How do you get that the ideal gas molar volume is "22.4 L"22.4 L???
1 Answer
Well, the volume you describe at
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 = nRTPV=nRT where:
PP is pressure in"atm"atm or"bar"bar .VV is volume in"L"L .V/nVn is the molar volume in"L/mol"L/mol .nn is the mols of ideal gas in"mols"mols .
R = "0.082057 L"cdot"atm/mol"cdot"K"R=0.082057 L⋅atm/mol⋅K is the universal gas constant, ifPP is in"atm"atm andVV is in"L"L . On the other hand,R = "0.083145 L"cdot"bar/mol"cdot"K"R=0.083145 L⋅bar/mol⋅K , ifPP is in"bar"bar .
TT is the temperature in"K"K .
At
color(blue)(barV_("atm")) -= V/n = (RT)/P¯¯¯Vatm≡Vn=RTP
= (("0.082057 L"cdot"atm/mol"cdot"K")("273.15 K"))/("1 atm")=(0.082057 L⋅atm/mol⋅K)(273.15 K)1 atm
== ulcolor(blue)"22.414 L/mol"
At
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"