Calculating Number of O2 and N2 molecules per m^3 of air

[FaraDazed]

Homework Statement

Part A: Calculate how many mocules per cubic metre in air at 27°C and at a pressure of 1.1 atm (1atm is 1.01 x 10^5)

Part B: What is mass in grams of 2.4m^3 of this air.

Assume that 75% of the air is N2 (with molar mass of 28 g/mol) and 25% O2 (with molar mass of 32 g/mol)

Homework Equations

$PV=nRT$

The Attempt at a Solution

Using the equation above, I solved for n, which I took it to mean the number of moles of air.

$n=\frac{PV}{RT}=\frac{(1.1 \times 1.01 \times 10^5)(1)}{8.315 \times (273.15 + 27)} = 44.515$

So then to work out for Nitrogen I found 75% of 44.515, which is 33.39 then multipled that by avagaros number to get the number of atoms, then divided by 2 to get the number of N2 molecules.

$(44.515)(0.75)=33.39 \\ 33.39 \times 6.02 \times 10^{23} = 2.01 \times 10^{25} \,\,\, atoms \\ 2.01 \times 10^{25} \times 0.5 =1.004 \times 10^{25} ,\,\,\ molecules$

Then did exactly the same process but for oxygen, using 44.515*0.25 and got a number of 3.35 x 10^24 molecules of O2.

Then for Part B:

I multiplied the number of N2 molecules in 1m^3 by 2.4 to get how many there would be in 2.4m^3, and then multiplied that by 28u to get the weight in Kg. Did the same process for the oxygen, and then added them together.

For N2
$(1.004 \times 10^{25})(2.4)=2.4096 \times 10^{25} \,\,\, molecules \\ (2.4096 \times 10^{25})(28 \times 1.66 \times 10^{-27})=1.12 kg$

For O2
$(3.3498 \times 10^{24})(2.4) = 8.3952 \times 10^{24} \,\,\,\, molecules \\ (8.3952 \times 10^{24})(32 \times 1.66 \times 10^{-27}) = 0.45 kg$

Therefore total mass in 2.4m^3 in grams
$1120+450 = 1570g$

 

[DrClaude]

So then to work out for Nitrogen I found 75% of 44.515, which is 33.39 then multipled that by avagaros number to get the number of atoms, then divided by 2 to get the number of N2 molecules.

Why do you assume that $n$ in the ideal gas equation of state is the number of moles of atoms in the gas?

 

[FaraDazed]

Why do you assume that $n$ in the ideal gas equation of state is the number of moles of atoms in the gas?

Because that is what i thought the 'n' represented in the equation $PV=nRT$ , the number of moles of the gas in question (in this case, air) , am I mistaken?

I meant, that n is the number of moles of air, which in this question assumes is made up of 75% N2 and 25% O2

 

[Chestermiller]

In the ideal gas law, n is the number of moles of molecules, not the number of moles of atoms. DrClaude was trying to hint to you that you shouldn't have divided by 2.

In part B, how many gram moles of air are there in 2.4 m³ of this gas? What is the weighted average molecular weight of air, given that it is 75% N2 and 25% O2?

Chet

 

[FaraDazed]

In the ideal gas law, n is the number of moles of molecules, not the number of moles of atoms. DrClaude was trying to hint to you that you shouldn't have divided by 2.

In part B, how many gram moles of air are there in 2.4 m³ of this gas? What is the weighted average molecular weight of air, given that it is 75% N2 and 25% O2?

Chet

Ah right, thanks. Yeah I was a bit unsure of that bit, whether it gave the number of atoms or number of molecules. I will have another go at it tomorrow now :)

 

[FaraDazed]

For N2
$(44.515)(0.75)=33.39 \\ 33.39 \times 6.02 \times 10^{23} = 2.01 \times 10^{25} \,\,\, molecules \\$

$(44.515)(0.25)=11.13 \\ 11.13 \times 6.02 \times 10^{23} = 6.7 \times 10^{24} \,\,\, molecules \\$

In part B, how many gram moles of air are there in 2.4 m³ of this gas? What is the weighted average molecular weight of air, given that it is 75% N2 and 25% O2?

Chet

So I know my values for the amount of molecules was wrong first time around, but my method for part B was also wrong?

The average molecular weight of air would be 29g/mol. But if my answer for the number of O2/N2 molecules is now correct, I have already taken into account the fact its 75:25 when calculating the number of molecules.

I will go ahead and do the same but with my new values, just in case it is correct.

For N2
$(2.4)(2.01 \times 10^{25})= 4.824 \times 10^{25} \,\,\,\, molcules / 2.4m^3 \,\,\, of \,\,\,\, air \\ (4.824 \times 10^{25})(28 \times 1.66 \times 10^{-27}) = 2.242 kg = 2242g$

For O2
$(2.4)(6.7 \times 10^{24}) = 1.608 \times 10^{25} molcules / 2.4m^3 \,\,\, of \,\,\,\, air \\ (1.608 \times 10^{25})(32 \times 1.66 \times 10^{-27}) = 0.854 kg = 854g$

Therefore total weight is 854+2242=3096g . Even though I know that air actually weights more than most people think, I do think my value is a bit too high.

Appreciate any help, thanks :)

 

[DrClaude]

Therefore total weight is 854+2242=3096g . Even though I know that air actually weights more than most people think, I do think my value is a bit too high.

That number is correct.

 

[FaraDazed]

That number is correct.

Thanks for the confirmation. Its put my mind to rest :)

 

[Chestermiller]

Thanks for the confirmation. Its put my mind to rest :)

A simpler way of doing it is: (44.5)(2.4)(29)=3097 grams

Chet