Why Ideal-Gas Predictions Become Less Accurate at High Pressure

Compress a sealed gas sample while keeping its temperature constant. At low pressure, its behaviour may follow the ideal gas equation very closely. Keep compressing, though, and the measured pressure can become noticeably higher than the ideal-gas prediction.

Why does the equation stop working so well?

The problem is that the ideal gas model treats molecules as if they have negligible volume and do not exert significant forces on one another. At high pressure, molecules are packed much closer together. Their actual size and short-range repulsive interactions can no longer be ignored.

At sufficiently high pressure, these effects usually cause a real gas to exert a greater pressure than the ideal gas equation predicts for the same amount of gas, volume, and temperature.

But there is an interesting complication: intermolecular attractions can produce the opposite effect at lower or intermediate pressures. Understanding both effects explains why real gases do not always deviate from ideal behaviour in the same direction.

01The experiment: when does the ideal gas model fail?

Imagine compressing a fixed amount of gas inside a sealed cylinder while maintaining a constant temperature.

At first, the gas behaves almost exactly as the ideal gas equation predicts. But as the pressure increases, experimental measurements begin to disagree with the prediction.

This is the problem scientists needed to explain.

Setting up the investigation

Consider a modern pressure-volume-temperature (PVT) experiment.

A sealed chamber contains a known amount of gas. A movable piston sits at one end of the chamber, allowing the gas volume to be reduced. A pressure sensor connects to the gas chamber, while a temperature-controlled jacket surrounds the apparatus.

The measurements are collected as follows:

• The amount of gas, \(n\), remains constant because the chamber is sealed.
• The temperature, \(T\), is maintained at a constant value using the temperature-controlled jacket.
• The piston position determines the gas volume, \(V\).
• A pressure sensor measures the absolute pressure, \(P\).

The piston is moved in small steps. After each movement, the gas is allowed to return to the set temperature before its pressure and volume are recorded.

What should the ideal gas model predict?

The ideal gas equation is

\[ PV=nRT \]

where:

• \(P\) is absolute pressure, measured in kPa or Pa.
• \(V\) is gas volume, measured in L or \(\mathrm{m^3}\).
• \(n\) is the amount of gas, measured in mol.
• \(R\) is the ideal gas constant.
• \(T\) is absolute temperature, measured in K.

For HSC calculations, a convenient value is

\[ R=8.314\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}} \]

when pressure is in kPa and volume is in litres.

Alternatively, use \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\) with pressure in Pa and volume in \(\mathrm{m^3}\).

Remember that gas-law calculations require:

• Temperature in kelvin, not degrees Celsius.
• Absolute pressure, not gauge pressure. If a gauge reading is supplied, add atmospheric pressure to obtain absolute pressure.

For this experiment, \(n\), \(R\), and \(T\) are constant. Therefore,

\[ PV=\text{constant} \]

The ideal gas model predicts that halving the volume should double the pressure. Compressing the gas further should continue following this relationship.

This is the prediction we can test against experimental measurements.

If you need to refresh how the ideal gas equation relates pressure, volume, temperature, and amount, see Ideal Gas Law and Gas Relationships for HSC Chemistry.

Comparing experimental results with ideal predictions

One useful way to investigate deviations is to calculate the compressibility factor, \(Z\).

It compares the experimentally measured behaviour of a gas with the behaviour predicted by the ideal gas equation:

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

The ideal gas model predicts

\[ Z=1 \]

A real gas can have a value above or below 1.

Consider the following illustrative experimental data for a fixed sample containing \(0.100\ \mathrm{mol}\) of gas at \(300\ \mathrm K\).

Illustrative data only: These readings represent a hypothetical gas sample, not measurements from a particular published experiment.

For this sample,

\[ nRT=(0.100)(8.314)(300) \]

\[ nRT=249.42\ \mathrm{kPa\,L} \]

Notice what happens as pressure increases.

• At \(100\ \mathrm{kPa}\), \(Z\) is approximately 1. The gas behaves almost ideally.
• At \(500\ \mathrm{kPa}\), \(Z\) falls below 1. The measured pressure is lower than the ideal prediction for the same volume and temperature.
• At \(1000\ \mathrm{kPa}\), \(Z\) rises above 1.
• At \(2000\ \mathrm{kPa}\), \(Z=1.15\). The measured pressure is approximately 15% higher than the ideal prediction.

The important observation is that the deviation changes direction.

