Question #babc8

1 Answer
Jul 2, 2016

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 #"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.