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 3 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:

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 $$V = \pu{24.8 L}.$$

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?

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:

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 $$V = \pu{24.8 L}.$$

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?

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. 22.4 L at STP

2. 22.4 L at NTP

3. 22.4 L at SATP

4. 22.4 L at both STP and NTP

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

STP : 0°C and 1 bar

NTP: 20°C and 1 atm

SATP: 25°C and 1 bar

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

So, the volume occupied by a mole of gas should be:

1) At STP

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

2) At NTP

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

3) At SATP

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

Nowhere, does the magical "22.4 L" figure in the three cases I've analyzed.

But since I've seen the "one mole occupied 22.4 L at STP/NTP" dictum so many times...I'm wondering if I've missed something.

So, 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 22.4 L" idea is so widespread, in spite of not being close (enough) to the values that I obtained?

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.

• 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).

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 $$V = \pu{24.8 L}.$$

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?

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. 22.4 L at STP

2. 22.4 L at NTP

3. 22.4 L at SATP

4. 22.4 L at both STP and NTP

Even Chem.SE is rife with the "fact" that a mole of ideal gas occupies 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 3 standard conditions; namely: Standard Temperature and Pressure (STP), Normal Temperature and Pressure (NTP) and Standard Ambient Temperature and Pressure (SATP).

Knowing that,

STP : 0°C and 1 bar

NTP: 20°C and 1 atm

SATP: 25°C and 1 bar

And using the equation, $$V = \frac {nRT}{P}$$ Where $n$ = 1, 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:

So, the volume occupied by a mole of gas should be:

1) At STP

$T = 273.0 K$, $P = 1 bar$, $R= 8.3144598$ x $\mathrm{10^{-2}LbarK^{-1}mol^{-1}}$. Plugging in all the values, I got $$V = 22.698475 L$$ which to a reasonable approximation, gives, $$V = 22.7 L$$

2) At NTP

$T = 293.0 K$, $P = 1 atm$, $R= 8.2057338$ x $\mathrm{10^{-2}LatmK^{-1}mol^{-1}}$. Plugging in all the values, I got $$V = 24.04280003 L$$ which to a reasonable approximation, gives, $$V = 24 L$$

3) At SATP

$T = 298.0 K$, $P = 1 bar$, $R= 8.3144598$ x $\mathrm{10^{-2}LbarK^{-1}mol^{-1}}$. Plugging in all the values, I got $$V = 24.7770902 L$$ which to a reasonable approximation, gives, $$V = 24.8L$$

Nowhere, does the magical "22.4 L" figure in the three cases I've analyzed.

But since I've seen the "one mole occupied 22.4 L at STP/NTP" dictum so many times...I'm wondering if I've missed something.

So, 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 22.4 L" idea is so widespread, in spite of not being close (enough) to the values that I obtained?

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. 22.4 L at STP

2. 22.4 L at NTP

3. 22.4 L at SATP

4. 22.4 L at both STP and NTP

Even Chem.SE is rife with the "fact" that a mole of ideal gas occupies 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 3 standard conditions; namely: Standard Temperature and Pressure (STP), Normal Temperature and Pressure (NTP) and Standard Ambient Temperature and Pressure (SATP).

Knowing that,

STP : 0°C and 1 bar

NTP: 20°C and 1 atm

SATP: 25°C and 1 bar

And using the equation, $$V = \frac {nRT}{P}$$ Where $n$ = 1, 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:

So, the volume occupied by a mole of gas should be:

1) At STP

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

2) At NTP

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

3) At SATP

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

Nowhere, does the magical "22.4 L" figure in the three cases I've analyzed.

But since I've seen the "one mole occupied 22.4 L at STP/NTP" dictum so many times...I'm wondering if I've missed something.

So, 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 22.4 L" idea is so widespread, in spite of not being close (enough) to the values that I obtained?