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Use normal composition of air to determine its density

  1. Mar 31, 2005 #1
    A room is 12 ft by 12ft by 8ft at 40F. Use normal composition of air to determine its density and assume a pressure of 1.00 atm.

    How do I get started?
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  3. Apr 1, 2005 #2


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    By removing room size, as it is irrelevant to the problem :smile:

    1 mole of oxygen at STP is 32g in 22.4l
    1 mole of nitrogen is 28g in 22.4l

    Knowing composition of the air you should be able to calculate mass of '1 mole' of air, that will give you density. Next step is pV = nRT to include 40 F in your calculations (unless 40 F is STP - I have no idea what 40 F is :smile: )

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  4. Apr 1, 2005 #3


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    It's a tricky question because you don't even use the dimension given.
    First find the composition of air. For easiness I'll just say 80% nitrogen 20% oxygen; those are g/g. Then you find the moles in 1 g of air.

    [tex]n = \frac{0.80g}{28g/mol} + \frac{0.20g}{32g/mol}[/tex]

    n = 0.03482 moles

    Then you take your gas law which says PV = nRT. You know the pressure, you want the volume, you know the moles, you know the gas constant, and you know the temperature (convert it to kelvin!).

    [tex]V = \frac{nRT}{P}[/tex]

    [tex]V = \frac{(0.03482)(0.082057)(277.594444)}{1}[/tex]

    V = 0.79318 L

    The gas constant is 0.082 when using atmospheres. 8.314 is only with pascals. The temp was converted to kelvin using unit converter.
    Above it was already assumed that the mass was 1g, so just take the mass and divide by this new volume.

    [tex]\rho = \frac{m}{v}[/tex]

    [tex]\rho = \frac{1g}{0.79318L}[/tex]

    density = 1.26 g/L

    I did not use the correct composition for air. You'll have to redo what I did using the correct composition in order to get the right answer

    By a quick google search I got the data "The density of air under standard conditions is only 1.239 milligrams per cubic centimeter". That's 1.239 g/L, so that's pretty close to my rough estimate.
  5. Apr 1, 2005 #4
    :( Maybe I should include the entire question as I've been unable to do the second part...

    So here it is:
    The combustion of octane is expressed by C8H8+25/2 O2->8 CO2+9 H2O. w/ delta H of -5471kJ.
    Estimate the mass of octane that would need to be burned to produce enough heat to raise the temperature of the air in a 12*12*8 ft^3 room from 40F to 78F on a mild winter day. Use normal composition of air to determine its density and assume a pressure of 1.00 atm.

    I would need the heat capacity of air, would I? I've been working on this for hours and I can't get the correct answer...
  6. Apr 1, 2005 #5


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    What kind of octane is that...?It should be
    [tex] C_{8}H_{18} [/tex]...!!

  7. Apr 1, 2005 #6
    can anyone offer help?
  8. Apr 1, 2005 #7


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    [tex] pV=\nu R T [/tex]. (Mendeleev-Clapeyron)(air assumed ideal,made up of diatomic molecules).

    [tex] T=278K,p=101,325 Pa,R= 8,314 J \ (kmol)^{-1} K^{-1} [/tex]
    and find the # of kmols of air before burning octane.Then use Avogadro-s # to find the # of molecules of air in the room (denoted below by "N").

    Now.Assume that heat is added to the air in the room at constant volume (no work).Then

    [tex] \Delta U = \frac{5}{2} N k_{B} \Delta T [/tex]...

    U know "N",u know Boltzmann's constant and the variation of temp (in K),then u can find [itex] \Delta U [/itex]...

    Write the equation

    [tex] C_{8}H_{18} +\frac{25}{2} O_{2}\rightarrow 8CO_{2}\uparrow+9H_{2}O\uparrow + 5471 KJ [/tex]

    U can find the # of moles of burnt octane and then the mass.

    Last edited: Apr 1, 2005
  9. Apr 1, 2005 #8


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    Similar to what shawn did...

    [tex][n(avg.molar mass)]/V=RT/P=density[/tex], this may or may not correspond to the standard condition density value of air, it all depends on the initial value of density in the room which is somewhat impertinent to the original question.

    I'm guessing that you'll need the specific heat to solve this question (off the top of my head I am not able to come up with any other methods). Solve for [itex]q[/itex].

    Arrange the molar enthalpy so that it becomes [itex] \Delta H /moles of octane [/itex], the form it is in now is in terms of moles/reaction, unless you neglected to place all of the information in the first place. In any case, I believe that the value should be the same. Solve for moles of octane, than convert to mass and you're done.
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