Question

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

Answer 1

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

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The ideal gas law can be used for situations of high-enough temperature `T` and low-enough pressure `P`:

`PV = nRT`

where:

• `P` is pressure in `"atm"` or `"bar"`.
• `V` is volume in `"L"`. `V/n` is the molar volume in `"L/mol"`.
• `n` is the mols of ideal gas in `"mols"`.

• `R = "0.082057 L"cdot"atm/mol"cdot"K"` is the universal 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 the temperature in `"K"`.

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

`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 `"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"`