Deviation from ideal gas law — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Causes of real gas deviation from ideal gas behavior, the van der Waals equation for real gases, the compressibility factor Z, and analysis of deviation under varying temperature and pressure conditions for AP exam questions.

You should already know: The ideal gas law $PV=nRT$, kinetic molecular theory postulates for ideal gases, the relationship between molecular structure and intermolecular force strength.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Deviation from ideal gas law?

The ideal gas law $PV=nRT$ is derived from kinetic molecular theory (KMT) postulates that make two simplifying assumptions: gas molecules have no volume of their own, and they experience no intermolecular attractive forces. When these assumptions do not hold for real gases, the relationship between $P$, $V$, and $T$ deviates from $PV=nRT$, which is defined as deviation from ideal gas law (also called non-ideal gas behavior or real gas deviation).

Per the AP Chemistry Course and Exam Description (CED), this topic contributes ~2-4% of total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. Deviation is most pronounced when gas molecules are packed close together: at high pressure, low temperature, or near the condensation point of the gas. At these conditions, the two simplifying KMT assumptions break down, leading to measurable differences between predicted and observed gas properties. This topic tests both conceptual understanding of intermolecular forces and quantitative application of corrected gas laws.

2. Sources of Deviation from Ideal Behavior

The two core KMT assumptions that fail for real gases lead to two distinct, quantifiable sources of deviation, which dominate under different conditions.

First assumption failure: Ideal gas molecules have no volume. In reality, all gas molecules have finite molecular volume. At low pressure, molecules are very far apart, so the total volume of the molecules themselves is negligible compared to the total container volume, and this effect is insignificant. As pressure increases, molecules are pushed closer together, so molecular volume becomes a large fraction of the total container volume. The actual free volume available for molecules to move through is smaller than the measured container volume $V$, so this effect causes the observed product $PV$ to be larger than the ideal value $nRT$.

Second assumption failure: Ideal gas molecules have no intermolecular attractions. In reality, all molecules experience intermolecular attractions (London dispersion, dipole-dipole, hydrogen bonding). As a molecule moves toward the container wall to exert pressure, neighboring molecules pull it inward, reducing the force it exerts on the wall. This makes the observed pressure $P$ lower than the ideal pressure, so $PV$ is smaller than $nRT$. At high temperature, molecular kinetic energy is large enough to overcome intermolecular attractions, so this effect is negligible. At low temperature, this effect dominates deviation.

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. Step 1: At moderate pressure and room-to-cool temperature, intermolecular attraction strength is the primary driver of deviation, since molecular volume is still relatively low.
2. Step 2: Rank intermolecular force strength: He (very weak London dispersion forces) < O₂ (weak LDF) < CO₂ (stronger LDF than O₂, nonpolar) < CH₃OH (polar, hydrogen bonding, strong intermolecular attractions).
3. Step 3: CH₃OH is also much closer to its condensation point (boiling point 64.7°C) than the other options, which are permanent gases at 0°C, so deviation is further increased.
4. Step 4: The gas with the greatest deviation is CH₃OH.

Exam tip: When asked to rank deviation for different gases at the same $P$ and $T$, always compare intermolecular force strength first; only compare molecular size if intermolecular force strength is similar between the options.

3. Compressibility Factor Z

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: $$Z=\frac{PV}{nRT}$$ For an ideal gas, $PV=nRT$, so $Z=1$ at all pressures and temperatures. For real gases, $Z\ne 1$, and the value of $Z$ tells us which source of deviation dominates:

• If $Z<1$: The observed $PV$ is smaller than the ideal $nRT$, so the intermolecular attraction effect dominates.
• If $Z>1$: The observed $PV$ is larger than the ideal $nRT$, so the molecular volume effect dominates.

For all real gases, $Z$ approaches 1 as pressure approaches 0, because low pressure brings real behavior close to ideal. As pressure increases from 0, $Z$ typically dips below 1 (attraction dominates at low-to-moderate pressure) then rises above 1 at high pressure (molecular volume dominates at high pressure). Higher temperatures shift the $Z$ curve closer to $Z=1$, because increased kinetic energy reduces the impact of intermolecular attractions.

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. Step 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. Step 2: Substitute into the $Z$ formula: $Z=\frac{PV}{nRT}$.
3. Step 3: Calculate: $Z=\frac{(200)(0.10)}{(1)(0.0821)(273)}=\frac{20}{22.41}\approx 0.89$.
4. Step 4: Compare to 1: $Z=0.89<1$, so intermolecular attraction dominates at this condition.

Exam tip: AP MCQs regularly test the relationship between Z direction and dominant effect — never mix up $Z<1$ and $Z>1$; memorize "less than one, attraction done" to remember.

4. The van der Waals Equation for Real Gases

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: $$\left(P+a\frac{n^2}{V^2}\right)(V-nb)=nRT$$ Breakdown of correction terms:

• $nb$: The volume correction. The constant $b$ is proportional to the size of the gas molecule, so we subtract $nb$ from the measured container volume $V$ to get the actual free volume available for molecules to move through. Larger molecules have larger $b$ values.
• $a\frac{n^2}{V^2}$: The pressure correction. The constant $a$ is proportional to the strength of intermolecular forces between gas molecules, so we add this term to the measured pressure $P$ to get the ideal pressure that would exist if there were no attractions. Molecules with stronger intermolecular forces have larger $a$ values.

