# How to calculate with the gas laws

### Avogadro's law

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

### The law of Boyle and Mariotte

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.

### Gay-Lussac's law

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:
$T ( K)$ $=$ $t ( ^oC )$ $+$ $273.15$

### Charles's law

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:
$T ( K)$ $=$ $t (^oC)$ $ +$ $273.15$

### The ideal gas law

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)= $t(^oC)$ $+$ $273.15$

_ _ _ _

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
$R$ $=$ $8.3 \frac{J}{mol \cdot K}$

### Units

1 atm = 1.013 bar
1 bar = $10^5$ Pa
1 $m^3$ = $10^3$ L

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