of tiny particles moving randomly through mostly empty space. This is a useful starting point, but the model makes specific assumptions about those particles.

Assumption 1: Particle volume is negligible

Gas particles have physical size, but the ideal-gas model treats their combined volume as negligible compared with the volume occupied by the gas.

At low pressure, particles are spread relatively far apart. The space they occupy themselves is a small fraction of the available volume.

At high pressure, particles are crowded closer together. Their own volume becomes a more significant part of the picture, so treating them as point-like particles becomes less accurate.

Assumption 2: Intermolecular forces are negligible

Real gas particles attract and repel one another. The ideal-gas model ignores these forces except during collisions.

This assumption matters because intermolecular forces can change how particles move and how frequently and forcefully they collide with the walls of their container.

At low temperatures, particles have less average kinetic energy. Attractive forces can therefore have a greater influence on their motion.

Two other assumptions

The kinetic theory of an ideal gas also assumes that:

• Particles move continuously and randomly between collisions.
• Collisions between particles, and between particles and the walls, are perfectly elastic. Total kinetic energy is conserved in each collision.

In this model, gas pressure results from particles colliding with the walls. Absolute temperature is related to the particles’ average translational kinetic energy.

The important distinction is that real particles do have volume and intermolecular forces. An ideal gas is a model in which those effects are small enough to ignore.

03When do real gases depart from ideal behaviour?

Low pressure: particles are further apart

When pressure is reduced by expanding a gas at a fixed temperature, the particles become more widely separated.

Their own volume is less significant compared with the volume occupied by the gas. Attractive forces also have less influence because particles spend less time close to one another.

Both assumptions of the ideal-gas model become more reasonable.

High temperature: attractions matter less

Increasing the temperature increases the particles’ average kinetic energy.

The particles still attract one another, but those attractions have less influence relative to their kinetic energy. This makes the assumption of negligible intermolecular attractions more reasonable.

High temperature does not eliminate intermolecular forces. It reduces their relative influence on the gas’s behaviour.

High pressure: particle size and repulsions matter more

Compressing a gas brings particles closer together. Their finite size becomes harder to ignore, and short-range repulsive interactions can become significant.

The ideal-gas model assumes that particles occupy negligible volume, so it can become unreliable under these conditions.

Low temperature: attractions become more significant

Cooling a gas reduces its average kinetic energy. Attractive forces can then have a greater influence on its behaviour.

If the gas approaches condensation, the ideal-gas approximation can become particularly poor because the particles are increasingly influenced by one another.

Does the identity of the gas matter?

Yes. Consider helium and ammonia at the same moderate pressure and temperature.

Helium atoms are very small and experience relatively weak dispersion forces. Ammonia molecules have stronger intermolecular attractions, including hydrogen bonding.

Helium is therefore generally expected to behave more ideally under these conditions. Its particle volume is less significant, and its intermolecular attractions are weaker.

This is why pressure and temperature alone do not tell the whole story. The nature of the particles matters, too.

04Using the ideal-gas equation

The central equation for HSC calculations is

\[ PV=nRT \]

where:

• \(P\) is the absolute pressure of the gas, in kilopascals (kPa).
• \(V\) is the volume, in litres (L).
• \(n\) is the amount of gas, in moles (mol).
• \(R\) is the universal gas constant.
• \(T\) is the absolute temperature, in kelvin (K).

For pressure in kPa and volume in L, use

\[ R=8.31\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}} \]

You can also use \(R=8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}\) with pressure in pascals and volume in cubic metres. These are equivalent unit conventions because \(1\ \mathrm{kPa\,L}=1\ \mathrm{J}\).

Convert temperature before substituting

Gas-law calculations require kelvin, not degrees Celsius.

\[ T(\mathrm K)=t(^{\circ}\mathrm C)+273.15 \]

For example, \(25.0^{\circ}\mathrm C=298.15\ \mathrm K\).

Kelvin is an absolute temperature scale. A temperature of \(20^{\circ}\mathrm C\) is not twice as hot as \(10^{\circ}\mathrm C\), so Celsius values cannot be used directly in gas-law ratios.

Also, use absolute pressure, not gauge pressure. If a question gives gauge pressure, convert it to absolute pressure before using the equation.

