Question #babc8
1 Answer
Here's what's going on here.
Explanation:
That is the correct value given the current definition of STP conditions.
http://goldbook.iupac.org/S06036.html
You see, many textbooks and online resources still use the old definition of STP, which implied
- a pressure of
" ""1 atm" - a temperature of
" "0^@"C"
However, this definition was changed by IUPAC more than 30 years ago to
- a pressure of
" "10^5"Pa" = "100 kPa" - a temperature of
" "0^@"C"
If you use the ideal gas law equation for these new conditions, given that
color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = " 101.325 kPa")color(white)(a/a)|)))
you will get
PV = nRT implies V/n = (RT)/P
This will be equal to
V/n = (0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(100/101.325color(red)(cancel(color(black)("atm")))) = "22.72 L mol"^(-1)
This means that one mole of gas kept under STP conditions occupies
1 color(red)(cancel(color(black)("mole"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = "22.7 L"
Therefore, the molar volume of a gas at STP is
Notice that using
V/n = (0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 0)color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm")))) = "22.42 L mol"^(-1)
This is why the molar volume of a gas at STP is said to be