A container with 3.1 moles will hold how many liters of gas at STP?

1 Answer
May 30, 2017

We're asked to find the volume occupied by #3.1# #"mol"# of gas at standard temperature and pressure conditions.

To solve this, we can use the ideal-gas equation:

#PV=nRT#

where

  • #P# is the pressure of the gas in atmospheres (#"atm"#)

  • #V# is the volume occupied by the gas in liters (#"L"#)

  • #n# is the number of moles of the gas present (#"mol"#)

  • #R# is a constant called the universal gas constant, equal to #0.08206("L"·"atm")/("mol"·"K")#, and

  • #T# is the absolute temperature of the system (absolute temperature simply means its Kelvin temperature, #"K"#)

At standard temperature and pressure, the pressure is #1# #"atm"#, and temperature is #0^"o""C"#, or #273.15# #"K"#.

Let's now plug in all our known variables, and rearrange the ideal-gas equation to solve for the volume, #V#:

#V = (nRT)/P = ((3.1cancel("mol"))(0.08206("L"·cancel("atm"))/(cancel("mol")·cancel("K")))(273.15cancel("K")))/(1cancel("atm"))#

#= color(red)(69"L"#

You may have also learned that #1# #"mol"# of any gas at stp occupies a volume of #22.4# #"L"# ( Avogadro's law). We could also have solved it knowing that:

#3.1cancel("mol")((22.4"L")/(1cancel("mol"))) = color(red)(69"L"#

If you ever wondered why this is so, it's because the product of #0.08206("L"·"atm")/("mol"·"K")# and #273.13"K"# is #22.4# (it's pretty cool, right?). Either way works, but you can only use the conversion factor #22.4"L" = 1"mol"# if the gas is at standard temperature and pressure.