Obviously, to calculate the volume/space inhabited by a mole that (an ideal) gas, you"ll have to specify temperature ($T$) and also pressure ($P$), find the gas continuous ($R$) value through the appropriate units and also plug them all in the best gas equation $$PV = nRT.$$
The problem? It appears to it is in some kind of typical "wisdom" everywhere the Internet, that one mole of gas occupies $22.4$ liters that space. Yet the standard problems (STP, NTP, or SATP) mentioned lack consistency end multiple sites/books. Common claims: A mole of gas occupies,$pu22.4 L$ in ~ STP$pu22.4 L$ at NTP$pu22.4 L$ in ~ SATP$pu22.4 L$ in ~ both STP and NTP
Even Chem.SE is rife through the "fact" the a mole of right gas occupies $pu22.4 L$, or some extension thereof.
You are watching: 1 mole of a gas occupies 22.4 l at
Being so utterly frustrated through this situation, I made decision to calculate the volumes inhabited by a mole of right gas (based on the right gas equation) because that each that the three standard conditions; namely: standard Temperature and also Pressure (STP), typical Temperature and also Pressure (NTP) and also Standard approximately Temperature and Pressure (SATP).
Knowing that,STP: $pu0 ^circ C$ and $pu1 bar$NTP: $pu20 ^circ C$ and $pu1 atm$SATP: $pu25 ^circ C$ and $pu1 bar$
And making use of the equation, $$V = frac nRTP,$$where $n = pu1 mol$, by default (since we"re talking about one mole the gas).
See more: 16 Ounces Is How Many Grams, 16 Ounce To Gram Conversion Calculator
I"ll draw proper values the the gas constant $R$ indigenous this Wikipedia table:
The volume inhabited by a mole the gas have to be:
At STPeginalignT &= pu273.0 K,&P &= pu1 bar,&R &= pu8.3144598 imes 10^-2 l bar K^-1 mol^-1.endalignPlugging in every the values, I gained $$V = pu22.698475 L,$$ which to a reasonable approximation, gives$$V = pu22.7 L.$$
At NTPeginalignT &= pu293.0 K,&P &= pu1 atm,&R &= pu8.2057338 imes 10^-2 l atm K^-1 mol^-1.endalignPlugging in all the values, I acquired $$V = pu24.04280003 L,$$ which to a reasonable approximation, gives $$V = pu24 L.$$
At SATPeginalignT &= pu298.0 K,&P &= pu1 bar,&R &= pu8.3144598 imes 10^-2 together bar K^-1 mol^-1.endalignPlugging in every the values, I got $$V = pu24.7770902 L,$$ which come a reasonable approximation, provides $$V = pu24.8 L.$$
Nowhere does the miracle "$pu22.4 L$" figure in the three situations I"ve analyzed appear. Due to the fact that I"ve checked out the "one mole occupies $pu22.4 L$ in ~ STP/NTP" dictum so numerous times, I"m wonder if I"ve to let go something.
My question(s):Did ns screw up v my calculations?(If i didn"t screw up) Why is it that the "one mole occupies $pu22.4 L$" idea is so widespread, regardless of not gift close (enough) to the values that ns obtained?