# The density of air #20# #km# above Earths surface is #92# #g##/##m^3.# The pressure of the atmosphere is #42# #mm# #Hg#, and the temperature is #-63# #°C#. What is the average molar mass of the atmosphere at this altitude?

##
If the atmosphere at this altitude consists of only #O_2# and #N_2# , what is the mole fraction of each gas?

If the atmosphere at this altitude consists of only

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The idea here is that you need to start from the standard form of the ideal gas law equation

#color(blue)(PV = nRT) " " " "color(purple)((1))#

and try to manipulate it by incorporating *density* and *molar mass*.

As you know, **molar mass** is defined as mass per unit of mole. This means that you an write the number of moles of substance by using its molar mass and its mass

#color(blue)(M_M = m/n implies n = m/M_M)#

Plug this into equation color(purple)((1)) to get

#PV = m/M_M *RT #

Rearrange this equation to solve for *average molar mass* of air.

#M_M = m/V * (RT)/P " " " "color(purple)((2))#

Now focus on density, which is defined as mass per unit of volume.

#color(blue)(rho = m/V)#

Plug this into equation

#M_M = rho * (RT)/P#

Plug in your values and solve for *grams per cubic meter* to *grams per liter*, the pressure from *mmHg* to *atm*, and the temperature from *degrees Celsius* to *Kelvin*.

#92"g"/color(red)(cancel(color(black)("m"^3))) * (1color(red)(cancel(color(black)("m"^3))))/(10^3"L") = "0.092 g/L"#

This will get you

#M_M = 0.092"g"/color(red)(cancel(color(black)("L"))) * (0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + (-63))color(red)(cancel(color(black)("K"))))/(42/760color(red)(cancel(color(black)("atm"))))#

#M_M = "28.723 g/mol"#

Now, the *average molar mass* of air is calculated by taking the **weighted average** of the molar masses of its constituent gasses.

More specifically, each gas will contribute to the average molar mass **proportionally** to the **mole fraction** it has in the mixture.

In your case, air is said to contain oxygen gas,

#chi_(O_2) + chi_(N_2) = 1#

The mixture contains two gases, so their mole fractions **must add up** to give

#chi_(O_2) xx M_(O_2) + chi_(N_2) xx M_(N_2) = "28.72 g/mol"#

The molar masses of oxygen gas and nitrogen gas are

#M_(O_2) = "31.999 g/mol" " "# and#" "M_(N_2) = "28.0134 g/mol"#

Use these two equations to get

#chi_(O_2) = 1 - chi_(N_2)#

#(1 - chi_(N_2)) * 31.999 color(red)(cancel(color(black)("g/mol"))) + chi_(N_2) * 28.0134 color(red)(cancel(color(black)("g/mol"))) = 28.723 color(red)(cancel(color(black)("g/mol")))#

#31.999 - 3.9856 * chi_(N_2) = 28.723#

#chi_(N_2) = 3.276/3.9856 = 0.822#

This means that you have

#chi_(O_2) = 1- 0.822 = 0.178#

Now, you need to round these answers to two sig figs, the number of sig figs you have for your given values

#M_"M air" = color(green)("29 g/mol")#

#chi_(O_2) = color(green)(0.18)#

#chi_(N_2) = color(green)(0.82)#