Deviation from Ideal Gas Law

AP Chemistry· AP Chemistry CED — Intermolecular Forces and Properties· 14 min read

1. What Is Deviation from Ideal Gas Law?★★☆☆☆⏱ 3 min

The ideal gas law $P V = n R T$ is derived from kinetic molecular theory (KMT) that makes two key simplifying assumptions: gas molecules have no volume of their own, and they experience no intermolecular attractive forces.

📘 Definition

Deviation from ideal gas behavior

A measurable difference between observed pressure, volume, and temperature properties of a real gas and the values predicted by the ideal gas law, occurring when the two core KMT assumptions break down.

Example:

Deviation is most pronounced at high pressure, low temperature, or near a gas's condensation point, where molecules are packed close together.

This topic contributes ~2-4% of your total AP Chemistry exam score, appearing in both multiple-choice and free-response sections. It tests both conceptual understanding of intermolecular forces and quantitative application of corrected gas laws.

2. Sources of Deviation from Ideal Behavior★★☆☆☆⏱ 3 min

The two failed KMT assumptions lead to two distinct sources of deviation, which dominate under different conditions:

Molecular volume effect: Real gas molecules have finite volume. As pressure increases, molecules are pushed closer together, so molecular volume becomes a large fraction of total container volume. Free volume available for movement is smaller than measured container volume $V$ , so observed $P V$ is larger than the ideal value $n R T$ .
Intermolecular attraction effect: Real gas molecules experience intermolecular attractions. As a molecule moves toward the container wall, neighboring molecules pull it inward, reducing the force exerted on the wall. Observed pressure $P$ is lower than ideal pressure, so $P V$ is smaller than $n R T$ . This effect dominates at low temperature, where kinetic energy is too low to overcome attractions.

📐 Worked Example

Which of the following gases shows the greatest deviation from ideal behavior at 5 atm and 0°C? (A) He (B) O₂ (C) CH₃OH (g) (D) CO₂

1
At moderate pressure and cool temperature, intermolecular attraction strength is the primary driver of deviation, since molecular volume is still relatively low.
2
Rank intermolecular force strength: He (very weak London dispersion forces) < O₂ (weak LDF) < CO₂ (stronger LDF than O₂) < CH₃OH (polar, hydrogen bonding, strong intermolecular attractions).
3
CH₃OH is also much closer to its condensation point (boiling point 64.7°C) than the other options, further increasing deviation.
4
The gas with the greatest deviation is CH₃OH.

3. Compressibility Factor Z★★★☆☆⏱ 3 min

The compressibility factor $Z$ is a dimensionless quantity that directly quantifies the magnitude and direction of deviation from ideal gas behavior. It is defined as:

Z = \frac{PV}{nRT}

Ideal gas: $Z = 1$ at all conditions
Real gas $Z < 1$ : Observed $P V$ < ideal $n R T$ , so intermolecular attraction effect dominates
Real gas $Z > 1$ : Observed $P V$ > ideal $n R T$ , so molecular volume effect dominates

For all real gases, $Z$ approaches 1 as pressure approaches 0 (low pressure brings behavior close to ideal). As pressure increases from 0, $Z$ typically dips below 1 then rises above 1 at high pressure. Higher temperatures shift the $Z$ curve closer to $Z = 1$ .

📐 Worked Example

At 0°C, 200 atm, 1 mole of hydrogen gas has a measured volume of 0.10 L. Calculate $Z$ and identify which source of deviation dominates.

1
List given values: $n = 1$ mol, $P = 200$ atm, $V = 0.10$ L, $R = 0.0821$ L·atm/(mol·K), $T = 273$ K.
2
Substitute into the $Z$ formula:
3
$Z = \frac{PV}{nRT}$
4
$Z = \frac{(200)(0.10)}{(1)(0.0821)(273)} = \frac{20}{22.41} \approx 0.89$
5
Compare to 1: $Z = 0.89 < 1$ , so intermolecular attraction dominates at this condition.

4. The van der Waals Equation for Real Gases★★★★☆⏱ 5 min

The van der Waals equation modifies the ideal gas law to correct for both sources of deviation, adding two gas-specific empirical correction terms. The full form is:

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

$nb$ : Volume correction. $b$ is proportional to molecular size, so we subtract $nb$ from measured container volume $V$ to get actual free volume. Larger molecules have larger $b$ .
$a \frac{n ^{2}}{V ^{2}}$ : Pressure correction. $a$ is proportional to intermolecular force strength, so we add this term to measured pressure $P$ to get ideal pressure with no attractions. Stronger intermolecular forces have larger $a$ .

📐 Worked Example

Calculate the pressure of 3.00 mol of hydrogen sulfide (H₂S) in a 10.0 L container at 273 K using the van der Waals equation. Given $a = 4.48$ atm·L²/mol², $b = 0.0427$ L/mol, $R = 0.0821$ L·atm/(mol·K).

