The Combined Gas Law

Notice that this single equation contains Boyle's, Charles's, and Gay-Lussac's laws as special cases: if temperature is held constant ($T_1=T_2$), the $T$'s cancel and you're left with $P_1{V}_1=P_2{V}_2$ — Boyle's law. If pressure is held constant, the $P$'s cancel, leaving Charles's law. If volume is held constant, the $V$'s cancel, leaving Gay-Lussac's law. You never actually need to memorize all four equations separately — the combined gas law is the master equation, and the simple laws are just what's left over when one variable doesn't change.

You can build the combined gas law from the simple laws by imagining the change happening in two stages. Start with a sample at $P_1$, $V_1$, $T_1$. Stage 1: hold the temperature at $T_1$ and change the pressure to $P_2$. Boyle's law gives an in-between volume $V_x=\frac{P_1{V}_1}{P_2}$. Stage 2: hold the pressure at $P_2$ and change the temperature to $T_2$. Charles's law scales that volume by the temperature ratio: $V_2=V_x\times \frac{T_2}{T_1}=\frac{P_1{V}_1{T}_2}{P_2{T}_1}\text{.}$ Rearranging gives $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$. The gas does not care which path it took; only the starting and ending states matter. Another way to say it: for a sealed sample, the quantity $PV/T$ never changes.

1. Convert both temperatures to Kelvin immediately.
2. Make sure pressure and volume use consistent units on both sides (they don't have to be kPa/L specifically, but $P_1$ and $P_2$ must share units, and $V_1$ and $V_2$ must share units).
3. Identify which of the six quantities ($P_1,V_1,T_1,P_2,V_2,T_2$) is unknown.
4. Rearrange algebraically to isolate the unknown, then substitute and calculate.
5. Sanity-check the direction of the answer: does it make physical sense (e.g., did volume increase when you'd expect it to, given the changes in P and T)?

A fast way to set up (and check) any combined gas law problem is the correction-factor method: start with the old value of the unknown and multiply it by one ratio for each variable that changed, choosing each ratio by reasoning instead of algebra. For a new volume, $V_2=V_1\times (\text{pressure ratio})\times (\text{temperature ratio})\text{.}$ If the pressure went UP, the gas is squeezed, so the pressure ratio must be less than 1: use $\frac{P_1}{P_2}$. If the temperature went UP, the gas expands, so the temperature ratio must be greater than 1: use $\frac{T_2}{T_1}$. This is exactly what the algebra gives, but thinking it through catches upside-down fractions. Why may pressure and volume be in any unit, but temperature must be in kelvins? Converting kPa to atm multiplies both $P_1$ and $P_2$ by the same factor, so their ratio does not change. Converting °C to K ADDS 273.15, and adding the same number to the top and bottom of a fraction changes its value. For example, $\frac{20}{10}=2$, but $\frac{293.15}{283.15}\approx 1.04$.

A weather balloon is released with only part of its full volume filled with helium. As it climbs, the outside pressure drops enormously while the temperature falls more gently, so the balloon keeps expanding (the pressure factor wins) until it bursts, usually somewhere around 30 km up. In a diesel engine, the piston squeezes air to about one-twentieth of its volume so quickly that the air also heats up; by the combined gas law the pressure rises by far more than a factor of 20, and the hot air ignites the injected fuel without a spark plug. In the lab, gas volumes are often measured at room conditions and then "corrected to STP" with the combined gas law so that results from different days and different labs can be compared fairly. Even a sealed bag of chips puffs up on an airplane, because the cabin is kept at only about 75 kPa while the bag was sealed near 100 kPa.