Pressure, Thermal Equilibrium and Ideal Gas Law — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: absolute vs gauge pressure, the zeroth law of thermodynamics, thermal equilibrium definition, absolute temperature scale, combined gas law for closed systems, and both molar and molecular forms of the ideal gas law.

You should already know: Force and area definitions from mechanics, basic kinetic theory of matter, SI unit conversion rules.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 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 Pressure, Thermal Equilibrium and Ideal Gas Law?

This topic is core to Unit 2 Thermodynamics, which makes up 18–22% of the total AP Physics 2 exam score, with this specific subtopic contributing roughly 6–8% of total exam points. Concepts from this topic appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with fluid mechanics or heat engine problems. Pressure is defined as the magnitude of perpendicular force per unit area exerted by a substance on a boundary; for gases, this force arises from constant elastic collisions of gas molecules with the container wall. Thermal equilibrium is the steady state two systems reach when in thermal contact, with no net heat transfer between them. The ideal gas law is the equation of state for a hypothetical ideal gas, which approximates real gas behavior at low pressures and high temperatures where intermolecular forces and molecular volume are negligible. It unifies all historical empirical gas laws into a single relationship applicable to any gas sample under appropriate conditions.

2. Pressure and the Zeroth Law of Thermal Equilibrium

Pressure is a scalar quantity (we only measure magnitude, as force is always perpendicular to the container surface) defined as: $$P=\frac{F_{\perp }}{A}$$ where $F_{\perp }$ is the perpendicular force on area $A$. The SI unit of pressure is the pascal ($1\text{Pa}=1\text{N}/{\text{m}}^2$), though non-SI units like atmospheres (atm) are common in problems; always convert to consistent units for calculation. A critical distinction for real-world problems is between absolute pressure and gauge pressure: gauge pressure measures pressure relative to atmospheric pressure, so the relationship is: $$P_{\text{abs}}=P_{\text{gauge}}+P_{\text{atm}}$$ Most practical pressure gauges (tire gauges, blood pressure cuffs) measure gauge pressure by zeroing out atmospheric pressure, so you must add atmospheric pressure to get the absolute pressure required for all gas law calculations.

Thermal equilibrium describes the steady state of two systems connected by a diathermal (heat-permeable) wall: after sufficient time, no net heat flows between the systems, and all macroscopic properties are constant. The zeroth law of thermodynamics formalizes this: If system A is in equilibrium with system B, and system A is in equilibrium with system C, then system B is in equilibrium with system C. This law establishes that temperature is a universal property that determines whether thermal equilibrium occurs. For all gas law calculations, we must use absolute (Kelvin) temperature, which has a true zero at absolute zero (0 K = -273.15°C, the theoretical point of zero molecular kinetic energy). The conversion is $T(\text{K})=T{(}^{\circ }\text{C})+273.15$, rounded to 273 for most AP problems with whole-number Celsius values.

Worked Example

A car tire gauge reads 220 kPa when measured at sea level, where atmospheric pressure is 101 kPa. The tire is rated for a maximum absolute pressure of 325 kPa to avoid blowout. If the tire warms to thermal equilibrium with 35°C air on the highway, is the tire safe, and what absolute temperature is required for gas law calculations?

1. Recall the relationship between gauge and absolute pressure: $P_{\text{abs}}=P_{\text{gauge}}+P_{\text{atm}}$.
2. Calculate the absolute pressure of the tire: $P_{\text{abs}}=220\text{kPa}+101\text{kPa}=321\text{kPa}$.
3. Convert Celsius temperature to absolute Kelvin: $T(\text{K})=35+273=308\text{K}$.
4. Compare the absolute pressure to the maximum rating: 321 kPa < 325 kPa, so the tire is within the safety limit.

Exam tip: AP exam questions almost always give temperatures in Celsius for context, but require absolute temperature for all gas law calculations. Convert to Kelvin first, before plugging values into any formula, no exceptions.

3. Combined Gas Law for Closed Systems

Before the ideal gas law was formalized, four separate empirical gas laws described gas behavior under changing conditions for a fixed amount of gas:

• Boyle's Law (constant temperature): Pressure is inversely proportional to volume: $P\propto 1/V$
• Charles's Law (constant pressure): Volume is proportional to absolute temperature: $V\propto T$
• Gay-Lussac's Law (constant volume): Pressure is proportional to absolute temperature: $P\propto T$
• Avogadro's Law (constant P, T): Volume is proportional to moles of gas: $V\propto n$

All of these can be combined into a single relationship that applies to any change of state for a gas, called the combined gas law: $$\frac{P_1{V}_1}{n_1{T}_1}=\frac{P_2{V}_2}{n_2{T}_2}$$ For a closed system (no gas added or removed, so $n_1=n_2=\text{constant}$), this simplifies to the commonly used form: $$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$$ Intuition: For a fixed amount of gas, if you reduce volume, molecules collide with the walls more often, increasing pressure. If you increase temperature, molecules move faster, so if pressure is held constant, volume expands to keep collision frequency constant. This formula is ideal for problems where you know all but one variable for an initial and final state of a gas sample.

