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

1 Answer
Aug 7, 2017

Well, the volume you describe at "273.15 K"273.15 K is the molar volume at STP, assuming standard pressure is "1 atm"1 atm. Modern IUPAC (i.e. post-1982), however, states "1 bar"1 bar for the standard pressure.

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 TT and low-enough pressure PP:

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 Latm/molK is the universal gas constant, if PP is in "atm"atm and VV is in "L"L. On the other hand, R = "0.083145 L"cdot"bar/mol"cdot"K"R=0.083145 Lbar/molK, if PP is in "bar"bar.

  • TT is the temperature in "K"K.

At "1 atm"1 atm and 0^@ "C"0C, the pre-1982 definition of STP, the molar volume V/nVn is then:

color(blue)(barV_("atm")) -= V/n = (RT)/P¯¯¯VatmVn=RTP

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

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

At "1 bar" and 0^@ "C", the post-1982 definition of STP, the molar volume V/n is then:

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"