In Kinetic Molecular Theory you met a gas as a swarm of tiny molecules flying in all directions, with pressure coming from their hits on the walls. Here that picture becomes one equation that ties together pressure $P$, volume $V$, temperature $T$ and the amount of gas. You will rearrange it often, as in Solving for a Variable.
Why it works
Think about the hits on the walls. Twice as many molecules make twice as many hits, so $P \propto n$. Hotter molecules move faster, so they hit harder and more often, and the pressure grows in proportion to the absolute temperature. Give the same molecules twice the volume, and their hits spread over more wall and come less often, so $P \propto 1/V$. Put together, $P \propto nT/V$. The constant that makes it an equation is the gas constant $R$. The kelvin scale starts at absolute zero, where this pressure would drop to zero; that is why the law needs kelvins.
Key idea
$$PV = nRT = Nk_BT$$ with $P$ in pascals, $V$ in cubic meters, $T$ in kelvins ($T = T_C + 273$), $n$ in moles, $N$ the number of molecules, $R = 8.31\,\text{J/(mol}\cdot\text{K)}$ and $k_B = 1.38 \times 10^{-23}\,\text{J/K}$.
Worked examples
Worked example: How much gas?
A $10.0\,\text{L}$ tank holds gas at $200\,\text{kPa}$ and $27\,^\circ\text{C}$. How many moles are inside?
- SI units: $P = 2.00 \times 10^{5}\,\text{Pa}$, $V = 10.0\,\text{L} = 0.0100\,\text{m}^3$, $T = 27 + 273 = 300\,\text{K}$.
- $n = \dfrac{PV}{RT} = \dfrac{(2.00 \times 10^{5})(0.0100)}{(8.31)(300)}$.
- $n = \dfrac{2000\,\text{J}}{2493\,\text{J/mol}} = 0.802\,\text{mol}$. (Pascals times cubic meters are joules.)
Answer: $0.802\,\text{mol}$
Worked example: A warm can
A sealed, rigid can of air at $150\,\text{kPa}$ warms from $20\,^\circ\text{C}$ to $40\,^\circ\text{C}$. What is the new pressure?
- $n$ and $V$ are fixed, so $P/T$ is fixed: $P_2 = P_1 \dfrac{T_2}{T_1}$.
- Kelvins: $T_1 = 293\,\text{K}$, $T_2 = 313\,\text{K}$.
- $P_2 = (150\,\text{kPa})\dfrac{313}{293} = 160\,\text{kPa}$, a rise of about $7\%$, not a doubling.
Answer: $160\,\text{kPa}$
Worked example: Counting molecules
How many molecules are in $1.0\,\text{L}$ of air at $101\,\text{kPa}$ and $0\,^\circ\text{C}$?
- Use the molecule form: $N = \dfrac{PV}{k_BT}$, with $V = 1.0 \times 10^{-3}\,\text{m}^3$ and $T = 273\,\text{K}$.
- $$N = \dfrac{(1.01 \times 10^{5})(1.0 \times 10^{-3})}{(1.38 \times 10^{-23})(273)} = 2.7 \times 10^{22}$$
Answer: $2.7 \times 10^{22}$ molecules
Common mistake
Using Celsius. Warming a gas from $20\,^\circ\text{C}$ to $40\,^\circ\text{C}$ does not double its pressure, because $40\,^\circ\text{C}$ is not twice as hot as $20\,^\circ\text{C}$: it is $313\,\text{K}$ against $293\,\text{K}$. Check: convert every temperature to kelvins before it goes into the gas law, and remember that a temperature of $0\,^\circ\text{C}$ would make $P = 0$ if Celsius were allowed.
Real gases follow $PV = nRT$ closely at everyday pressures and temperatures. It fails near the point where a gas condenses into a liquid, when the molecules' own size and attraction start to matter.