This tells us something important: real gases do not necessarily have pressures greater than ideal predictions at every pressure.

At moderate pressures, attractive forces can be significant. At sufficiently high pressures, finite molecular volume and repulsive interactions generally become dominant.

To understand why, we need to look at the assumptions behind the ideal gas model.

02Why high pressure changes gas behaviour

The ideal gas model makes three important assumptions about gas particles:

1. The particles have negligible volume compared with the total volume of the container.
2. There are no significant intermolecular attractive or repulsive forces between particles.
3. Collisions between particles and the container walls are elastic.

These assumptions make calculations manageable and work extremely well for many gases under ordinary conditions.

However, they become less accurate when molecules are forced close together.

1. Molecules have a finite volume

Imagine a group of people standing in a large hall.

If there are only a few people, their individual body sizes occupy a very small fraction of the total floor space. You could approximately treat them as points without making much difference to your estimate.

Now imagine squeezing the same group into a small lift.

Their actual sizes become much more important. The amount of space available for movement is no longer approximately equal to the total space inside the lift.

Gas molecules behave similarly.

At low pressure, the distance between molecules is relatively large. The volume occupied by the molecules themselves is negligible compared with the container volume.

At high pressure, the molecules are much closer together. Their finite size becomes a significant limitation on how closely the gas can be compressed.

The analogy has a limit: molecules are not solid people with fixed personal space. Their interactions arise from their molecular and electronic structure. However, the analogy captures why particle size becomes important as density increases.

This is called the excluded-volume effect.

The ideal gas model assumes that the entire container volume is available for molecular motion. In reality, molecules occupy space, reducing the volume available to other molecules.

As pressure increases, this difference becomes increasingly important.

The result is that the real gas becomes harder to compress than the ideal model predicts.

At the same temperature and container volume, the real gas therefore generally exerts a higher pressure once excluded-volume effects dominate.

2. Intermolecular forces affect pressure

The second important limitation is the assumption that molecules do not exert forces on one another.

Real molecules experience intermolecular attractions and repulsions.

These interactions affect how gas molecules move and how they transfer momentum to the container walls.

There are two effects to understand.

Intermolecular attractions

At moderate separations, molecules attract one another.

Consider a molecule approaching the container wall. Other molecules inside the gas attract it back towards the interior.

This reduces the momentum transfer to the wall compared with the ideal-gas prediction.

Consequently, the measured pressure can be lower than the ideal pressure.

This produces a negative deviation, where

\[ Z<1 \]

Short-range repulsions

When molecules are forced extremely close together, repulsive interactions become very strong.

These interactions resist further compression.

The gas therefore requires a greater pressure than the ideal model predicts to maintain the same temperature and volume.

This produces a positive deviation, where

\[ Z>1 \]

Which effect dominates?

The relative importance of these two effects depends on the gas, its temperature, and its pressure.

This table describes typical behaviour, not a universal pressure boundary. Different gases have different intermolecular interactions, and temperature also affects the extent of non-ideal behaviour.

The key idea is that increasing pressure brings molecules closer together. Eventually, the ideal assumptions become poor approximations.

03How scientists corrected the ideal gas model

The failure of the ideal gas model at high density was an important problem in the development of molecular theory.

In 1873, Dutch physicist Johannes Diderik van der Waals published his doctoral thesis, On the Continuity of the Gaseous and Liquid States.

He recognised that the simple gas model could not fully describe the behaviour of dense gases and liquids. His work introduced corrections that accounted for two features missing from the ideal model:

• The finite volume of molecules.
• The attractive forces between molecules.

His later Nobel lecture discussed the development of his equation of state and the remaining differences between theoretical predictions and experimental results. You can read van der Waals’s 1910 Nobel lecture and the Royal Society of Chemistry’s historical summary.

The van der Waals equation

The van der Waals equation modifies the ideal gas equation:

\[ \left(P+a\frac{n^2}{V^2}\right)(V-nb)=nRT \]

The new constants \(a\) and \(b\) represent corrections for intermolecular attractions and finite molecular volume.

The attraction correction, \(a\)

The term

\[ a\frac{n^2}{V^2} \]

corrects for the reduction in measured pressure caused by intermolecular attractions.

It is added to the measured pressure because attractions reduce the pressure exerted on the container walls.

The volume correction, \(b\)

The term

\[ nb \]

accounts for the volume excluded by the molecules themselves.

