# How do you get that the ideal gas molar volume is #"22.4 L"#???

##### 1 Answer

Well, the volume you describe at **molar volume at STP**, assuming standard pressure is

For the purposes of your chemistry class, use the STP defined in your textbook, as that is how you would be graded.

The **ideal gas law** can be used for situations of high-enough temperature

#PV = nRT# where:

#P# ispressurein#"atm"# or#"bar"# .#V# isvolumein#"L"# .#V/n# is the molar volume in#"L/mol"# .#n# is themols of ideal gasin#"mols"# .

#R = "0.082057 L"cdot"atm/mol"cdot"K"# is theuniversal gas constant, if#P# is in#"atm"# and#V# is in#"L"# . On the other hand,#R = "0.083145 L"cdot"bar/mol"cdot"K"# , if#P# is in#"bar"# .

#T# is thetemperaturein#"K"# .

At *pre-1982* definition of STP, the molar volume

#color(blue)(barV_("atm")) -= V/n = (RT)/P#

#= (("0.082057 L"cdot"atm/mol"cdot"K")("273.15 K"))/("1 atm")#

#=# #ulcolor(blue)"22.414 L/mol"#

At *post-1982* definition of STP, the molar volume

#color(blue)(barV_("bar")) -= V/n = (RT)/P#

#= (("0.083145 L"cdot"bar/mol"cdot"K")("273.15 K"))/("1 bar")#

#=# #ulcolor(blue)"22.711 L/mol"#