Question #eb368

1 Answer
Aug 22, 2015

You start from the idea lgas law equation and work your way from there.

Explanation:

The molar volume of a gas simply means the volume occupied by 1 mole of an ideal gas under certain conditions for temperature and pressure.

More often than not, the molar volume of a gas is given for a pressure of #"100 kPa"# and a temperature of #0^@"C"# - these values for pressure and temperature describe the Standard Temperature and Pressure conditions.

So, let's say that you want to determine what the molar volume of a gas at STP is. Start from the ideal gas law equation

#PV = nRT" "#,where

#P# - the pressure of the gas;
#V# - the volume it occupies;
#n# - the number of moles of gas;
#R# - the universal gas constant, usually given as #0.082("atm" * "L")/("mol" * "K")#
#T# - the temperature of the gas expressed in Kelvin.

Rearrange this equation to have #V/n# on one side

#V = (nRT)/P#

#V/n = (RT)/P#

Use the STP values to get - don't forget to convert the pressure from kPa to atm!

#V/n = (0.082(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"))))#

#V/n = 22.7"L"/"mol"#

To get the volume occupied by one mole, simple replace #n# with #"1 mole"#

#V/(1color(red)(cancel(color(black)("mole")))) = 22.7"L"/color(red)(cancel(color(black)("mole"))) = color(green)("22.7 L")#

The molar volume of a gas at STP is equal to #"22.7 L"#.

You can calculate the molar volume of a gas at any pressure and temperature. For example, at a pressure of #"2 atm"# and a temperature of #100^@"C"#, you get

#V/n = (0.082(color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 100)color(red)(cancel(color(black)("K"))))/(2color(red)(cancel(color(black)("atm"))))#

#V/n = 15.3 "L"/"mol"#

This means that under these conditions for pressure and temperature, the molar volume of a gas is

#V = "15.3 L"#

SIDE NOTE Many textbooks and online resources still give the molar volume of a gas as being equal to 22.4 L at STP.

That happens because the old definition of STP is being used, meaning that the condtions are #0^@"C"# and #"1 atm"#.

If you use these values, you will indeed get

#V/n = (0.082(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"))))#

#V = "22.4 L"#