Instead of assuming that the full container volume \(V\) is available for molecular motion, the model uses

\[ V_{\text{available}}=V-nb \]

Here, \(a\) and \(b\) are substance-specific constants. When pressure is measured in kPa and volume in L, their units are:

• \(a\): \(\mathrm{kPa\,L^2\,mol^{-2}}\)
• \(b\): \(\mathrm{L\,mol^{-1}}\)

The equation is a more realistic model over a wider range of conditions, although it is still an approximation.

Its most useful contribution to this discussion is that it explains why the ideal gas equation becomes less accurate as molecules get closer together.

The attraction correction accounts for reduced pressure, while the volume correction accounts for the space molecules occupy.

At sufficiently high densities, the volume and repulsive effects become particularly important.

HSC course scope: The van der Waals equation is included here as an explanatory extension, not as a claim that students must memorise it for the HSC. As at November 2026, the 2017 Chemistry Stage 6 syllabus remains in use, with the 2025 Chemistry 11-12 syllabus scheduled to begin implementation in 2028. Check your school’s course program before treating this equation as examinable content. See NESA’s syllabus information.

04Worked examples

05Questions and solutions

Question 1

A sample of gas contains \(0.0500\ \mathrm{mol}\) of particles and occupies \(0.500\ \mathrm L\) at \(300\ \mathrm K\).

An experiment measures its absolute pressure as \(260\ \mathrm{kPa}\).

Calculate:

a) The pressure predicted by the ideal gas equation.

b) The compressibility factor.

c) Whether the gas shows a positive or negative deviation from ideal behaviour.

Solution 1

The ideal pressure is \(249.4\ \mathrm{kPa}\), and the compressibility factor is approximately \(1.04\). The gas therefore shows a small positive deviation.

Part a: Calculate the ideal pressure.

Use

\[ PV=nRT \]

Rearrange:

\[ P_{\text{ideal}}=\frac{nRT}{V} \]

Substitute:

\[ P_{\text{ideal}}= \frac{(0.0500)(8.314)(300)}{0.500} \]

\[ P_{\text{ideal}}=249.42\ \mathrm{kPa} \]

Part b: Calculate the compressibility factor.

Use the experimental pressure:

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

\[ Z=\frac{(260)(0.500)} {(0.0500)(8.314)(300)} \]

\[ Z=\frac{130}{124.71} \]

\[ Z=1.04 \]

Part c: Identify the deviation.

Because \(Z>1\), the real gas exerts a greater pressure than the ideal model predicts under these conditions.

The measured pressure is approximately 4.2% higher than the ideal prediction.

This indicates a positive deviation, although the difference is relatively small.

Question 2

A high-pressure vessel contains a gas at an absolute pressure of \(1.50\ \mathrm{MPa}\), a volume of \(0.400\ \mathrm L\), and a temperature of \(320\ \mathrm K\).

Independent experimental measurements give a compressibility factor of \(Z=1.13\).

Calculate:

a) The actual amount of gas in the vessel.

b) The amount of gas a student would calculate if they incorrectly assumed ideal behaviour.

c) The percentage error in the student’s answer, relative to the actual amount.

Solution 2

The actual amount of gas is approximately \(0.200\ \mathrm{mol}\). The ideal gas calculation gives \(0.226\ \mathrm{mol}\), overestimating the amount by approximately 13%.

Part a: Calculate the actual amount of gas.

We start with

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

Rearrange for \(n\):

\[ n=\frac{PV}{ZRT} \]

Convert the pressure:

\[ 1.50\ \mathrm{MPa}=1500\ \mathrm{kPa} \]

Substitute:

\[ n= \frac{(1500)(0.400)} {(1.13)(8.314)(320)} \]

\[ n=0.1996\ \mathrm{mol} \]

Therefore, the actual amount is approximately

\[ n_{\text{actual}}=0.200\ \mathrm{mol} \]

Part b: Calculate the ideal-gas estimate.

If the student ignores the compressibility factor, they use

\[ n_{\text{ideal}}=\frac{PV}{RT} \]

Substitute:

\[ n_{\text{ideal}}= \frac{(1500)(0.400)} {(8.314)(320)} \]

\[ n_{\text{ideal}}=0.2255\ \mathrm{mol} \]

Part c: Calculate the percentage error.

