Question #624ba

1 Answer
Mar 17, 2016

Answer:

#"3.00 moles"#

Explanation:

The idea here is that one mole of any ideal gas that is kept under Standard Temperature and Pressure, STP, conditions will occupy #"22.4 L " -># this is known as the molar volume of a gas at STP.

This can be derived using the ideal gas law equation

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#, where

#P# - the pressure of the gas
#V# - the volume occupied by the gas
#n# - the number of moles of gas present
#R# - the universal gas constant, usually given as #0.0821("atm" * "L")/("mol" * "K")#
#T# - the absolute temperature of the gas

STP conditions are defined as a pressure of #"1 atm"# and a temperature of #"273.15 K"#. Plug these values into the ideal gas law equation and solve for #V/n#

#PV = nRT implies V/n = (RT)/P#

#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 tells you that every mole of an idea lgas that is being kept under STP conditions will occupy #"22.4 L"#.

http://slideplayer.com/slide/1716615/

So, if #"22.4 L"# correspond to one mole of any ideal gas, it follows that #"67.2 L"# will correspond to

#67.2color(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)"3.00 moles"color(white)(a/a)|)))#

SIDE NOTE It's worth noting that the current STP conditions are defined as a pressure of #"100 kPa"# and a temperature of #"273.15 K"#.

Under these conditions for pressure and temperature, one mole of any ideal gas occupies #"22.7 L"#.