At $0^o C$ and 1 atm: $V$ $=$ $n\cdot22.4$ where n is the number of moles and V the volume of the gas expressed in litres
At constant temperature, applying to a transformation from a state 1 to a state 2: $P_1 \cdot V_1$ $=$ $P_2 \cdot V_2$ where pressures respectively and volumes are expressed in the same units.
At constant pressure, applying to a transformation from a state 1 to a state 2: $\frac{V_1}{T_1}$ $=$ $\frac{V_2}{T_2}$ where volumes are expressed in the same units and temperatures are Kelvin temperatures: $T ( K)$ $=$ $t ( ^oC )$ $+$ $273.15$
At constant volume, applying to a transformation from a state 1 to a state 2: $\frac{P_1}{T_1}$ $=$ $\frac{P_2}{T_2}$ where pressures are expressed in the same units and temperatures are Kelvin temperatures: $T ( K)$ $=$ $t (^oC)$ $ +$ $273.15$
Applying to a transformation from a state 1 to a state 2: $\frac{P_1 \cdot V_1}{T_1}$ $=$ $\frac{P_2 \cdot V_2}{T_2}$ where pressures respectively and volumes are expressed in the same units and temperatures are Kelvin temperatures: T ( Kelvin)= $t(^oC)$ $+$ $273.15$
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Applying to a gived state of a gas: $P \cdot V$ $=$ $n \cdot R \cdot T$ where n is the number of moles, T the Kelvin temperature T ( K)$ $=$ $t ( $^oC $) + 273.15 If V is expressed in litres and P in atmospheres, then $R$ $=$ $0.082 \frac{L \cdot atm}{mol \cdot K}$ If V is expressed in $m^3$ and P in Pascal, then $R$ $=$ $8.3 \frac{J}{mol \cdot K}$
1 atm = 1.013 bar 1 bar = $10^5$ Pa 1 $m^3$ = $10^3$ L