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Density of an ideal gas as a function of height

  1. Jan 16, 2017 #1
    1. The problem statement, all variables and given/known data
    If air has a density of ρ0 on the surface, calculate its density as a function of the height y for two scenarios:
    (a) the temperature is constant at T0;
    (b) the temperature decreases linearly T(y) = T0 − ay.
    Express your results using the given variables together the gravitational acceleration g, gas constant R and air molar mass M.

    I think I am making an incorrect assumption or just going about this incorrectly. Let me know if you guys see the mistake. I am just focusing on part a as of now.
    2. Relevant equations

    1) F = ma
    2) pV = nRT = (m/M)RT

    3. The attempt at a solution

    1) First, using the ideal gas law, we know:
    ρ=(pM)/(RT)
    2) Since M, R, and T are constant if we can figure out how pressure is changing with respect to height (y) then we can find how density is as well.

    3) Assuming a "slab" of air with a height of dy and a cross-sectional area of A at surface level we can apply Newton's second law:
    Fp_up - Fp_down - mg = ma
    4) Fp_up = pressure acting up on slab Fp_down = pressure acting down on slab and a = 0. This gives:
    Fp_up = Fp_down + mg
    5) Since p = (F/A) ⇒ F = pA:
    pA = (p + dp)A + ρ(Ady)g
    6) I think this is where my mistake is but since we're considering a slab at surface level ρ in the above equation would be equal to ρ0

    7) Form here I was able to solve the differential equation in step #5 and get:
    p(y) = P00gy
    8) I then substituted this in for p in the equation in step #1 and in turn got ρ as a function of y.
     
  2. jcsd
  3. Jan 16, 2017 #2

    Orodruin

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    Note that ##\rho## changes with height as pressure changes with height. What does this tell you about the differential equation?
     
  4. Jan 16, 2017 #3
    That is has two different functions in it. What I ended up doing was using ρ=(pM)/(RT) to find dp/dy and then substitute that into the step 5 equation which gives me a linear first order differential equation consisting only of ρ'(y) and ρ(y). I did a similar approach in part b.

    1) Using ρ=(pM)/(RT) I could find dp/dy=(R/M)(T'(y)ρ(y) + T(y)ρ'(y)).
    2) I then substituted this into the equation:
    dp/dy=-gρ​
    3) SInce we know T(y) we can express it as a first order differential equation with variable coefficients. I used an integrating factor that does not work out very nicely. Is there something I missed or is this just unavoidable?
     
  5. Jan 16, 2017 #4

    haruspex

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    Please show that in detail.
    But I suggest it will be easier working with an equation for p.
    At 5) you have dp=ρg dy, and at 1), pM=ρRT. Eliminating ρ, what do they give you?
     
  6. Jan 16, 2017 #5
    1) I used equation one, pM=ρRT, and solved for p yielding p=(RT/M)ρ.
    2) From here I found:
    dp/dy=(RT/M)ρ'(y). ​
    3) Using:
    pA=(p+dp)A+ρ(Ady)g ⇒ dp=-ρgdy ⇒ dp/dy=-gρ​
    4) I replaced dp/dy with (RT/M)ρ' and got (RT/M)ρ'=-gp.
    5) Since this is a linear first order homogeneous diff eq, ρ'+(gM/RT)ρ=0, I used the characteristic polynomial, r+(gM/RT)=0.
    6) I ended up getting this:
    ρ(y)=ρoe(-gM/RT)y
     
  7. Jan 16, 2017 #6

    haruspex

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    Ok, that was a little convoluted but you got to the right place.
    (I would have written dp=-ρg dy, pM=ρRT, so dp/p=-(gM/RT)dy, p=Ce-(gM/RT)y etc.)
    So what about the case where T is linear with height?
     
  8. Jan 16, 2017 #7
    For this part I used dp/dy=-gρ because the forces were not dependent upon temperature. I also used p=(ρRT/M) to get dp/dy=(R/M)(T'(y)ρ(y)+T(y)ρ'(y)) since both temperature and density are dependent upon height. We know T(y)=To-ay so T'(y)=-a. Combinding dp/dy=(R/M)(T'(y)ρ(y)+T(y)ρ'(y)) and dp/dy=-gρ I was able to come up with this differential equation (sorry for it being a little messy):
    ρ'(y)+((Mg-aR)/(RTo-ay))ρ(y)=0​
     
  9. Jan 17, 2017 #8

    haruspex

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    Should that be R(T0-ay)?
    Anyway, can you solve that? For a hint see my last step in post #6.
     
  10. Jan 17, 2017 #9

    Charles Link

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    @Ian Baughman I think you might find it useful to know that fluids and gases experience a force per unit volume given by ## \vec{f}_v=-\nabla P ##. The gravitational field supplies a force per unit volume of ## \vec{f}_g=-\delta g \hat{z} ## where ## \delta ## is the mass density of the fluid or gas. In order to have equilibrium, the net forces must be equal to zero: ## -\nabla P-\delta g \hat{z}=0 ##. From this you can get a differential equation for the pressure as a function of height. You are already on the right track, but I thought you might find this of some use.
     
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