Question #1499b

1 Answer
Oct 13, 2015

rho = "1.18 g/L"ρ=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"_2N2, and oxygen gas, "O"_2O2.

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%80% nitrogen and 20%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"MM air=8010028.0134 g/mol+2010032.0 g/mol

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

According to the ideal gas law equation, you have

PV = n * RT" "PV=nRT , where

PP - the pressure of the gas;
VV - the volume of the gas;
nn - the number of moles of gas;
RR - the ideal gas constant, equal to 0.082"atm L"/"mol K"0.082atm Lmol K
TT - 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_Mn=mMM

Plug this into the ideal gas law equation to get

PV = m/M_M * RTPV=mMMRT

Rearrange this equation to get m/VmV on one side

PV * M_M = m * RT implies m/V = (P * M_M)/(RT)PVMM=mRTmV=PMMRT

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

rho = (P * M_M)/(RT)ρ=PMMRT

Plug in your values to find the value of rhoρ

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")