Stack all three laws and they collapse into one: $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$, the combined gas law — P-one V-one over T-one equals P-two V-two over T-two. Same fixed amount of gas, same subscript convention (1 before, 2 after), and now any of the three knobs may move. Pin any one of them and the matching pair cancels, handing you Boyle, Charles or Gay-Lussac back. Learn this one and you own all four.

What the combined law cannot tell you is HOW MUCH gas is in the vessel — it compares two states and the amount cancels itself out. For that you need $PV = nRT$, the ideal gas law, read aloud P V equals n R T. $P$ is the absolute pressure, $V$ the volume, $n$ the amount of gas in moles, $T$ the absolute temperature, and $R$ the universal gas constant — the exchange rate between the mechanical side and the molecular side. On this paper $R = 8.314\ \mathrm{kPa \cdot L/(mol \cdot K)}$, so pressures arrive in kPa and volumes in L, and every unit cancels: kPa·L on the left, and mol times kPa·L/(mol·K) times K on the right. Push the units through your rearrangement every time. If they refuse to cancel into the answer's units, the rearrangement is wrong, no appeal — and units that do work out never prove you right. The check is a one-way street.

One shortcut is worth memorising. At STP — 101.325 kPa and 273.15 K — $PV = nRT$ has already been solved for you: one mole of any ideal gas fills $V_m = 22.4\ \mathrm{L}$, so $V = n\,V_m$. Two cautions. It is STP only: at SATP (25 °C, 100 kPa) the molar volume is 24.8 L/mol, about 10 % larger, and swapping them is a favourite exam trap. And $V_m$ does not care WHICH gas it is — Avogadro's insight, and the reason a mole of hydrogen and a mole of sulfur dioxide fill the same balloon.