# Question babc8

Jul 2, 2016

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 $\text{ ""1 atm}$
• a temperature of $\text{ "0^@"C}$

However, this definition was changed by IUPAC more than 30 years ago to

• a pressure of $\text{ "10^5"Pa" = "100 kPa}$
• a temperature of $\text{ "0^@"C}$

If you use the ideal gas law equation for these new conditions, given that

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 atm " = " 101.325 kPa}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

you will get

$P V = n R T \implies \frac{V}{n} = \frac{R T}{P}$

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 $\text{22.7 L}$.

Notice that using $\text{1 atm}$ 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 $\text{22.4 L}$. But keep in mind that the value for the pressure used to get $\text{22.4 L}$ is not current anymore.