Squeeze a sealed syringe and the air inside takes up less room. Leave a party balloon in a hot car and it swells. Heat a closed can and the pressure inside climbs until the can may burst. In each case the amount of gas stays the same while its pressure, volume or temperature changes. The gas laws for changing conditions predict the new state from the old one.

In this lesson you will use Boyle’s, Charles’s and Gay-Lussac’s laws and the combined gas law for a fixed amount of gas, and you will correct a measured gas volume to standard conditions, saying which standard you use (Figure 1).

1. State 1 and state 2P, V, T before and after
2. $+ 273.15$every T to K; same P and V units
3. Consistent units
4. $\dfrac{PV}{T}$cross out what is constant
5. One lawBoyle, Charles, Gay-Lussac or combined
6. $\times \dfrac{T_2}{T_1}$one factor per change
7. Answercheck the direction; 3 s.f.

$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ with T in K.No R is needed: the units cancel.STP: $273.15\,\text{K},\ 1\,\text{atm}$.

Learning Objectives

After this lesson, you should be able to:

1. Use the combined gas law (and Boyle, Charles, Gay-Lussac) for a fixed amount of gas.
2. Convert gas volumes to standard conditions, stating which standard is used.

Key Ideas

In The Ideal Gas Law you used $PV = nRT$ for one set of conditions. Here a sample of gas goes from a first state (subscript 1, before) to a second state (subscript 2, after). No gas gets in or out, so $n$ is constant, and $R$ is a constant. Rearranging $PV = nRT$ to $\frac{PV}{T} = nR$ shows that $\frac{PV}{T}$ has the same value in both states:

This is the combined gas law. Because the same quantity appears on both sides, the units of $P$ and $V$ cancel: use any pressure unit and any volume unit, as long as both states use the same one. You do not need a value of $R$. The temperature is different: it must be in K, because only the Kelvin scale starts at zero (Key Idea 2).

Key idea: for a fixed amount of gas, $\frac{PV}{T}$ is constant. Write state 1 and state 2 side by side, put both temperatures in K, and keep the same pressure and volume units in both states.

1. Boyle’s Law: Pressure and Volume

If the temperature also stays constant, $T_1 = T_2$ cancels from the combined gas law and leaves Boyle’s law:

Pressure and volume are inversely proportional: double the pressure and the volume halves. On a graph of $V$ against $P$ every state of the sample lies on one curve (Figure 2). Solving for the new volume gives $V_2 = V_1 \times \frac{P_1}{P_2}$: when the pressure goes up, the pressure ratio is less than 1 and the volume goes down.

Key idea: Boyle’s law holds at constant $T$ and $n$. Higher pressure, smaller volume.

2. Charles’s Law: Volume and Absolute Temperature

If the pressure stays constant (a balloon or a freely moving piston), $P_1 = P_2$ cancels and leaves Charles’s law:

The volume is proportional to the Kelvin temperature. Extend the straight line of volume against temperature to lower temperatures and it reaches zero volume at 0 K, which is −273.15 °C, absolute zero (Figure 3, panel (a)). A real gas liquefies long before that, but the line shows where the Kelvin scale starts.

This is why Celsius temperatures never go into a gas-law ratio. The zero of the Celsius scale is just the freezing point of water, so a ratio of Celsius temperatures means nothing. Warming a gas from 22.0 °C to 85.0 °C multiplies its volume by 1.21 (the Kelvin ratio), not by 3.86 (the Celsius ratio), as Figure 3, panel (b) shows.

Key idea: Charles’s law holds at constant $P$ and $n$, with $T$ in K. Hotter gas, larger volume.

3. Gay-Lussac’s Law and the Combined Gas Law

In a rigid container the volume cannot change. With $V_1 = V_2$, the combined gas law becomes Gay-Lussac’s law: the pressure is proportional to the Kelvin temperature.

Each of the three simple laws is the combined gas law with one more variable held constant: cross it out of both sides (Figure 4). When pressure and temperature both change, keep the full combined law. Solving it for the new volume gives one factor for each change:

Check each factor against what the change should do. A higher pressure should shrink the gas, so its pressure factor must be less than 1; a higher temperature should expand it, so its temperature factor must be more than 1.

Combined (only $n$ constant): $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$

Boyle ($T$ also constant): $P_1V_1 = P_2V_2$

Charles ($P$ also constant): $\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$

Gay-Lussac ($V$ also constant): $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$

Key idea: cross out what is held constant. Then check that each pressure or temperature factor pushes the answer in the direction you expect.

4. Standard Conditions

A gas volume means little unless you know its pressure and temperature, so measured volumes are often corrected to a standard set of conditions. This lesson uses STP as Whitten defines it: 273.15 K (0 °C) and exactly 1 atm, which is 760 torr or 101.325 kPa. To correct a volume to STP, use the combined gas law with state 2 at STP.

Note: IUPAC defines standard pressure as 1 bar (100 kPa) at the same 273.15 K. One mole of ideal gas occupies 22.414 L at 1 atm and 22.711 L at 1 bar, about 1.3% more (Figure 5). Every problem in this lesson says which standard it uses; do the same in your answers.

