Unit 9 · Topic 9.2

9.2 The Ideal Gas Law

The ideal gas law, $PV = nRT = Nk_BT$ , connects a gas's pressure, volume, amount and temperature. It works well for most real gases at everyday conditions and lets you predict how one quantity changes when you change another. Graphs of these quantities also reveal absolute zero.

Key terms

ideal gas
ideal gas law
mole
kelvin (absolute temperature)
absolute zero
universal gas constant

The ideal gas model

An ideal gas is a simplified model with four assumptions. The atoms move in random directions. Each atom's own volume is tiny compared with the container. Collisions are perfectly elastic. And the atoms push on each other only when they collide, with no attraction or repulsion at a distance.

Real gases behave almost ideally when they're thin and warm, because the atoms are far apart and moving fast. They stray from the model at very high pressures or very low temperatures, when atoms are crowded together and their small attractions start to matter.

The ideal gas law

$PV = nRT = Nk_BT$

P is pressure in pascals, V is volume in m³, T is temperature in kelvins, n is the number of moles, and N is the number of atoms. R = 8.31 J/(mol·K) is the universal gas constant. The two forms match because N = $nN_A$ , with Avogadro's number $N_A = 6.02 \times 10^{23}$ per mole, and $k_B = R/N_A$ .

For a sealed container, n stays fixed, so $\dfrac{PV}{T}$ is constant. That gives a quick ratio method:

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

Any quantity held constant cancels. At constant volume, P is proportional to T. At constant temperature, P is inversely proportional to V. At constant pressure, V is proportional to T.

Reading gas graphs

Graphs are a big part of this topic, especially in lab questions.

Graph	Held constant	Shape
P vs. T (kelvins)	V and n	straight line through the origin
V vs. T (kelvins)	P and n	straight line through the origin
P vs. V	T and n	curve that falls off like 1/V (an isotherm)
P vs. 1/V	T and n	straight line through the origin, slope nRT

Finding absolute zero

If you measure the pressure of a fixed volume of gas at several Celsius temperatures and plot P against T, the points fall on a straight line. Extend the line down to where the pressure would be zero, and it hits the temperature axis at about −273 °C. Different gases and different amounts all point to the same spot.

That temperature is absolute zero, 0 K, where the model says the atoms would have no kinetic energy. Real gases turn liquid before they get there, which is why you extrapolate rather than measure. This is why the kelvin scale starts at absolute zero, and why the gas law only works in kelvins.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
How many moles?
A 0.025 m³ container holds gas at 1.0 × 10⁵ Pa and 300 K. How many moles of gas are inside, and about how many atoms?
Show the solutionHide the solution
1. Step 1: Solve PV = nRT for n: $n = \dfrac{PV}{RT} = \dfrac{(1.0 \times 10^5)(0.025)}{(8.31)(300)}$ .
2. Step 2: n ≈ 1.0 mol.
3. Step 3: Atoms: N = $nN_A$ ≈ (1.0)(6.02 × 10²³) ≈ 6.0 × 10²³.
Answer: About 1.0 mol, or about 6.0 × 10²³ atoms
Example 2Calculator allowed
Heating a rigid tank (classic trap)
A rigid, sealed tank holds gas at 2.0 atm and 27 °C. The tank is heated to 127 °C. What is the new pressure?
Show the solutionHide the solution
1. Step 1: Rigid and sealed means V and n stay the same, so P/T is constant: $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ .
2. Step 2: Convert to kelvins: T₁ = 300 K and T₂ = 400 K.
3. Step 3: $P_2 = P_1 \dfrac{T_2}{T_1} = (2.0 \text{ atm})\dfrac{400}{300} \approx 2.7$ atm.
4. Step 4: The trap: using Celsius gives 2.0 × 127/27 ≈ 9.4 atm, which is far too big. Pressure units can stay in atm because they cancel in the ratio; temperature can't stay in °C.
Answer: About 2.7 atm
Example 3Calculator allowed
Extrapolating to absolute zero
A student keeps a fixed volume of gas sealed and measures its pressure: 100.0 kPa at 0 °C and 136.6 kPa at 100 °C. Assuming the graph of P against T is a straight line, at what temperature would the pressure be zero?
Show the solutionHide the solution
1. Step 1: Slope = (136.6 − 100.0) kPa / (100 − 0) °C = 0.366 kPa/°C.
2. Step 2: Starting at 100.0 kPa at 0 °C, the pressure drops 0.366 kPa for every degree colder. Zero pressure needs a drop of 100.0 kPa: 100.0 / 0.366 ≈ 273 degrees.
3. Step 3: So the line reaches P = 0 at about −273 °C.
Answer: About −273 °C, which is absolute zero

Common mistakes

Plugging Celsius into PV = nRT or into a ratio. Convert to kelvins every time.
Mixing up n and N. Use R with moles (n) and $k_B$ with the number of atoms (N).
Forgetting volume units. PV = nRT needs m³ and Pa; 1 L = 1.0 × 10⁻³ m³.
Expecting P vs. V at constant temperature to be a straight line. It's a curve; plot P against 1/V to get a line.

On the exam

Lab questions often ask you to linearize gas data, for example plotting P against 1/V and using the slope to find nRT or the temperature.
Factor-of-change questions show up often: say what's held constant, then use the proportion it leaves.
Be ready to explain the gas law with kinetic theory: hotter atoms hit harder and more often, so pressure rises at fixed volume.

Connected topics

Videos

AP Physics 2 - Unit 9 - Lesson 2 - Ideal Gas Law
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
AP Physics 2 - Ideal Gases
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Thermodynamics part 2: Ideal gas law | Thermodynamics | Physics | Khan Academy
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Check yourself

4 questions on 9.2 The Ideal Gas Law. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

What volume does 2.0 mol of an ideal gas occupy at a temperature of 300 K and a pressure of $1.0 \times 10^5$ Pa?

Question 2 of 4Calculator allowed

A 3.0 L tank holds an ideal gas at a pressure of $2.0 \times 10^5$ Pa and a temperature of 400 K. About how many atoms are in the tank? (1 L = $1.0 \times 10^{-3}$ m³)

Question 3 of 4Calculator allowed

An ideal gas in a sealed, rigid container is heated from 27 °C to 127 °C. By what factor does its pressure change?

Question 4 of 4Calculator allowed

A fixed amount of ideal gas has its pressure doubled while its volume is tripled. What happens to its absolute temperature?

0 of 4 answered