What volume does one mole of an ideal gas occupy? 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,


$\pu{22.4 L}$ at STP
$\pu{22.4 L}$ at NTP
$\pu{22.4 L}$ at SATP
$\pu{22.4 L}$ 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:





The volume occupied by a mole of gas should be:


At STP
\begin{align}
T &amp;= \pu{273.0 K},&amp;
P &amp;= \pu{1 bar},&amp;
R &amp;= \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
$$V = \pu{22.7 L}.$$
At NTP
\begin{align}
T &amp;= \pu{293.0 K},&amp;
P &amp;= \pu{1 atm},&amp;
R &amp;= \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 
$$V = \pu{24 L}.$$
At SATP
\begin{align}
T &amp;= \pu{298.0 K},&amp;
P &amp;= \pu{1 bar},&amp;
R &amp;= \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 
$$V = \pu{24.8 L}.$$




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):


Did I screw up with my calculations?
(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?
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 &amp;=\frac{nRT}{P}\\
  &amp;=\frac{\pu{1  mol} \times \pu{8.2057338 \times 10^-2 L * atm//K * mol}
    \times \pu{273.15 K}}{\pu{1 atm}}\\
  &amp;=\pu{22.41396 L}\\
  &amp;\approx \pu{22.4 L}
\end{align}