The molar gas volume: Avogadro's law

What are standard conditions of temperature and pressure?

Standard conditions (s.t.p.): $0^o\; C$ and $1013 \; mbar$ $=\;1 atm$ $= \;760\; mmHg)$

Be careful: Recently I.U.P.A.C. has defined S.T.P. to $0^o\; C$ and $1000 \; mbar$ ! For chemists, that makes very little difference.

What is the volume (s.t.p.) of $1\; mol\; CO_2$ ?

$1,962\; g$ have a volume of $1\; L$ $1\; mol = 44\; g$ have a volume of $\frac{1\cdot 44}{1,962}$ $=$ $22,4 \; L$

Avogadro has verified that all gases have the same molar volume (s.t.p.). Formulate his law!

At standard temperature and pressure, one mol of any gas takes a volume of $22,4\; L$

Write down a formula for Avogadro's law !

Given n the number of mol of a gas (s.t.p.), V its volume expressed in L: $V\;=\;n\cdot 22,4$

Write down Avogadro's law with volumic mass $\rho$ !

Take one mol of gas (s.t.p.), its mass is the molar mass $M$ and its volume $22,4\; L$ :

Given M the molar mass of a gas and $\rho$ its volumic mass expressed in $\frac{g}{L}$ (s.t.p.): $\rho$ $=$ $\frac{M}{22,4}$

(relative) Density of a gas (s.t.p.) is defined by: $d_{gaz}$ $=$ $\frac{\rho_{gas}}{\rho_{air}}$ Now, at standard conditions: $\rho_{air}$ $=$ $1,293 \frac{g}{L}$ Write down Avogadro's law using density $d$

For any gas (s.t.p.): $d=\frac{\frac{M}{22,4}}{1,293}$

Given M the molar mass of a gas and d its density (s.t.p.): $d$ $=$ $\frac{M}{29}$