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.

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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 STPeginalignT &= 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 NTPeginalignT &= 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 SATPeginalignT &= 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?