Which equation should you choose?

The combined gas law is a shortcut derived from \(PV=nRT\). It works when the amount of gas is unchanged and the gas is treated as ideal. If a reaction produces or consumes gas, you cannot use it without accounting for the change in \(n\).

05Worked examples

06Common mistakes

• Substituting Celsius into \(PV=nRT\). Celsius does not start at absolute zero, so convert every gas-law temperature to kelvin first.

• Mixing units. Using \(R=8.31\) with pressure in kPa and volume in L is consistent. Mixing those units with volume in \(\mathrm{m^3}\) without converting will give an incorrect result.

• Assuming every gas follows \(PV=nRT\) exactly. The equation is exact for the ideal-gas model, but real gases only approximate it. The accuracy depends on the gas and its conditions.

• Assuming high pressure only makes attractions stronger. Crowding also makes particle volume and repulsive interactions more important. These effects can influence the gas in different directions.

• Using pressure alone to judge ideal behaviour. Pressure depends on both temperature and particle density. A low pressure caused by cooling in a rigid container does not mean the particles have moved further apart.

07Optional extension: Why can real gases deviate in different directions?

The ideal-gas model neglects two effects that can compete.

First, intermolecular attractions can pull particles towards one another. Under suitable conditions, this reduces the pressure relative to the ideal-gas prediction for the same amount of gas, volume, and temperature.

Second, real particles occupy finite volume and experience short-range repulsions. At high densities, these effects can increase the pressure relative to the ideal-gas prediction.

Consequently, it is not correct to claim that every real gas always has a higher pressure, or always has a lower pressure, than an ideal gas under the same conditions. The result depends on which non-ideal effects dominate.

A useful advanced quantity is the compressibility factor, \(Z\):

\[ Z=\frac{PV}{nRT} \]

It compares the measured gas behaviour with the ideal-gas prediction.

• \(Z=1\): the gas matches the ideal-gas equation.
• \(Z<1\): attractive effects dominate the deviation.
• \(Z>1\): finite particle volume and repulsive effects dominate the deviation.

This is an optional extension, not a requirement for routine HSC ideal-gas calculations.

08Questions and solutions

Question 1

A 0.125 mol sample of gas is at 100 kPa and 310 K. Calculate its volume, assuming ideal behaviour.

Solution 1

The volume is \(3.22\ \mathrm L\). Since the amount, pressure, and temperature are known, rearrange \(PV=nRT\) to find \(V\).

\[ \begin{aligned} V&=\frac{nRT}{P}\\ &=\frac{(0.125)(8.31)(310)}{100}\\ &=3.22\ \mathrm L \end{aligned} \]

The units are consistent with \(R\) in \(\mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\). The calculated volume is the ideal-gas prediction for the stated conditions.

Question 2

A fixed amount of gas occupies 1.50 L at 95.0 kPa and \(18.0^{\circ}\mathrm C\). It is heated to \(65.0^{\circ}\mathrm C\), while its pressure increases to 110 kPa.

Assuming ideal behaviour, calculate the final volume. Explain why using the Celsius temperatures directly would be incorrect.

Solution 2

The final volume is \(1.50\ \mathrm L\), to three significant figures. The amount of gas is unchanged, so the combined gas law applies.

First, convert both temperatures:

\[ T_1=291.15\ \mathrm K \]

\[ T_2=338.15\ \mathrm K \]

Rearrange the combined gas law:

\[ V_2=\frac{P_1V_1T_2}{P_2T_1} \]

Substitute the values:

\[ \begin{aligned} V_2&=\frac{(95.0)(1.50)(338.15)} {(110)(291.15)}\\ &=1.50\ \mathrm L \end{aligned} \]

The temperature increase tends to increase the volume, while the pressure increase tends to decrease it. In this case, their effects almost cancel.

Using Celsius directly would give incorrect temperature ratios because \(0^{\circ}\mathrm C\) is not the point of zero thermal motion. Gas-law relationships require an absolute temperature scale.

Question 3

At the same moderate pressure and temperature, helium and ammonia are compared.

Which gas is expected to behave more ideally? Explain your answer using both particle volume and intermolecular attractions.

Solution 3

Helium is expected to behave more ideally because its particle volume is very small and its intermolecular attractions are relatively weak.

