Question

Question #babc8

Answer 1

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 `"22.7 L"`.

Notice that using `"1 atm"` for the pressure will get you

`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 `"22.4 L"`. But keep in mind that the value for the pressure used to get `"22.4 L"` is not current anymore.