Solving a House Air Mass Calculation Problem

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Homework Help Overview

The discussion revolves around calculating the total mass of air inside a house with a given volume at a specific temperature and pressure. The problem involves concepts from thermodynamics and gas laws, particularly focusing on the ideal gas law and Charles's law.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of Charles's law and the ideal gas law to determine the mass of air in the house. Some question the appropriateness of using Charles's law due to the fixed volume of the house.

Discussion Status

There is ongoing exploration of different methods to solve the problem, with some participants providing calculations based on the ideal gas law. Others offer checks on density and assumptions made in the calculations. The conversation reflects a mix of confirmations and clarifications without reaching a definitive conclusion.

Contextual Notes

Participants discuss assumptions regarding the composition of air and the implications of temperature changes on air mass. There is mention of a follow-up question regarding changes in air mass with temperature variations, indicating a continuation of the problem-solving process.

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Homework Statement



A house has a volume of 840 m^3.
What is the total mass of air inside the house at 20 degrees celsius? Assume that the pressure is 1.00 atm.

Homework Equations


Charles law Volume is directly proportional to Temperature
1 mol of gas at 0 degrees celsius has V = 22.4L


The Attempt at a Solution


using charles law i determined that the new volume at 20 degrees celsius would be = (20*22.4 L ) or (22.4 *10^-3 m^3)*(2) = .448 m^3

number of moles = 840 m^3 / .448 m^3 = 1875 mol
1 mol of air has a mass of about 29 g = .029 kg

so m= (1875 mol) (.029kg/mol) = 54.37 kg

my method apparenlt is faulty because the answer is rejected
Please help me solve this
 
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I don't think Charles law applies since the house has fixed volume. You just need to use the ideal gas law PV = nRT
 
so using PV=nRT

n = PV / RT = (1.013*10^5 Pa)*(840 m^3) / (8.314)(293.15K) = 3.49131*10^4

then 3.49131*10^4 mol (0-029 kg/mol)= 1012 kg

does that seem correct?
 
Looks about right, the density of air at sea level (20 C) is 1.2Kg/m^3 IIRC.
 
I used that the assumption which was also used in one fo the examples in my book that air is 20 % O and 80% N ..so adding together the percent compositions yielded 29 g ..

Why did you offer me the density figure of air ?
 
Just saying that your density would be about 1.2Kg/m^3, just as a sanity check.
 
oh ok. I was making this problem much more complicated than it turned out. Thanks for leading me through !
 
uh oh. There's a follow up question : If the temperature drops to -19 C, what mass of air enters or leaves the house?
it wants delta m
I used the same equation and substituted T with delta T which would be -39 K ..
but that's not working ...
 
P1V1/T1 = P2V2/T2

Volume is constant so, you know P1 and T1 and T2, so work out P2. Then use P2 to find out how much air is still in the house at P2, T2, V using PV=nRT remember that n=M/m
 
  • #10
so here it goes :

P1 = 1.013*10^5 Pa
T1 = 293.15 K
T2 = 254.15 K

P1/T1 = P2/T2
and i got P2= 8.78233*10^4
then: PV=nRT
(8.78233*10^4 Pa)(840m^3)=m(8.314)(254.15K)
n= 3.49131*10^4

im confused here after... please explain how to incorporate n=M/m

can i not just multiply n by 0.029 kg which is mass of 1 mol of air
and then subtract that amount from the inital mass to get delta m ?
 
  • #11
Doesn't that formula assume that n is constant? i.e. closed system.

I think you just assume that the pressure is constant, but haven't do any thermodynamics for a while so you have constant pressure and constant volume for there it's rather easy to find the change in n.
 
  • #12
hmmm ... so there was no need to find P2?

And yes I looked up in my notes the ideal gas law is only valid if the number of particles (n) stays constant
 
Last edited:

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