At low pressure, both correction terms are negligible, so the van der Waals equation reduces to the ideal gas law, matching expected behavior.

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. Step 1: Rearrange the equation to solve for $P$: $P=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}$.
2. Step 2: Calculate $nRT=(3.00)(0.0821)(273)=67.33$ atm·L. Calculate volume correction: $V-nb=10.0-(3.00)(0.0427)=10.0-0.1281=9.8719$ L.
3. Step 3: Calculate pressure correction: $a\frac{n^2}{V^2}=4.48\frac{(3.00{)}^2}{(10.0{)}^2}=4.48\frac{9}{100}=0.403$ atm.
4. Step 4: Solve for $P$: $P=\frac{67.33}{9.8719}-0.403=6.82-0.403=6.42$ atm.

Exam tip: Always confirm that the units of $a$ and $b$ match your units for $P$, $V$, and $R$; AP problems almost always use atm and liters, which matches $R=0.0821$ L·atm/(mol·K).

5. Common Pitfalls (and how to avoid them)

• 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 $Z=P{V}_{\text{measured}}/nRT$, so $Z>1$ means measured $PV$ 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: Claim that all real gases have $Z<1$ at all pressures above 1 atm. Why: Students generalize from the 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 calculate or reason based on conditions, don't assume a fixed relationship.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

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

Worked Solution: The van der Waals constant $a$ increases with increasing intermolecular force strength. We rank the gases by intermolecular force strength: Ne (monatomic, very weak London dispersion forces) < CH₄ (nonpolar, weak LDF, larger molar mass than Ne) < HCl (polar, dipole-dipole interactions, stronger forces than CH₄) < CH₃COOH (polar, hydrogen bonding, strongest forces). This matches the order in option A. The correct answer is A.

Question 2 (Free Response)

A researcher tests non-ideal behavior for 1.50 mol of ethane (C₂H₆) in a 5.00 L container at 298 K. (a) Calculate the pressure of ethane assuming ideal behavior. (b) Van der Waals constants for ethane are $a=5.51$ atm·L²/mol² and $b=0.0651$ L/mol. Calculate the pressure of ethane using the van der Waals equation. (c) Explain why the van der Waals pressure differs from the ideal pressure, and identify which correction term is larger in magnitude for this condition.

Worked Solution: (a) Use the ideal gas law $P=\frac{nRT}{V}$: $$P=\frac{(1.50\text{ mol})(0.0821\text{ Lcdotpatm/(molcdotpK)})(298\text{ K})}{5.00\text{ L}}=\frac{36.70}{5.00}=7.34\text{ atm}$$

(b) Rearrange the van der Waals equation: $P=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}$: $$V-nb=5.00\text{ L}-(1.50\text{ mol})(0.0651\text{ L/mol})=4.902\text{ L}$$ $$\frac{nRT}{V-nb}=\frac{36.70}{4.902}=7.49\text{ atm}$$ $$a\frac{n^2}{V^2}=5.51\frac{(1.50{)}^2}{(5.00{)}^2}=0.496\text{ atm}$$ $$P=7.49-0.496=6.99\text{ atm}$$

(c) The van der Waals pressure (6.99 atm) is lower than the ideal pressure (7.34 atm) because intermolecular attractions between ethane molecules reduce the force exerted on the container wall, lowering observed pressure. The magnitude of the pressure correction (0.496 atm) is larger than the magnitude of the volume correction (7.49 - 7.34 = 0.15 atm), so the attraction (a) correction dominates here.

Question 3 (Application / Real-World Style)

Industrial reactors for ammonia synthesis operate at 200 atm and 700 K (427°C). A 1000 L reactor holds 500 mol of H₂ gas at 700 K. Use the van der Waals equation to calculate the pressure of H₂, given $a=0.247$ atm·L²/mol² and $b=0.0266$ L/mol. Compare your result to the ideal gas prediction and explain what this means for reactor design.

Worked Solution:

1. Rearrange van der Waals: $P=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}$
2. Calculate terms: $nRT=(500)(0.0821)(700)=28735$ atm·L. $V-nb=1000-(500)(0.0266)=986.7$ L. $a{n}^2/V^2=0.247\ast (500{)}^2/(1000{)}^2=0.0618$ atm.
3. Solve: $P=(28735/986.7)-0.0618\approx 29.12-0.0618\approx 29.06$ atm (for 500 mol; scaling to 200 atm total operating pressure confirms the trend). Wait, correction for 200 atm context: for 3440 mol H₂ in 1000 L, P ideal = (3440 * 0.0821 * 700)/1000 ≈ 198 atm, van der Waals P = (34400.0821700)/(1000 - 34400.0266) - 0.247(3440)^2/(1000)^2 ≈ 198/(1000 - 91.5) - 2.9 ≈ 218 - 2.9 = 215 atm.
4. The real pressure is ~15 atm higher than the ideal prediction of 200 atm. In context: At this high industrial operating pressure, the volume of the hydrogen molecules themselves significantly reduces the free volume in the reactor, leading to a higher actual pressure than the ideal gas law predicts. Engineers must account for this deviation when designing reactor vessels to withstand the higher actual pressure.

7. Quick Reference Cheatsheet

8. 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 Unit 3. This topic also directly feeds into gas stoichiometry problems for high-pressure industrial reactions, which often 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 that are tested across the entire AP Chemistry curriculum.

Follow-on topics for further study: Intermolecular forces, Ideal gas law, Phase diagrams, Vapor pressure