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?

Question

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

Henry Antrim · Accepted Answer

Here's what I got.

The idea behind this is that you should begin by solving the ideal gas law equation in its standard form. #color(blue)(PV = nRT) " " " "color(purple)((1))# and make an effort to adjust it by adding molar mass and density. Since mass per unit of mole is the definition of molar mass, you can write the number of moles of a substance by using both its mass and molar mass. #color(blue)(M_M = m/n implies n = m/M_M)# Enter this into the formula color(purple)((1)) to obtain #PV = m/M_M *RT # Rearrange this equation to solve for #M_M#, which for this problem will represent the average molar mass of air. #M_M = m/V * (RT)/P " " " "color(purple)((2))# Let's now concentrate on density, which is defined as mass per volume unit. #color(blue)(rho = m/V)# Plug this into equation #color(purple)((2))# to get #M_M = rho * (RT)/P# Plug in your values and solve for #M_M# - keep in mind that you will need to convert the density of air from 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"# You'll get this #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"# The weighted average of the molar masses of the constituent gases of air is now used to determine the average molar mass of air. More precisely, according to its mole fraction in the mixture, each gas will contribute a certain amount to the average molar mass. In your case, air is said to contain oxygen gas, #"O"_2#, and nitrogen gas, #"N"_2#. This means that you can write #chi_(O_2) + chi_(N_2) = 1# The mixture contains two gases, so their mole fractions must add up to give #1#. #chi_(O_2) xx M_(O_2) + chi_(N_2) xx M_(N_2) = "28.72 g/mol"# The nitrogen gas and oxygen gas molar masses are #M_(O_2) = "31.999 g/mol" " "# and #" "M_(N_2) = "28.0134 g/mol"# Apply these two formulas to obtain #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 indicates that you've #chi_(O_2) = 1- 0.822 = 0.178# You must now round these responses to the nearest two sig figs, or the total number of sig figs you have for the supplied values. #M_"M air" = color(green)("29 g/mol")# #chi_(O_2) = color(green)(0.18)# #chi_(N_2) = color(green)(0.82)#

Cooper Anderson · Answer

undefined The average molar mass of the atmosphere at this altitude is approximately 29.8 g/mol.

Serenity Allen · Answer

undefined To find the average molar mass of the atmosphere at this altitude, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)

We need to rearrange this equation to solve for the number of moles (n):

n = PV / RT

First, we need to convert pressure from mm Hg to atm:

1 atm = 760 mm Hg

So, 42 mm Hg = 42 / 760 atm

Now, let's convert the temperature from Celsius to Kelvin:

Temperature in Kelvin = Temperature in Celsius + 273.15

So, -63 °C = -63 + 273.15 K

Now, we can plug in the values into the equation:

n = (42 / 760) * V / (0.0821 * (-63 + 273.15))

The volume (V) is not given, but since we are dealing with the density of air, we can consider a volume of 1 m^3 for simplicity.

n = (42 / 760) * 1 / (0.0821 * (-63 + 273.15))

Now, we can calculate the number of moles (n).

n ≈ 0.002836 moles

Now, we can use the definition of molar mass:

Molar mass (M) = mass / number of moles

Given that the density of air is 92 g/m^3, the mass of 1 m^3 of air is 92 g.

So, molar mass (M) = 92 g / 0.002836 moles

M ≈ 32458 g/mol

Thus, the average molar mass of the atmosphere at this altitude is approximately 32.458 kg/mol.