Combined gas equation

Boyle's Law ($PV = \text{constant}$ at fixed $T$) and Charles' Law ($V/T = \text{constant}$ at fixed $P$) each freeze one variable to study the other two. The combined gas equation removes that restriction and lets all three — pressure, volume, temperature — change at once, for the same fixed mass of gas:

where: $P_1, V_1, T_1$ are the pressure, volume, and temperature of the gas in its initial state, and $P_2, V_2, T_2$ are the same three quantities in its final state. Temperature must always be in kelvin — Celsius values plugged in directly will silently wreck the answer. A few things make this equation easy to use once you see the pattern:

• It is a ratio equation, not a proportionality between just two quantities. Each side is $PV/T$, and the equation says this combination stays the same before and after the change, for a fixed amount of gas.
• It contains Boyle's and Charles' Laws as special cases. Set $T_1 = T_2$ and the temperatures cancel, leaving $P_1V_1 = P_2V_2$ — Boyle's Law. Set $P_1 = P_2$ and the pressures cancel, leaving $V_1/T_1 = V_2/T_2$ — Charles' Law. This is a good way to check you have written the equation correctly.
• STP problems are just a special case of the general method. "Find the volume at STP" simply means one of the two states has $P = 1\,\text{atm}$ and $T = 273\,\text{K}$ already filled in — nothing about the method changes.

The picture below shows why the equation must hold: going from state 1 to state 2 can always be imagined as two steps, first a Boyle's-Law step at fixed temperature, then a Charles'-Law step at fixed pressure.

A fixed mass of gas occupies 200 cm³ at a pressure of 2 atm and temperature 47 °C. What volume will it occupy at a pressure of 1 atm and temperature 0 °C?

Step 1. Convert both temperatures to kelvin and list the known values. $P_1 = 2\,\text{atm}$, $V_1 = 200\,\text{cm}^3$, $T_1 = 47 + 273 = 320\,\text{K}$; and $P_2 = 1\,\text{atm}$, $T_2 = 0 + 273 = 273\,\text{K}$, with $V_2$ unknown.

Step 2. Write the combined gas equation and substitute. $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$ becomes $\dfrac{2 \times 200}{320} = \dfrac{1 \times V_2}{273}$.

Step 3. Simplify the left-hand side first, since it contains no unknowns. $\dfrac{400}{320} = 1.25$.

Step 4. Solve for $V_2$ by multiplying both sides by 273. $V_2 = 1.25 \times 273 = 341.25\,\text{cm}^3$. Notice both effects fought each other here: dropping the pressure alone would have increased the volume, while dropping the temperature alone would have decreased it. The combined equation weighs both correctly and gives $V_2 = 341.25\,\text{cm}^3$, meaning the net effect was a modest increase from the starting 200 cm³.

Competitive stretch

If the same 200 cm³ sample at 2 atm and 320 K were instead cooled to STP (1 atm, 273 K) directly, the same four steps apply — only the target state changes, and the method never needs to change with it. This is the habit worth building: always identify which six quantities you know, write them in two neat rows ($P_1, V_1, T_1$ and $P_2, V_2, T_2$), and only then substitute — this is also how ICSE examiners expect the working to be laid out.

• Using Celsius instead of kelvin. The single most common error — always convert with $T(\text{K}) = T(°\text{C}) + 273$ before substituting.
• Writing STP temperature as 0 K instead of 273 K. STP means $P = 1\,\text{atm}$ and $T = 273\,\text{K}$, not $T = 0\,\text{K}$.
• Mixing units of pressure or volume between the two states. $P_1$ and $P_2$ must share one unit, and $V_1$ and $V_2$ must share one unit, or the numeric answer will be wrong even with correct method.
• Reaching for the combined equation when only one variable changes. It still gives the right answer, but Boyle's or Charles' Law alone is simpler and safer when only two of the three quantities are actually varying.

Ask the tutor: Want to try a three-variable gas problem of your own, or see the Boyle's-Law-then-Charles'-Law derivation worked through with actual numbers instead of symbols?