# 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 **not current** anymore.