# Pressure, Thermal Equilibrium and Ideal Gas Law

> AP Physics 2 · Unit 2 Thermodynamics
> Source: https://www.owlsprep.com/study/ap-physics-2-u2-pressure-thermal-equilibrium-and-ideal/

This module covers absolute vs gauge pressure, the zeroth law of thermodynamics, thermal equilibrium, absolute temperature, the combined gas law, and both forms of the ideal gas law for AP Physics 2.

**Prerequisites:** Force and area definitions from mechanics; Basic kinetic theory of matter; SI unit conversion rules

## Learning objectives

- Distinguish between absolute pressure and gauge pressure
- State the zeroth law of thermodynamics and thermal equilibrium
- Convert temperature between Celsius and Kelvin
- Apply the combined gas law to closed systems
- Use both molar and molecular forms of the ideal gas law to solve problems

## Pressure and the Zeroth Law of Thermal Equilibrium

**Pressure** — The magnitude of the perpendicular force per unit area exerted by a substance on a boundary. For gases, pressure arises from elastic collisions of molecules with the container wall. Pressure is a scalar quantity.

*Notation:* P

*Example:* Gas molecules colliding with the inside wall of a balloon create outward pressure.

The SI unit of pressure is the pascal, where $1\ \text{Pa} = 1\ \text{N}/\text{m}^2$. A critical distinction for problem-solving is between **absolute pressure** and **gauge pressure**: gauge pressure measures pressure relative to atmospheric pressure.

$$P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}$$

**Thermal Equilibrium** — The steady state two systems reach when in thermal contact, with no net heat transfer between them, and all macroscopic properties remain constant.

The zeroth law of thermodynamics formalizes the concept of temperature: 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 establishes that temperature is the property that determines thermal equilibrium. For all gas law calculations, you must use absolute (Kelvin) temperature, with conversion: $T(\text{K}) = T(^\circ\text{C}) + 273.15$, rounded to 273 for most AP problems.

**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 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.

## Combined Gas Law for Closed Systems

Four separate empirical gas laws describe gas behavior for different fixed conditions. All of these can be combined into a single general relationship, called the combined gas law, that applies to any change of state for a gas sample.

- Boyle's Law (constant temperature): $P \propto 1/V$
- Charles's Law (constant pressure): $V \propto T$
- Gay-Lussac's Law (constant volume): $P \propto T$
- Avogadro's Law (constant P, T): $V \propto n$

$$\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$$

For a closed system (no gas added or removed, so $n_1 = n_2$), this simplifies to the commonly used form:

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

This formula is ideal for problems where you know all but one variable for an initial and final state of a gas sample. Intuition: reducing volume increases collision frequency and pressure; increasing temperature increases molecular speed, so volume expands at constant pressure.

**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

   $$T_1 = 27 + 273 = 300\ \text{K}, \quad T_2 = 127 + 273 = 400\ \text{K}$$
2. The sample is fixed, so $n$ is constant, use the simplified combined gas law

   $$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$
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 increases by a factor of ~5.3, which matches the result.

> **Exam tip:** When using the combined gas law, pressure and volume units only need to be consistent across initial and final states. Temperature must always be in Kelvin, no exceptions, even if units cancel out.

## The Ideal Gas Law (Two Forms)

The combined gas law tells us that $\frac{PV}{nT}$ is a universal constant for all ideal gases, called the universal gas constant $R$. This gives the most common molar form of the ideal gas law, the equation of state for an ideal gas.

$$PV = nRT$$

Where $n$ is the number of moles of gas, and $R$ has two common values depending on units: $R = 8.314\ \text{J}/(\text{mol·K})$ for SI units (pressure in Pa, volume in m³), and $R = 0.0821\ \text{L·atm}/(\text{mol·K})$ for pressure in atm and volume in liters. A second form, used for counting individual molecules in kinetic theory problems, replaces moles with number of molecules $N$.

**Derivation:** Derive the molecular form of the ideal gas law

*Starting from:* Molar form $PV = nRT$ and $n = N/N_A$

1. Substitute $n = N/N_A$ into the molar form, where $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ is Avogadro's number:
2. $$PV = \frac{N}{N_A}RT = N \left(\frac{R}{N_A}\right) T$$
3. The term $\frac{R}{N_A}$ is Boltzmann's constant, $k_B = 1.38 \times 10^{-23}\ \text{J/K}$.

*Conclusion:* This gives the molecular form of the ideal gas law:

$$PV = Nk_B T$$

An ideal gas is defined as a gas where molecular volume is negligible, there are no intermolecular forces, and all collisions are elastic. Real gases follow this law closely at low pressure and high temperature, the standard assumption for 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. Use the molar form of the ideal gas law, rearranged to solve for pressure

   $$P = \frac{nRT}{V}$$
2. We use $R = 8.314\ \text{J}/(\text{mol·K})$ since we have SI units (volume in m³, pressure requested in Pa)
3. Substitute values

   $$P = \frac{(200\ \text{mol})(8.314\ \text{J}/(\text{mol·K}))(290\ \text{K})}{1.5\ \text{m}^3} \approx 3.22 \times 10^5\ \text{Pa}$$
4. Calculate gauge pressure from absolute 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.