The percentage error relative to the actual amount is

\[ \%\text{ error}= \frac{n_{\text{ideal}}-n_{\text{actual}}} {n_{\text{actual}}}\times100 \]

\[ =\frac{0.2255-0.1996}{0.1996}\times100 \]

\[ =13.0\% \]

The ideal calculation overestimates the amount of gas.

This makes sense because \(Z>1\). For the same pressure, volume, and temperature, the ideal equation attributes more moles to the sample than are actually present.

Question 3

A fixed amount of gas is compressed at a constant temperature of \(350\ \mathrm K\).

The sample contains \(0.0500\ \mathrm{mol}\) of gas. The following experimental measurements are recorded.

a) Calculate the compressibility factor at each state.

b) Identify which state shows the greatest deviation from ideal behaviour.

c) Explain why the deviation changes direction as the pressure increases.

Solution 3

The compressibility factors are approximately:

• State A: \(Z=0.990\)
• State B: \(Z=0.956\)
• State C: \(Z=1.15\)

State C has the greatest deviation, with a positive deviation from ideal behaviour.

Part a: Calculate the compressibility factors.

First calculate the constant \(nRT\):

\[ nRT=(0.0500)(8.314)(350) \]

\[ nRT=145.495\ \mathrm{kPa\,L} \]

Now use

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

For state A:

\[ Z_A=\frac{(200)(0.720)}{145.495} \]

\[ Z_A=0.990 \]

For state B:

\[ Z_B=\frac{(800)(0.174)}{145.495} \]

\[ Z_B=0.956 \]

For state C:

\[ Z_C=\frac{(1600)(0.105)}{145.495} \]

\[ Z_C=1.15 \]

Part b: Identify the greatest deviation.

Compare each value with the ideal value of 1.

State C has the greatest absolute difference from 1.

Its compressibility factor is approximately 15% above the ideal value.

Part c: Explain the change in deviation.

At states A and B, \(Z<1\). This indicates that attractive intermolecular forces have a greater influence on the measured pressure than the excluded-volume and repulsive effects.

At state C, \(Z>1\). The gas is now sufficiently compressed that finite molecular volume and short-range repulsive interactions have become dominant.

The important point is that intermolecular attractions do not suddenly disappear at high pressure. Rather, the competing effects change in relative importance.

This explains why a real gas can initially show a negative deviation and later develop a positive deviation as it is compressed.

Question 4

A student wants to investigate why a gas deviates from ideal behaviour at high pressure.

They place a sealed gas sample inside a syringe surrounded by an insulating foam sleeve. They push the plunger in quickly and record the pressure immediately after compression.

The measured pressure is higher than the pressure predicted by the ideal gas equation.

The student concludes that the experiment proves gas molecules occupy a significant volume.

Explain why this conclusion is not justified.

Describe how the experimental design should be improved to determine whether the increased pressure is caused by non-ideal behaviour rather than a temperature change.

Solution 4

The conclusion is not justified because the experiment does not maintain a constant temperature during compression.

The increased pressure could be caused by heating the gas rather than by deviations from ideal behaviour.

When the student pushes the plunger in quickly, work is done on the gas. This increases its internal energy and can raise its temperature.

Even an ideal gas experiences an increase in pressure when compressed and heated at the same time.

For a fixed volume and amount of gas, the ideal gas equation gives

\[ P\propto T \]

Therefore, an increase in temperature can explain the higher measured pressure without requiring any non-ideal molecular effects.

The insulating foam makes the problem worse because it slows heat transfer between the gas and its surroundings. It does not maintain a constant temperature.

Improvements to the experiment

The student should:

1. Place the gas chamber inside a thermostated jacket to maintain a constant temperature.
2. Compress the gas slowly in small increments rather than pushing the plunger quickly.
3. Allow the gas to reach thermal equilibrium after each compression.
4. Use a temperature probe to confirm that the gas has returned to the required temperature before recording pressure.
5. Keep the amount of gas constant and ensure the apparatus does not leak.
6. Use calibrated pressure and volume measurements, with absolute pressure recorded.

After these changes, the student can calculate

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

for each set of measurements.

If \(Z\) differs significantly from 1 after temperature, amount, and measurement errors have been controlled, the results provide evidence of non-ideal gas behaviour.

The original experiment failed because it confused two different effects: increasing pressure through heating and increasing pressure through real-gas interactions.

Question 5

A sealed vessel contains \(0.100\ \mathrm{mol}\) of a real gas at a constant temperature of \(300\ \mathrm K\).

