Question #94805

1 Answer
Mar 27, 2016

Answer:

#"1.27 moles"#

Explanation:

The thing to remember about Standard Temperature and Pressure (STP) is that one mole of any ideal gas occupies #"22.4 L"# under these conditions - this is known as the molar volume of a gas at STP.

Notice that the problem provides you with the volume of the gas and asks for the number of moles of gas that would occupy this volume under STP conditions.

This means that you're essentially looking for a conversion factor that can take you from volume to moles and vice versa.

Now, STP conditions are usually defined as a pressure of #"1 atm"# and a temperature of #0^@"C"#, or #"273.15 K"#.

You can use the ideal gas law to write

#color(purple)(|bar(ul(color(white)(a/a)color(black)(PV = nRT implies V/n = (RT)/P)color(white)(a/a)|)))#

Here #R# is the universal gas constant and is equal to #0.0821("atm" * "L")/("mol" * "K")#.

Plug in the values for pressure and temperature given by the definition of STP to get

#V/n = (0.0821(color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * 273.15color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm")))) = "22.4 L mol"^(-1)#

This is the molar volume of a gas at STP.

So, if one mole of any ideal gas occupies #"22.4 L"#, it follows that #"28.5 L"# would contain

#28.5color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.4color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)"1.27 moles"color(white)(a/a)|)))#

The answer is rounded to three sig figs.

SIDE NOTE The definition of STP has been changed to correspond to a pressure of #"100 kPa"# and a temperature of #0^@"C"#.

Under these conditions, one mole of any ideal gas occupies #"22.7 L"#, not #"22.4 L"#.

However, many textbooks and online resources still use the old definition of STP, so stick with that one to be on the safe side.