> **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.

## Concept Check

**Check your understanding**

Test your understanding with these AP-style multiple choice questions

1. 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?

   - 1.0 L
   - 2.3 L
   - 4.0 L
   - 14 L

   *Why:* Correct: You converted to absolute temperature and applied Charles's law correctly. The wrong answer 14 L comes from using Celsius directly instead of Kelvin, the most common mistake.

2. A ping pong ball with volume $3.3 \times 10^{-5}\ \text{m}^3$ has an internal absolute pressure of $1.2 \times 10^5\ \text{Pa}$ at 20°C. What is the approximate number of air molecules inside the ball?

   - $9.8 \times 10^{20}$
   - $9.8 \times 10^{18}$
   - $9.8 \times 10^{23}$
   - $1.2 \times 10^5$

   *Why:* Correct: You used the molecular form $PV = Nk_B T$ and converted temperature correctly to Kelvin.

## Common pitfalls

- **Wrong:** Using Celsius temperature directly in gas laws without converting to Kelvin
  - Why it fails: Most problems give temperatures in Celsius for context, and students forget gas laws depend on absolute temperature.
  - Correct: Add 273 to any Celsius temperature as the first step in any gas law calculation, before plugging values into formulas.
- **Wrong:** Using gauge pressure directly as absolute pressure in the ideal gas law
  - Why it fails: Most practical gauges measure gauge pressure, so students forget to add atmospheric pressure to get the absolute pressure required by the law.
  - Correct: Always check if a given pressure is gauge or absolute; add atmospheric pressure to any gauge pressure before calculation.
- **Wrong:** Mixing units for R in the ideal gas law, e.g., using R = 8.314 with pressure in atm and volume in liters
  - Why it fails: Students memorize R but forget its value depends on the units of P and V.
  - Correct: 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:** Forgetting that n is constant when using the simplified combined gas law
  - Why it fails: Students use the simplified version by default even when gas is added or removed from the system.
  - Correct: 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:** 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 it fails: Students rush to plug into the formula without checking proportionality.
  - Correct: Always check your answer against intuition: if volume decreases, pressure should increase, so adjust your algebra if the result contradicts intuition.
- **Wrong:** Using the ideal gas law for high-pressure, low-temperature real gases and expecting an exact result
  - Why it fails: Students assume the ideal gas law applies to all gases in all problems.
  - Correct: 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.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Pressure Definition | $P = F_\perp / A$ | P is scalar, SI unit: Pa = N/m² |
| Gauge vs Absolute Pressure | $P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}$ | Always use absolute pressure in gas laws |
| Temperature Conversion | $T(\text{K}) = T(^\circ\text{C}) + 273.15$ | Always convert to Kelvin for all gas law calculations |
| Combined Gas Law | $\frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2}$ | Simplify to $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ for fixed n |
| Ideal Gas Law (Molar Form) | $PV = nRT$ | $R = 8.314$ for Pa, m³; $R = 0.0821$ for atm, L |
| Ideal Gas Law (Molecular Form) | $PV = Nk_B T$ | $k_B = 1.38 \times 10^{-23}$ J/K, N = number of molecules |
| Boyle's Law (constant n, T) | $P_1V_1 = P_2V_2$ | P inversely proportional to V |
| Charles's Law (constant n, P) | $V_1/T_1 = V_2/T_2$ | V directly proportional to absolute T |
| Gay-Lussac's Law (constant n, V) | $P_1/T_1 = P_2/T_2$ | P directly proportional to absolute T |
| Avogadro's Law (constant P, T) | $V_1/n_1 = V_2/n_2$ | V directly proportional to moles of gas |
| Zeroth Law of Thermodynamics | If A ⇌ B, A ⇌ C, then B ⇌ C | Establishes temperature as the property that determines thermal equilibrium |

## What's next

This subtopic is the foundational equation of state for all thermodynamics processes you will study next in AP Physics 2 Unit 2. Without a solid understanding of how pressure, volume, temperature, and moles of gas relate to each other, you cannot correctly analyze work done by expanding or contracting gases, heat transfer between systems, or the efficiency of heat engines. All of these are high-weight topics that regularly appear on the AP Physics 2 free-response section, so mastering this material is critical for earning a high score. This topic also connects directly to kinetic theory of gases, where you will use the ideal gas law to relate temperature to average molecular kinetic energy.

- [Unit 2 Thermodynamics Overview](https://www.owlsprep.com/study/ap-physics-2-u2-overview/)
- [Heat and Energy Transfer](https://www.owlsprep.com/study/ap-physics-2-u2-heat-and-energy-transfer/)
- [Thermal Processes](https://www.owlsprep.com/study/ap-physics-2-u2-thermal-processes/)

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