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Mass, Density and Volume of atmosphere

  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data:

    The density of the earth's atmosphere varies with altitude, and can be approximated by an exponential: ρ(h)=ρ0e^(-h/h0) where ρ0 = 1.3 kg/m3 (the approximate density at sea level) and h0 = 8.2 km (this is determined empirically). Calculate the mass of 15 km of the atmosphere above a 1m2 area at sea level.

    2. Relevant equations
    Mass= density. volume volume= area . hight
    3. The attempt at a solution
    I tried to solve it in two different ways but both gives me incorrect answer, The correct answer is 8900kg
    My First attempt
    : I substitute the height in the equation then i got p(h)=1.3xe^(-15/8.2) p(h)=0.19936 kg/m^3
    Mass= p(h) x volume =0.19936x area x height = 0.19936x1x15x10^3=2990.4kg but the answer is not correct?!!!!!!!!!
    My second attempt: I expand the equation to get mass=V.p(h)=h.A.p0.e^(-h/h0) then integrate it to get mass= [h^2/2 . A.p0.e^(-h/h0)+ h.p0/h0 .e^(-h/h0)] from h=0 to h=15000
    mass= 23478038kg also not correct!!!
     
  2. jcsd
  3. Oct 25, 2016 #2

    jbriggs444

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    It would be easier to follow your work if you defined all your variable names before using them.
    So you have mass = V.p(h). V being the volume and p(h) being the density at the current height h.

    You proceed to equate V to h.A where A is the surface area under a column of height h. But that's wrong. If you are integrating, the total height h does not contribute to the volume of an incremental element. Only the incremental height ##dh## contributes. You need to be adding up the masses of incremental volumes in the column.

    Keeping track of units could have shown you that the result was dimensionally inconsistent.
     
  4. Oct 25, 2016 #3
    you mean that i should not integrate ? but in my first attempt i didn't integrate and the answer was also incorrect
     
  5. Oct 25, 2016 #4

    jbriggs444

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    No, I mean that you need to be careful what you integrate.

    When you integrate, you are essentially adding up the masses of a whole bunch of incremental volumes starting at the base of the column and working your way to the top. What is the formula for the mass of an incremental volume of height ##dh## at altitude ##h## with area ##A## and density ##\rho(h)##?
     
    Last edited: Oct 25, 2016
  6. Oct 25, 2016 #5
    mass=area.p0.e^(-dh/h0)dh right?
     
  7. Oct 25, 2016 #6

    jbriggs444

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    No, that is not it.

    Try it one step at a time. Do not substitute for the density yet. What is the formula for the incremental mass in terms of ##dh##, ##h##, ##A## and ##\rho(h)##?
     
    Last edited: Oct 25, 2016
  8. Oct 25, 2016 #7
    the formula for volume for each dh is Area.dh and that for the density p(h)=p0.e^(-dh/h0)
    right till now?
     
  9. Oct 25, 2016 #8

    jbriggs444

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    No. That is not the formula for density.
     
  10. Oct 25, 2016 #9
    p(h)= -p(0)/h0 .e^(-dh/h0) ??
     
  11. Oct 25, 2016 #10

    jbriggs444

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    By now you have seen and liked the complete solution that was posted by another helper and subsequently removed.

    The problem with the above formula for density is that it is not a function of h. "h" and "dh" are different things.
     
  12. Oct 25, 2016 #11
    yeah i put h in the equation instead of dh and that what made all my wrong going wrong
     
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