STP (this lesson, Whitten): $273.15\,\text{K},\ 1\,\text{atm}$

$1\,\text{atm} = 760\,\text{torr} = 101.325\,\text{kPa}$; molar volume $22.414\,\text{L/mol}$

IUPAC standard: $273.15\,\text{K},\ 1\,\text{bar}$

$1\,\text{bar} = 100\,\text{kPa}$; molar volume $22.711\,\text{L/mol}$

Key idea: STP here is 273.15 K and 1 atm (760 torr, 101.325 kPa). The IUPAC standard is 273.15 K and 1 bar. Always name the standard.

Try it: change the conditions of a gas

Choose a law, set state 1 and state 2, and predict whether the new value will be larger, smaller or the same before the answer appears. Try the Celsius ratio to see why it fails, set state 2 to STP or to the IUPAC standard, and use the worked-example buttons to replay each example.

5. A Strategy for Changing Conditions

Follow the roadmap in Figure 1:

1. Make a table of state 1 and state 2 with units, and mark the unknown.
2. Convert every temperature to K. Put both pressures in the same unit and both volumes in the same unit.
3. Choose the law: cross out what is constant ($T$ for Boyle, $P$ for Charles, $V$ for Gay-Lussac; nothing but $n$ for the combined law).
4. Solve for the unknown and substitute. For a standard volume, state 2 is the named standard.
5. Check the direction of every factor, and round the answer to 3 significant figures.

Exam tip: predict before you calculate. Decide whether the new volume, pressure or temperature should be larger or smaller; if your answer moves the other way, a ratio is upside down.

Worked Example 1: Compressing a Gas (Boyle’s Law)

Problem: A cylinder with a movable piston holds 6.80 L of helium at 1.35 atm. At constant temperature the piston pushes in until the pressure is 2.25 atm. What is the new volume, in L?

Step 1. List the two states. The temperature and the amount of gas are constant, so use Boyle’s law. Both pressures are in atm.

Step 2. Predict. The pressure goes up, so the volume must go down.

Step 3. Solve $P_1V_1 = P_2V_2$ for $V_2$ and substitute. The atm cancel, leaving L.

Final Answer: 4.08 L

Check: the pressure factor, 0.600, is less than 1, so the volume went down, as predicted.

1. State 1$1.35\,\text{atm},\ 6.80\,\text{L}$
2. $\times \dfrac{1.35\,\text{atm}}{2.25\,\text{atm}}$Steps 1 to 3: P up to 2.25 atm: factor below 1
3. New volume$V_2 = 4.080\,\text{L} = 4.08\,\text{L}$smaller, as predicted

Your turn: Boyle’s law

Same steps as Worked Example 1: list both states, predict, then use $P_1V_1 = P_2V_2$ (the unknown may be a volume or a pressure).

Worked Example 2: Warming a Balloon (Charles’s Law)

Problem: A balloon holds 2.75 L of air at 22.0 °C. It is warmed to 85.0 °C at constant pressure. What is its new volume, in L?

Step 1. Convert both temperatures to K.

Step 2. Predict. The pressure is constant and the gas gets hotter, so use Charles’s law; the volume must go up.

Step 3. Solve $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ for $V_2$ and substitute. The K cancel, leaving L.

Final Answer: 3.34 L

Check: the temperature factor, 1.21, is more than 1, so the volume went up, as predicted. Using the Celsius temperatures would give 10.6 L, nearly four times the starting volume, which is far too large for a modest warming.

1. Temperatures$22.0\,\text{°C} \rightarrow 85.0\,\text{°C}$
2. $+ 273.15$Step 1: both to kelvins
3. Temperatures in K$295.15\,\text{K} \rightarrow 358.15\,\text{K}$
4. $\times \dfrac{358.15\,\text{K}}{295.15\,\text{K}}$Steps 2 and 3: V₁ 2.75 L; factor above 1
5. New volume$V_2 \approx 3.337\,\text{L} \approx 3.34\,\text{L}$larger, as predicted

Your turn: Charles’s law

Same steps as Worked Example 2: both temperatures to K, predict, then $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (the unknown may be a volume or a temperature; give a temperature in K).

Worked Example 3: Pressure and Temperature Both Change

Problem: A weather balloon holds 485 mL of helium at 96.5 kPa and 28.5 °C. A test chamber changes the conditions to 185 kPa and −15.5 °C. What is the new volume, in mL?

Step 1. Convert both temperatures to K. Both pressures are in kPa.

Step 2. Predict. Only the amount of gas is constant, so use the combined gas law. The pressure goes up and the temperature goes down: both changes shrink the gas.

Step 3. Solve $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ for $V_2$ and substitute. The kPa and the K cancel, leaving mL.

Final Answer: 216 mL

Check: both factors are less than 1, so the volume went down, as predicted.

If the volume does not change, $V$ cancels and this is Gay-Lussac’s law, $\frac{P_1}{T_1} = \frac{P_2}{T_2}$: the same steps then give a new pressure or a new temperature.

