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

Aug 7, 2017

Well, the volume you describe at $\text{273.15 K}$ is the molar volume at STP, assuming standard pressure is $\text{1 atm}$. Modern IUPAC (i.e. post-1982), however, states $\text{1 bar}$ 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 $T$ and low-enough pressure $P$:

$P V = n R T$

where:

• $P$ is pressure in $\text{atm}$ or $\text{bar}$.
• $V$ is volume in $\text{L}$. $\frac{V}{n}$ is the molar volume in $\text{L/mol}$.
• $n$ is the mols of ideal gas in $\text{mols}$.
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$ is the universal gas constant, if $P$ is in $\text{atm}$ and $V$ is in $\text{L}$. On the other hand, $R = \text{0.083145 L"cdot"bar/mol"cdot"K}$, if $P$ is in $\text{bar}$.

• $T$ is the temperature in $\text{K}$.

At $\text{1 atm}$ and ${0}^{\circ} \text{C}$, the pre-1982 definition of STP, the molar volume $\frac{V}{n}$ is then:

$\textcolor{b l u e}{{\overline{V}}_{\text{atm}}} \equiv \frac{V}{n} = \frac{R T}{P}$

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

$=$ $\underline{\textcolor{b l u e}{\text{22.414 L/mol}}}$

At $\text{1 bar}$ and ${0}^{\circ} \text{C}$, the post-1982 definition of STP, the molar volume $\frac{V}{n}$ is then:

$\textcolor{b l u e}{{\overline{V}}_{\text{bar}}} \equiv \frac{V}{n} = \frac{R T}{P}$

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

$=$ $\underline{\textcolor{b l u e}{\text{22.711 L/mol}}}$