Question #1499b

1 Answer
Oct 13, 2015

Answer:

#rho = "1.18 g/L"#

Explanation:

The first thing to do here is figure out what the average molar mass of air is by using the molar masses of nitrogen gas, #"N"_2#, and oxygen gas, #"O"_2#.

This will then help you find a relaationship between the ideal gas law and the density of air.

So, if you know that air is #80%# nitrogen and #20%# oxygen, you can say that its molar mass will depend on those proportions

#M_"M air" = 80/100 * "28.0134 g/mol" + 20/100 * "32.0 g/mol"#

#M_"M air" = 22.411 + 6.4 = "28.81 g/mol"#

According to the ideal gas law equation, you have

#PV = n * RT" "#, where

#P# - the pressure of the gas;
#V# - the volume of the gas;
#n# - the number of moles of gas;
#R# - the ideal gas constant, equal to #0.082"atm L"/"mol K"#
#T# - the temperature of the gas.

Now, you should also know that the number of moles of a substance can be determined by using a sample of that substance and its molar mass.

#n = m/M_M#

Plug this into the ideal gas law equation to get

#PV = m/M_M * RT#

Rearrange this equation to get #m/V# on one side

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

The ratio between the mass of a substance and the volume it occupies is actually equal to its density, #rho#. Therefore,

#rho = (P * M_M)/(RT)#

Plug in your values to find the value of #rho#

#rho = (1.000color(red)(cancel(color(black)("atm"))) * 28.81"g"/color(red)(cancel(color(black)("mol"))))/(0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K")))) = "1.178 g/L"#

I'll leave the answer rounded to three sig figs

#rho = color(green)("1.18 g/L")#