1. Temperatures$28.5\,\text{°C} \rightarrow -15.5\,\text{°C}$
2. $+ 273.15$Step 1: both to kelvins
3. Temperatures in K$301.65\,\text{K} \rightarrow 257.65\,\text{K}$
4. $485\,\text{mL} \times \dfrac{96.5\,\text{kPa}}{185\,\text{kPa}}$Step 3: P up: factor below 1
5. After the pressure factor$253.0\,\text{mL}$rounded to 4 s.f.
6. $\times \dfrac{257.65\,\text{K}}{301.65\,\text{K}}$Step 3: T down: factor below 1
7. New volume$V_2 \approx 216.1\,\text{mL} \approx 216\,\text{mL}$smaller, as predicted

Your turn: the combined gas law

Same steps as Worked Example 3: both temperatures to K, predict the effect of each change, then solve the combined gas law for the unknown.

Worked Example 4: Correcting a Volume to STP

Problem: A student collects 3.65 L of hydrogen at 742 torr and 24.5 °C. What volume would the gas occupy at STP (273.15 K and 1 atm), in L?

Step 1. Write state 2 as the standard and convert the measured temperature to K. Use 760 torr for 1 atm so that both pressures are in torr.

Step 2. Predict. The pressure rises a little and the temperature drops, so the volume must go down.

Step 3. Solve the combined gas law for $V_2$ and substitute. The torr and the K cancel, leaving L.

Final Answer: 3.27 L at STP (273.15 K and 1 atm)

At the IUPAC standard (273.15 K and 1 bar), $P_2$ is 100 kPa. Converting 742 torr gives 98.93 kPa (rounded to 4 significant figures; the unrounded value gives the same answer), and the volume is $3.314\,\text{L} \approx 3.31\,\text{L}$, about 1.3% larger. The two standards give different answers, so always say which one you used.

1. Measured state$3.65\,\text{L},\ 742\,\text{torr},\ 24.5\,\text{°C}$
2. $+ 273.15$Step 1: state 2 is STP
3. Standard state$P_2 = 760\,\text{torr},\ T_2 = 273.15\,\text{K}$STP: 273.15 K and 1 atm
4. $\times \dfrac{742\,\text{torr}}{760\,\text{torr}}$ $\times \dfrac{273.15\,\text{K}}{297.65\,\text{K}}$Steps 2 and 3: T₁ 297.65 K; both factors below 1
5. Volume at STP$V_2 \approx 3.270\,\text{L} \approx 3.27\,\text{L}$smaller, as predicted

Your turn: a volume at standard conditions

Same steps as Worked Example 4: state 2 is the named standard, temperatures in K, the same pressure unit in both states. Read which standard each problem uses.

Common Mistakes

Throughout, keep at least 4 significant figures in intermediate values and round only the final answer to 3.

Common mistake: Using Celsius temperatures in a ratio. In Worked Example 2, $\frac{85.0}{22.0}$ gives 10.6 L instead of 3.34 L. In Worked Example 3 a Celsius ratio would even give a negative volume, which is impossible. Add 273.15 to every temperature first.

Common mistake: Turning a ratio upside down. In Worked Example 1, $V_1 \times \frac{P_2}{P_1}$ gives 11.3 L: a larger volume at a higher pressure, which cannot be right. In Worked Example 2 the inverted temperature ratio gives 2.27 L for a gas that was warmed. Predict the direction first, then check each factor.

Common mistake: Leaving out the temperature change. In Worked Example 3, using only the pressure factor gives 253 mL. Both the pressure and the temperature changed, so the combined gas law needs both factors.

Common mistake: Leaving out the pressure change. In Worked Example 3, using only the temperature factor gives 414 mL. Include every variable that changes.

Common mistake: Using 298.15 K as the standard temperature. Standard temperature is 273.15 K (0 °C), not room temperature. In Worked Example 4, using 298.15 K gives 3.57 L instead of 3.27 L.

Common mistake: Mixing pressure units. In Worked Example 4, $P_1$ is in torr, so 1 atm must be written as 760 torr. Writing 1 for $P_2$ while $P_1$ is in torr would make the volume 760 times too large.

Key Takeaway

For a fixed amount of gas, $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$. Hold $T$ constant and it becomes Boyle’s law; hold $P$ constant and it becomes Charles’s law; hold $V$ constant and it becomes Gay-Lussac’s law. Temperatures must be in K; pressures and volumes need only the same unit in both states. To correct a volume to standard conditions, make state 2 the standard and say which one: STP here is 273.15 K and 1 atm; the IUPAC standard is 273.15 K and 1 bar.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures and simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0. Data: Whitten, Davis, Peck and Stanley, Chemistry, 9th ed. (Brooks/Cole, Cengage Learning, 2010), Section 12-4 (pp. 406 to 409) for Boyle’s law; Section 12-5 (pp. 409 to 411) for Charles’s law, absolute zero and the Kelvin scale; Section 12-6 (p. 411) for standard temperature and pressure; Section 12-7 (pp. 412 to 413) for the combined gas law and Gay-Lussac’s law. Standard molar volumes (Figure 5): from R = N_Ak, the 2019 SI defining constants (CODATA 2022, NIST). The simulator and the practice problems are audited; see the lesson audit log.