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# Why $22.4litres$ called the molar volume of gas?

Last updated date: 23rd Jul 2024
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Hint :Molar volume of a gas is defined as the volume of one mole of the gas. Thus, the molar volume is also the volume occupied by $6.02{\text{ }} \times {\text{ }}{10^{23}}\;$ particles of gas . $STP$ refers to the standard temperature of ${0^ \circ }C$ and pressure of $1{\text{ }}atm\;$ . Room conditions refer to the temperature of $25^\circ C$ and the pressure of $1{\text{ }}atm\;$ .

Always the Standard temperature and pressure $\left( {STP} \right)$ is defined as $0^\circ C{\text{ }}\left( {273.15{\text{ }}K} \right)$ and $1{\text{ }}atm$ pressure. The molar volume of a gas is the volume of one mole of a gas at $STP$ . At $STP$ one mole
( $6.02{\text{ }} \times {\text{ }}{10^{23}}\;$ representative particles) of any gas occupies a volume of $22.4{\text{ }}L$ . Standard Molar Volume is the volume occupied by one mole of any gas at $STP$ .
The Molar Volume of an ideal gas at $STP$ , which we define to be  ${0^ \circ }C$ and $1{\text{ }}atm\;$ arbitrarily (because we're old-fashioned and stuck in 1982) is $22.411{\text{ }}L/mol.$
To calculate this we can use the Ideal gas law of $PV = nRT$ At $STP$ (Standard Temperature and Pressure), we CHOSE:
$P = 1{\text{ }}atm$
$V = ?$
$n = 1{\text{ }}mol$
$R = 0.082057{\text{ }}L \cdot atm/mol \cdot K$
$T{\text{ }} = {\text{ }}273.15{\text{ }}K \\ \\$
$V = nRTP$
$= (1mol)(0.082057atm \cdot Lmol \cdot K)273.15K1atm$
= $22.4{\text{ }}L$
This is the volume of one mole of ideal gas at STP.

Note :
We have not specified the identity of the gas; we have specified only that the pressure is $1{\text{ }}atm\;$ and the temperature is $273{\text{ }}K$ . This makes for a very useful approximation: any gas at $STP$ has a volume of $22.4{\text{ }}L$ per mole of gas; that is, the molar volume at $STP$ is $22.411{\text{ }}L/mol.$ .This molar volume makes a useful conversion factor in stoichiometry problems if the conditions are at $STP$ . If the conditions are not at $STP$ , a molar volume of $22.411{\text{ }}L/mol.$ is not applicable. However, if the conditions are not at $STP$ , the combined gas law can be used to calculate the volume of the gas at $STP$ ; then the $22.411{\text{ }}L/mol.$ molar volume can be used.