Calculating Oxygen Mass in a 4.0x7.0x2.5 Room

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Discussion Overview

The discussion revolves around calculating the total mass of oxygen in a room with specified dimensions, assuming standard temperature and pressure (STP) and that air contains 20.95% oxygen by volume. The focus is on applying the ideal gas law and related equations to derive the mass of oxygen present.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents a homework problem asking for the mass of oxygen in a room and expresses uncertainty about how to start.
  • Another participant introduces the ideal gas law (PV = nRT) as a relevant concept for solving the problem.
  • A subsequent reply suggests that the participant needs to find the number of moles (n) and mentions an equation that relates moles to mass.
  • Further, a participant proposes calculating the number of moles of air by assuming nitrogen as the remaining component and provides a method to find the mass of oxygen based on the volume of the room.
  • One participant calculates the volume of the room and applies the ideal gas law to find the number of moles of air, subsequently calculating the mass of oxygen based on its percentage in air.
  • A later reply humorously questions the units of the calculated mass, suggesting a variety of measurement units without providing clarity on the intended unit.

Areas of Agreement / Disagreement

Participants generally agree on the use of the ideal gas law and the method to calculate the mass of oxygen, but there is no consensus on the final unit of measurement for the mass calculated.

Contextual Notes

The discussion does not clarify the assumptions made regarding the average molecular mass of air or the specific conditions under which the calculations are valid. There are also unresolved steps in the calculations presented.

Who May Find This Useful

Students or individuals interested in applying the ideal gas law to practical problems, particularly in chemistry or physics contexts.

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Homework Statement



What is the total mass (in grams) of oxygen in a room measuring 4.0 by 7.0 by 2.5? Assume that the gas is at STP and that air contains 20.95% oxygen by volume.


The Attempt at a Solution


I don't know where to begin.
 
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Yes.
 
Then you have everything you need except the variable n, or the number of moles. There's also an equation that relates n (number of moles) to the mass. Do you know of this equation?
 
Then that gives you the number of moles of air -
Find the average molecular mass of air by assuming that the rest is N2 (mass=28)

Then you know the number of moles of O2 - and so the mass
 
V = (4*7*2.5) = 70m^3
P = 1atm
T = 273 K
R = .000082
n = PV/RT
n = 3126.95
1 mole of oxygen is 32 g/mol so
32 * 3126.95 = 1000062.53 but you only want 20.95% of that so
1000062.53 * .2095 = 20963.10
= 2.10 x 10^ 4
 
Zhalfirin88 said:
2.10 x 10^ 4

Pounds? Talents? Stones? Carats? Maunds? Catties? Wiarduneks?
 

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