Worked Example

A fixed sample of gas has an initial volume of 2.0 L, initial pressure of 1.0 atm, and initial temperature of 27°C. The gas is compressed to 0.5 L, and the temperature rises to 127°C. What is the final pressure of the gas?

1. Convert temperatures to absolute Kelvin first: $T_1=27+273=300\text{K}$, $T_2=127+273=400\text{K}$.
2. The sample is fixed, so $n$ is constant, and we use the simplified combined gas law: $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$.
3. Rearrange to solve for $P_2$: $P_2=P_1\frac{V_1{T}_2}{V_2{T}_1}$.
4. Substitute values: $P_2=(1.0\text{atm})\frac{(2.0\text{L})(400\text{K})}{(0.5\text{L})(300\text{K})}\approx 5.3\text{atm}$.
5. Check intuition: Volume decreased by a factor of 4, temperature increased by a factor of 4/3, so pressure should increase by a factor of ~5.3, which matches our result.

Exam tip: When using the combined gas law, pressure and volume units only need to be consistent across the initial and final states (you can use liters and atm instead of SI units). Temperature must always be in Kelvin, no exceptions, even if units cancel out.

4. The Ideal Gas Law (Two Forms)

The combined gas law tells us that $\frac{PV}{nT}$ is a constant for all ideal gases, called the universal gas constant $R$. This gives the first and most commonly used form of the ideal gas law, the molar form: $$PV=nRT$$ where $n$ is the number of moles of gas, and $R$ has two common values depending on your units: $R=8.314\text{J}/(\text{molcdotpK})$ for SI units (pressure in Pa, volume in m³) and $R=0.0821\text{Lcdotpatm}/(\text{molcdotpK})$ for pressure in atm and volume in liters.

A second form, used when counting individual molecules (common for kinetic theory problems), replaces moles with number of molecules $N$. Since $n=N/N_A$, where $N_A=6.022\times 10^{23}{\text{mol}}^{-1}$ is Avogadro's number, substituting gives: $$PV=\frac{N}{N_A}RT=N\left(\frac{R}{N_A}\right)T=N{k}_{B}T$$ where $k_B=1.38\times 10^{-23}\text{J/K}$ is Boltzmann's constant. The molecular form is: $$PV=N{k}_{B}T$$ An ideal gas is defined as a gas where molecular volume is negligible compared to total volume, there are no intermolecular forces, and all collisions are elastic. Real gases follow this law closely at low pressure and high temperature, which is the assumption for nearly all AP Physics 2 gas problems.

Worked Example

A 1.5 m³ scuba tank holds 200 moles of air at 290 K. What is the absolute pressure of the air inside the tank, in Pascals? What is the corresponding gauge pressure, if atmospheric pressure is $1.01\times 10^5\text{Pa}$?

1. We know $n$, $V$, and $T$, so use the molar form of the ideal gas law: $PV=nRT$, rearranged to $P=\frac{nRT}{V}$.
2. Volume is in SI units (m³) and pressure is requested in Pascals, so use $R=8.314\text{J}/(\text{molcdotpK})$.
3. Substitute values: $P=\frac{(200\text{mol})(8.314\text{J}/(\text{molcdotpK}))(290\text{K})}{1.5{\text{m}}^3}\approx 3.22\times 10^5\text{Pa}$.
4. Calculate gauge pressure: $P_{\text{gauge}}=P_{\text{abs}}-P_{\text{atm}}=3.22\times 10^5\text{Pa}-1.01\times 10^5\text{Pa}=2.21\times 10^5\text{Pa}$.
5. This result is reasonable: a typical scuba tank has a gauge pressure of ~2-3 atm, which matches our result of ~2.2 atm gauge.

