Question #48160

1 Answer
Apr 28, 2016

Answer:

#"1.25 g L"^(-1)#

Explanation:

The idea here is that you need to use the ideal gas law equation and the molar mass of nitrogen gas, #"N"_2#, to find a relationship between the mass of a sample of nitrogen gas and the volume it occupies at STP.

Now, STP conditions will most likely be given to you as a pressure of #"1 atm"# and a temperature of #0^@"C"#, or #"273.15 K"#.

The ideal gas law equation looks like this

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#, 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 #0.0821("atm" * "L")/("mol" * "K")#
#T# - the absolute temperature of the gas

As you know, a compound's molar mass tells you the mass of one mole of that substance. IN your case, nitrogen gas has a molar mass of

#M_M = "28.0134 g mol"^(-1)#

This means that every mole of nitrogen gas has a mass of #"28.0134 g"#.

You can thus replace the number of moleso f nitrogen gas, #n#, in the ideal gas law equation by the ratio between a mass #m# and the molar mass of the gas

#n = m/M_M#

Plug this into the ideal gas law equation to get

#PV = m/M_M * RT#

Rearrange to get

#PV *M_M = m * RT#

This will be equivalent to

#P * M_M = m/V * RT#

But since density, #rho#, is defined as mass per unit of volume, you can say that you have

#m/V = (P * M_M)/(RT)#

and therefore

#color(purple)(|bar(ul(color(white)(a/a)color(black)(rho = (P * M_M)/(RT))color(white)(a/a)|)))#

Now all you have to do is use the STP values for pressure and temperature and the molar mass of nitrogen gas to get

#rho = (1 color(red)(cancel(color(black)("atm"))) * "28.0134 g" color(red)(cancel(color(black)("mol"^(-1)))))/(0.0821(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 273,15color(red)(cancel(color(black)("K")))) = color(green)(|bar(ul(color(white)(a/a)"1.25 g L"^(-1)color(white)(a/a)|)))#

#color(white)(a)#
ALTERNATIVE APPROACH

As you know, one mole of any ideal gas kept at #"1 atm"# and #0^@"C"# occupies #"22.4 L"# -- this is known as the molar volume of a gas at STP.

Moreover, you know that one mole of nitrogen gas has a mass of #"28.0134 g"#. Since density tells you mass per unit of volume, all you have to do now is determine the mass of one liter of nitrogen gas at STP

#1 color(red)(cancel(color(black)("L N"_2))) * overbrace((1 color(red)(cancel(color(black)("mole N"_2))))/(22.4color(red)(cancel(color(black)("L N"_2)))))^(color(purple)("molar volume of a gas at STP")) * overbrace("28.0134 g"/(1color(red)(cancel(color(black)("mole N"_2)))))^(color(blue)("molar mass of N"_2)) = "1.25 g"#

Since this is how many grams you get per liter of nitrogen gas at STP, it follows that its density will be

#rho = color(green)(|bar(ul(color(white)(a/a)"1.25 g L"^(-1)color(white)(a/a)|)))#