The combined gas law: P, V, and T together

When everything changes at once

Real situations rarely change one variable at a time. A weather balloon climbs into thinner air and colder air simultaneously; a compressor squeezes gas while heating it. The three simple gas laws each freeze two variables, so none of them alone can handle the climb.

The fix is algebraic. Boyle says $V$ is proportional to $1/P$, Charles says $V$ is proportional to $T$, and Avogadro says $V$ is proportional to $n$. Multiply the proportions and, for a fixed amount of gas, $PV/T$ is constant:

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

A worked example, the balloon climb: 100 L at $P_1 = 1.00$ atm, $T_1 = 293$ K rises to $P_2 = 0.50$ atm, $T_2 = 243$ K. Rearranging, $V_2 = V_1(P_1/P_2)(T_2/T_1) = 100 \times (1.00/0.50) \times (243/293) = 100 \times 2.00 \times 0.829 \approx 166$ L. Halved pressure doubled the volume; cold shrank it back by 17 percent. The net result is expansion, but only the combined law gives the number.

Three laws in one equation

The combined law is a Swiss-army equation: each simple law is hiding inside it. Hold temperature constant ($T_1 = T_2$) and the temperatures cancel, leaving $P_1V_1 = P_2V_2$, Boyle's law. Hold pressure constant and you get $V_1/T_1 = V_2/T_2$, Charles's law. Hold volume constant and $P_1/T_1 = P_2/T_2$ appears, Gay-Lussac's law.

This means one memorized equation replaces three, as long as you remember which variable is being held fixed in each problem. It also gives a built-in error check: if your scenario has constant temperature, your combined-law setup should collapse to Boyle's form.

A worked example of collapsing: a piston compresses 6.00 L of gas at 1.00 atm and 300 K down to 2.00 L at the same 300 K. The full equation gives $P_2 = P_1V_1T_2/(V_2T_1) = 1.00 \times 6.00 \times 300/(2.00 \times 300) = 3.00$ atm, and the temperatures visibly cancel mid-calculation. Recognizing the Boyle's-law shortcut would have skipped the clutter, but the combined form never gives a wrong answer when the conditions are right.

Solving multi-change problems

Combined-law problems reward an orderly setup:

1. Label all six values: $P_1$, $V_1$, $T_1$, $P_2$, $V_2$, $T_2$, marking the unknown.
2. Convert temperatures to kelvins immediately.
3. Isolate the unknown algebraically before substituting numbers.
4. Check that each change pushes the answer the right way: lower pressure expands, higher temperature expands.

A worked example, the scuba fill: a tank holds 12.0 L at 1.00 atm and $20^\circ\text{C}$ (293 K). It is filled to 200 atm and the gas warms to $45^\circ\text{C}$ (318 K) during the fast fill. Wait, that is not our scenario; instead, ask what the pressure reads after the tank cools back to $20^\circ\text{C}$ with the same gas inside. Now $V$ is constant and $n$ is fixed: $P_2 = P_1T_2/T_1 = 200 \times 293/318 \approx 184$ atm. The 'lost' 16 atm never leaked; it was thermal pressure that vanished with the heat. Divers learn to judge a fill only after the tank cools.

Rising air parcels and mountain weather

The atmosphere is the grandest combined-law laboratory. A parcel of air forced over a mountain rises into lower pressure and expands, and expansion cools it, which is why mountaintops are cold and why the windward side of a range gets rain as the cooling parcel's vapor condenses.

A worked example: surface air at 1.00 atm and $25^\circ\text{C}$ (298 K) rises to 0.70 atm and cools to $5^\circ\text{C}$ (278 K). A 1.00 m^3 parcel becomes $V_2 = 1.00 \times (1.00/0.70) \times (278/298) ≈ 1.33$ m^3. It expanded by a third, doing work against the surrounding air, and that work is exactly where its lost thermal energy went. Meteorologists run this calculation continuously; it is the physics behind the lapse rate, the steady cooling of air with altitude.

Knowing when to upgrade to PV = nRT

The combined law's boundary is the amount of gas: $n$ must stay fixed. Leaks, additions, and chemical reactions that consume or produce gas all break it. When moles change, step up to the ideal gas law, $PV = nRT$, which is the combined law with $n$ promoted to a full variable.

Extreme conditions are the other boundary. Near liquefaction or at very high pressure, $PV/T$ drifts from constancy because particle attractions and particle volume stop being negligible. The van der Waals equation patches both.

A worked example of choosing correctly: a 10.0 L tank at 5.00 atm and 300 K develops a leak, dropping to 4.00 atm at 300 K. The combined law is useless here because gas escaped. The ideal gas law gives $n_1 = P_1V/RT = (5.00 \times 10.0)/(0.08206 \times 300) = 2.03$ mol and $n_2 = (4.00 \times 10.0)/(0.08206 \times 300) = 1.62$ mol, so about 0.41 mol escaped. Match the equation to the variables that actually change.

Common mistakes

Real-world applications

• Tracking weather balloon ascent through falling pressure and temperature

• Designing compressors that heat gas while squeezing it

• Predicting scuba tank pressure after filling and cooling

• Modeling air parcels rising and sinking in the atmosphere