Helium consists of small, single atoms that experience weak dispersion forces. Ammonia consists of larger molecules with stronger intermolecular attractions, including hydrogen bonding.

For helium, the assumptions of negligible particle volume and negligible intermolecular forces are generally better approximations under these conditions. For ammonia, the stronger attractions make the second assumption less accurate.

The trap is to argue that both substances are gases, so both must behave ideally. Being in the gas state does not guarantee that the ideal-gas assumptions are valid to the same degree.

Question 4

A gas occupies 0.990 L at 100 kPa. It is compressed isothermally to 5000 kPa, with the amount of gas unchanged.

Assuming ideal behaviour, calculate the predicted final volume. Then explain why a real gas may not follow this prediction exactly. Can you determine the direction of the deviation from the information given?

Solution 4

The ideal-gas model predicts a final volume of \(0.0198\ \mathrm L\). However, the actual volume cannot be determined uniquely from the information given.

Because the process is isothermal and the amount of gas is unchanged, \(PV\) is constant in the ideal-gas model:

\[ P_1V_1=P_2V_2 \]

Rearrange and substitute:

\[ \begin{aligned} V_2&=\frac{P_1V_1}{P_2}\\ &=\frac{(100)(0.990)}{5000}\\ &=0.0198\ \mathrm L \end{aligned} \]

The pressure has increased by a factor of 50, so the ideal-gas model predicts a volume that is one-fiftieth of the original.

At this much higher pressure, the particles are crowded together. Their own volume is less negligible, and short-range repulsive interactions may become important. Intermolecular attractions can also affect the result.

It is tempting to assume that the actual volume must be either greater or smaller than the calculated value. But the direction and magnitude of the deviation depend on the gas and its state. Without further information, the ideal-gas result is a prediction, not a guaranteed measurement.

Question 5

Two equal samples of the same real gas begin at 400 kPa and 400 K, with the same amount of gas in the same volume.

• Sample A is cooled in a rigid container until its pressure falls to 200 kPa.
• Sample B remains at 400 K and is expanded until its pressure also falls to 200 kPa.

Assume neither sample condenses.

A student argues that the samples must be equally close to ideal behaviour because their final pressures are identical.

Which sample should generally be closer to ideal behaviour? Explain your reasoning using particle density, kinetic energy, and the assumptions of the ideal-gas model.

Solution 5

Sample B should generally be closer to ideal behaviour. Although both samples finish at the same pressure, their temperatures and particle densities are different.

The key is to examine how each sample reached 200 kPa.

Sample A: cooling at constant volume

The container is rigid, so the amount of gas and its volume remain unchanged. Its particle density therefore stays the same.

Cooling reduces the particles’ average kinetic energy. As a result, intermolecular attractions can have a greater influence relative to that kinetic energy. The assumption of negligible attractions becomes less reliable.

The pressure has fallen, but the particles have not spread further apart. Lower pressure alone does not establish that the gas is more ideal.

Sample B: expansion at constant temperature

The temperature stays at 400 K, but the volume increases. The same number of particles now occupies more space, so the particle density decreases.

The particles are further apart on average, making intermolecular attractions less influential. Their own volume is also a smaller fraction of the volume occupied by the gas.

Both changes make the ideal-gas assumptions more reasonable.

Why the student’s argument fails

Pressure is not a direct measure of particle spacing by itself. It depends on temperature as well as particle density. A gas can reach the same pressure through cooling or expansion, but those processes produce different particle conditions.

The comparison is qualitative: the exact degree of non-ideal behaviour would require more information about the gas. Nevertheless, under the stated assumptions, Sample B has both the higher temperature and the lower particle density, so it is generally expected to behave more ideally.

09What to check next

When solving a gas-law problem, separate two decisions:

1. Which equation applies? Use \(PV=nRT\) for a single state, or the combined gas law when the amount of gas is unchanged.
2. How reliable is the model? Check the pressure, temperature, particle size, and intermolecular attractions.

If a chemical reaction produces the gas, the next step is to calculate the amount of gas from the balanced equation before using \(PV=nRT\). That connects gas-law calculations with stoichiometry, while keeping the assumptions behind the final answer clear.