1
Rearrange the equation to solve for $P$ :
2
$P = \frac{nRT}{V - nb} - a\frac{n^2}{V^2}$
3
Calculate $n R T = (3.00) (0.0821) (273) = 67.33$ atm·L. Calculate volume correction: $V - nb = 10.0 - (3.00) (0.0427) = 9.8719$ L.
4
Calculate pressure correction: $a \frac{n ^{2}}{V ^{2}} = 4.48 \frac{( 3.00 ) ^{2}}{( 10.0 ) ^{2}} = 0.403$ atm.
5
Solve for $P$ : $P = \frac{67.33}{9.8719} - 0.403 = 6.82 - 0.403 = 6.42$ atm.

✓ Quick check

Which of the following lists gases in order of increasing van der Waals constant $a$ (smallest to largest)?
- A) Ne < CH₄ < HCl < CH₃COOH
- B) CH₃COOH < HCl < CH₄ < Ne
- C) Ne < HCl < CH₄ < CH₃COOH
- D) CH₄ < Ne < CH₃COOH < HCl
Reveal answer
A) Ne < CH₄ < HCl < CH₃COOH —
The van der Waals constant $a$ increases with intermolecular force strength. Ranking by force strength: Ne (very weak LDF) < CH₄ (weak LDF) < HCl (dipole-dipole) < CH₃COOH (hydrogen bonding), so option A is correct.

5. Common Pitfalls

Wrong move:

Claiming $Z > 1$ means intermolecular attractions dominate deviation.

Why:

Students mix up which effect causes which direction of $Z$ deviation, confusing corrected and measured values.

Correct move:

Memorize that $Z = P V_{measured} / n R T$ , so $Z > 1$ means measured $P V$ is larger than ideal, which comes from molecular volume reducing free volume, so molecular volume dominates. $Z < 1 =$ attraction dominates.

Wrong move:

Misplacing correction terms in the van der Waals equation, adding $nb$ to $V$ or putting the $a$ term in the volume bracket.

Why:

Students misremember which correction applies to which variable.

Correct move:

Use the mnemonic 'P gets the a, V gets the b' to remember pressure correction for attraction, volume correction for molecular size.

Wrong move:

Ranking gas deviation at the same $P / T$ by only molecular size, ignoring intermolecular force strength.

Why:

Students remember two sources of deviation but prioritize the wrong one for moderate pressure conditions.

Correct move:

First compare intermolecular force strength (polarity, hydrogen bonding) when ranking deviation; only compare molecular size if intermolecular forces are similar.

Wrong move:

Claiming all real gases have $Z < 1$ at all pressures above 1 atm.

Why:

Students generalize from low-to-moderate pressure behavior where attraction often dominates.

Correct move:

Remember that at sufficiently high pressure, the molecular volume effect always becomes dominant, so $Z$ will always rise above 1 for any real gas.

Wrong move:

Assuming ideal gas calculated pressure is always higher than van der Waals calculated pressure for all real gases.

Why:

Students only see examples with strong attractive forces near room temperature and generalize incorrectly.

Correct move:

At very high pressure, the volume correction dominates, so van der Waals pressure will be higher than ideal pressure; always reason based on conditions, don't assume a fixed relationship.

6. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Compressibility Factor	$Z = \frac{P V}{n R T}$	$Z = 1$ = ideal; $Z < 1$ = attraction dominates; $Z > 1$ = molecular volume dominates
van der Waals Equation	$(P + a \frac{n ^{2}}{V ^{2}}) (V - nb) = n R T$	$P$ = measured pressure, $V$ = measured container volume
van der Waals $a$	Gas-specific empirical constant	Increases with intermolecular force strength; corrects pressure for attractions
van der Waals $b$	Gas-specific empirical constant	Increases with molecular size; corrects volume for molecular volume
Minimum deviation conditions	Low pressure, high temperature	Correction terms are negligible, behavior approaches ideal
Maximum deviation conditions	High pressure, low temperature	Largest deviation, near condensation point
Deviation ranking rule	Same $P / T$	Rank first by intermolecular force strength, then by molecular size

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Rank gas deviation multiple choice
2022 · FRQ
Van der Waals calculation and explanation

Going deeper

official syllabusAP Chemistry Course and Exam DescriptionUnit 3 learning objectives

What's Next

Mastery of non-ideal gas behavior is a critical prerequisite for understanding phase changes and intermolecular force comparisons, the next core topics in AP Chemistry Unit 3. This topic also directly applies to gas stoichiometry problems for high-pressure industrial reactions, which regularly appear in AP FRQ sections. Without understanding the sources of deviation and how to apply van der Waals corrections, you will struggle to explain why real gases condense into liquids at low temperature and high pressure, a key concept for phase diagrams and vapor pressure calculations. This topic also reinforces core ideas about intermolecular forces tested across the entire AP Chemistry curriculum.