Exam tip: Always match the value of R to your pressure and volume units. Using 8.314 with liters and atm will give a pressure off by three orders of magnitude, a common mistake that costs free-response points.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using Celsius temperature directly in the ideal gas law or combined gas law without converting to Kelvin. Why: Most problems give temperatures in Celsius for real-world context, and students forget gas laws depend on absolute temperature. Correct move: Add 273 to any Celsius temperature as the first step in any gas law calculation, before plugging values into formulas.
• Wrong move: Using gauge pressure directly as absolute pressure in the ideal gas law. Why: Most practical gauges measure gauge pressure, so students forget to add atmospheric pressure to get the absolute pressure required by the law. Correct move: Always check if a given pressure is gauge or absolute; add atmospheric pressure to any gauge pressure before calculation.
• Wrong move: Mixing units for R in the ideal gas law, e.g., using R = 8.314 with pressure in atm and volume in liters. Why: Students memorize R but forget its value depends on the units of P and V. Correct move: If using SI units (Pa, m³), use R = 8.314; if using atm and liters, use R = 0.0821; double-check units before calculation.
• Wrong move: Forgetting that n is constant when using the simplified combined gas law $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$. Why: Students use the simplified version by default even when gas is added or removed from the system. Correct move: Always include n on both sides of the combined gas law if the amount of gas changes; only drop n if the system is closed (fixed amount of gas).
• Wrong move: Treating the inverse relationship in Boyle's law as a direct relationship, e.g., calculating $P_2=P_1(V_2/V_1)$ instead of $P_2=P_1(V_1/V_2)$. Why: Students rush to plug into the formula without checking proportionality. Correct move: Always check your answer against intuition: if volume decreases, pressure should increase, so adjust your algebra if the result contradicts intuition.
• Wrong move: Using the ideal gas law for high-pressure, low-temperature real gases and expecting an exact result. Why: Students assume the ideal gas law applies to all gases in all problems. Correct move: On the AP exam, the problem will always state that you can treat the gas as ideal, so only use the ideal gas law when that assumption is given or implied.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A sealed balloon is heated from 27°C to 127°C at constant atmospheric pressure. The initial volume of the balloon is 3.0 L. What is the approximate final volume of the balloon? A) 1.0 L B) 2.3 L C) 4.0 L D) 14 L

Worked Solution: This is a constant pressure process for a sealed (fixed n) balloon, so we use Charles's law, which requires absolute temperature. Convert temperatures to Kelvin: $T_1=27+273=300\text{K}$, $T_2=127+273=400\text{K}$. Charles's law at constant P is $V_1/T_1=V_2/T_2$, so $V_2=V_1(T_2/T_1)=3.0\text{L}(400/300)=4.0\text{L}$. Wrong options come from using Celsius directly (option D is the result of using Celsius instead of Kelvin, the most common mistake). Correct answer: C.

Question 2 (Free Response)

A cylindrical container with a movable, frictionless piston holds 0.10 moles of ideal gas at 300 K. The piston has cross-sectional area 0.01 m², and the initial height of the gas column is 0.20 m. Atmospheric pressure outside the piston is $1.0\times 10^5\text{Pa}$, and the piston is initially in equilibrium. (a) Calculate the initial absolute pressure of the gas inside the piston, and the initial volume of the gas. (b) The gas is heated to 400 K, and the piston moves freely to reach a new equilibrium. What is the new height of the gas column? (c) After heating, 0.05 moles of gas leak out of the piston, while temperature remains 400 K and pressure remains constant. What is the new height of the gas column after the leak?

Worked Solution: (a) The piston is in equilibrium, so the pressure inside equals atmospheric pressure outside: $P=1.0\times 10^5\text{Pa}$. Initial volume is $V=Ah=(0.01{\text{m}}^2)(0.20\text{m})=0.0020{\text{m}}^3$. (b) Pressure and moles of gas are constant before the leak, so $V\propto T$, and $V=Ah$ so $h\propto T$. $h_2=h_1(T_2/T_1)=0.20\text{m}(400\text{K}/300\text{K})\approx 0.27\text{m}$. (c) Pressure and temperature are constant after the leak, so $V\propto n$, so $h\propto n$. $h_3=h_2(n_2/n_1)=0.27\text{m}(0.05\text{mol}/0.10\text{mol})=0.135\text{m}\approx 0.14\text{m}$.

Question 3 (Application / Real-World Style)

A standard ping pong ball has a volume of $3.3\times 10^{-5}{\text{m}}^3$, and at 20°C room temperature, the absolute pressure inside the ball is $1.2\times 10^5\text{Pa}$. Estimate the number of air molecules inside the ping pong ball.

Worked Solution: We use the molecular form of the ideal gas law, $PV=N{k}_{B}T$. Convert temperature to Kelvin: $T=20+273=293\text{K}$. Rearrange to solve for $N$: $N=\frac{PV}{k_{B}T}$. Substitute values: $N=\frac{(1.2\times 10^5\text{Pa})(3.3\times 10^{-5}{\text{m}}^3)}{(1.38\times 10^{-23}\text{J/K})(293\text{K})}\approx 9.8\times 10^{20}$ molecules. In context, a standard ping pong ball contains roughly $10^{21}$ air molecules, which matches the expected order of magnitude for gas volumes at room temperature and pressure.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational equation of state for all thermodynamics processes that you will study next in Unit 2. Without a solid understanding of how P, V, T, and n relate to each other, you cannot correctly analyze work done by expanding/contracting gases, heat transfer between systems, or the efficiency of heat engines—all major, high-weight topics that regularly appear on the AP Physics 2 free-response section. This topic also connects to kinetic theory of gases, where the ideal gas law is used to derive the relationship between temperature and average molecular kinetic energy, and to fluid statics from Unit 1, where pressure is a core governing concept.

• First Law of Thermodynamics
• Kinetic Theory of Gases
• Thermodynamic Processes and Cycles
• Heat Engines and Efficiency