The gas is compressed slowly through three equilibrium states. Its measured pressure and volume are:

All pressures are absolute, and the amount and temperature of gas remain constant throughout.

A student calculates the compressibility factor at state B and obtains approximately 1.00.

They confidently conclude:

“The gas behaves ideally at state B, so Boyle’s law must correctly predict the pressure at state C.”

Evaluate this claim.

Your response must:

a) Calculate \(Z\) at all three states.

b) Use Boyle’s law to predict the pressure at state C, starting from state B.

c) Compare the predicted pressure with the actual measured pressure.

d) Explain why obtaining \(Z=1\) at one state does not prove that the gas behaves ideally throughout the compression.

Solution 5

The student’s claim is incorrect.

Although \(Z\) is approximately 1 at state B, this only means that the ideal gas equation happens to match the measured pressure-volume relationship at that particular state.

It does not mean that the gas behaves ideally throughout the compression.

Part a: Calculate the compressibility factor at each state.

Start by calculating \(nRT\):

\[ nRT=(0.100)(8.314)(300) \]

\[ nRT=249.42\ \mathrm{kPa\,L} \]

Use

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

For state A:

\[ Z_A=\frac{(500)(0.479)}{249.42} \]

\[ Z_A=0.960 \]

For state B:

\[ Z_B=\frac{(1500)(0.1663)}{249.42} \]

\[ Z_B=1.00 \]

For state C:

\[ Z_C=\frac{(3000)(0.0981)}{249.42} \]

\[ Z_C=1.18 \]

The gas therefore moves from a negative deviation at A, through approximately ideal behaviour at B, to a positive deviation at C.

Part b: Predict the pressure at state C using Boyle’s law.

The student assumes that Boyle’s law applies between B and C.

For a fixed amount of gas at constant temperature,

\[ P_BV_B=P_CV_C \]

Rearrange:

\[ P_C=\frac{P_BV_B}{V_C} \]

Substitute the measurements:

\[ P_C= \frac{(1500)(0.1663)}{0.0981} \]

\[ P_C=2543\ \mathrm{kPa} \]

The ideal prediction is therefore approximately

\[ P_C=2.54\ \mathrm{MPa} \]

Part c: Compare the predicted and actual pressures.

The actual measured pressure at C is

\[ P_{\text{actual}}=3000\ \mathrm{kPa} \]

The ideal prediction is

\[ P_{\text{ideal}}=2543\ \mathrm{kPa} \]

The difference is

\[ 3000-2543=457\ \mathrm{kPa} \]

The actual pressure is approximately 18% higher than the Boyle’s law prediction.

Therefore, Boyle’s law does not accurately describe the compression from B to C.

Part d: Explain why the student’s reasoning fails.

The student’s mistake is assuming that a gas must behave ideally at every state if its compressibility factor equals 1 at one particular state.

However, \(Z=1\) only establishes that

\[ PV=nRT \]

at that particular state.

It does not establish that the ideal assumptions are physically correct.

At state B, attractive intermolecular effects and excluded-volume or repulsive effects can approximately cancel one another. Their combined effect produces a compressibility factor close to 1.

At state A, attractions have a greater influence, giving \(Z<1\).

At state C, the gas is more compressed, and excluded-volume and repulsive effects dominate, giving \(Z>1\).

This is the important distinction:

A real gas can produce the same numerical pressure-volume relationship as an ideal gas at one particular state without actually behaving ideally across a range of pressures.

Boyle’s law requires \(PV\) to remain constant throughout the isothermal process, not merely at one point.

The student has confused agreement with an equation at one state with validity of the underlying model across all states.

06Where this leads next

The ideal gas equation works because it simplifies the behaviour of molecules. It becomes less accurate at high pressure because molecules are no longer sufficiently far apart for their volume and intermolecular interactions to be ignored.

When analysing a real gas, remember:

• \(Z=1\): the gas matches the ideal gas equation at that state.
• \(Z<1\): attractive effects dominate, producing a negative deviation.
• \(Z>1\): excluded-volume and repulsive effects dominate, producing a positive deviation.

The next useful concept is the critical point and the relationship between gas compression and liquefaction. Once molecules become sufficiently close, intermolecular attractions also help determine whether a substance remains a gas or changes into a liquid.

For further practice with the relationships and graphs used to analyse gas behaviour, see How to Read Pressure-Volume and Temperature Gas Graphs.