What volume does one mole of an ideal gas occupy?

Question

This has been bugging me for a while now...

Obviously, to calculate the volume/space occupied by a mole of (an ideal) gas, you'll have to specify temperature ($T$) and pressure ($P$), find the gas constant ($R$) value with the right units and plug them all in the ideal gas equation $$PV = nRT.$$

The problem? It seems to be some sort of common "wisdom" all over the Internet, that one mole of gas occupies $22.4$ liters of space. But the standard conditions (STP, NTP, or SATP) mentioned lack consistency over multiple sites/books. Common claims: A mole of gas occupies,

1. $\pu{22.4 L}$ at STP
2. $\pu{22.4 L}$ at NTP
3. $\pu{22.4 L}$ at SATP
4. $\pu{22.4 L}$ at both STP and NTP

Even Chem.SE is rife with the "fact" that a mole of ideal gas occupies $\pu{22.4 L}$, or some extension thereof.

Being so utterly frustrated with this situation, I decided to calculate the volumes occupied by a mole of ideal gas (based on the ideal gas equation) for each of the three standard conditions; namely: Standard Temperature and Pressure (STP), Normal Temperature and Pressure (NTP) and Standard Ambient Temperature and Pressure (SATP).

Knowing that,

• STP: $\pu{0 ^\circ C}$ and $\pu{1 bar}$
• NTP: $\pu{20 ^\circ C}$ and $\pu{1 atm}$
• SATP: $\pu{25 ^\circ C}$ and $\pu{1 bar}$

And using the equation, $$V = \frac {nRT}{P},$$ where $n = \pu{1 mol}$, by default (since we're talking about one mole of gas).

I'll draw appropriate values of the gas constant $R$ from this Wikipedia table:

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The volume occupied by a mole of gas should be:

1. At STP \begin{align} T &= \pu{273.0 K},& P &= \pu{1 bar},& R &= \pu{8.3144598 \times 10^-2 L bar K^-1 mol^-1}. \end{align} Plugging in all the values, I got $$V = \pu{22.698475 L},$$ which to a reasonable approximation, gives $$V = \pu{22.7 L}.$$

2. At NTP \begin{align} T &= \pu{293.0 K},& P &= \pu{1 atm},& R &= \pu{8.2057338 \times 10^-2 L atm K^-1 mol^-1}. \end{align} Plugging in all the values, I got $$V = \pu{24.04280003 L},$$ which to a reasonable approximation, gives $$V = \pu{24 L}.$$

3. At SATP \begin{align} T &= \pu{298.0 K},& P &= \pu{1 bar},& R &= \pu{8.3144598 \times 10^-2 L bar K^-1 mol^-1}. \end{align} Plugging in all the values, I got $$V = \pu{24.7770902 L},$$ which to a reasonable approximation, gives $$V = \pu{24.8 L}.$$

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Nowhere does the magical "$\pu{22.4 L}$" figure in the three cases I've analyzed appear. Since I've seen the "one mole occupies $\pu{22.4 L}$ at STP/NTP" dictum so many times, I'm wondering if I've missed something.

My question(s):

1. Did I screw up with my calculations?
2. (If I didn't screw up) Why is it that the "one mole occupies $\pu{22.4 L}$" idea is so widespread, in spite of not being close (enough) to the values that I obtained?

Answer 1

The common saying is a hold over from when STP was defined to be $\pu{273.15 K}$ and $\pu{1 atm}$. However, IUPAC changed the definition in 1982 so that $\pu{1 atm}$ became $\pu{1 bar}$. I think the main issue is a lot of educators didn't get the memo and went right along either teaching STP as $\pu{1 atm}$ or continuing with the line they were taught ("$\pu{1 mol}$ of any gas under STP occupies $\pu{22.4 L}$") without realizing it didn't hold under the new conditions.

Just as a "proof" of this working for the old definition. \begin{align} V &=\frac{nRT}{P}\\ &=\frac{\pu{1 mol} \times \pu{8.2057338 \times 10^-2 L * atm//K * mol} \times \pu{273.15 K}}{\pu{1 atm}}\\ &=\pu{22.41396 L}\\ &\approx \pu{22.4 L} \end{align}

Answer 2

A big point of confusion is that it is still taught (at least in the mid-2000's) that STP is defined with respect to $\pu{273 K}$ and $\pu{1 atm}$ of pressure, or $\pu{1.01325 bar}$ of pressure, even though IUPAC changed their definition to be with respect to $\pu{1 bar}$ of pressure. By using the ideal gas law on the old STP definition, you get that the volume is $\approx 0.0821\cdot 273 \approx 22.4$ liters.