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Change in atmopheric pressure with height (Thermodynamics)

  • Thread starter Slepton
  • Start date
  • #1
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Homework Statement



Think of earth's atmosphere as an ideal gas of molecular weight [tex]\mu[/tex] in a uniform gravitational field. Let g denote the acceleration due to gravity.

If z denotes the height above the sea level, show that the change of atmospheric pressure p with height is given by

dp/dz = p([tex]\left([/tex]-[tex]\mu[/tex]g/RT)

where,
dp = change in pressure
dz = change in height
[tex]\mu[/tex] = molecular weight
R = Universal gas constant
T = temperature




Homework Equations



1) F = mg
2) P = F/A
3) PV = nRT



The Attempt at a Solution



With the relation between force and pressure, I derived an expression of p in terms of m and g. Using the ideal gas law, i obtained another equation for p. I am lost in my attempt to derive dp/dz. Please help.

Thank in advance,
Priam
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
Gas equation is given by pV = nRT, where is the number of mole.
n is given by [mass of the air column (ma) on unit area]/mu.
Hence pV = [(ma)/(mu)]RT .........(1)
Now p = rho*g*z.....(2)
and dp/dz = rho*g......(3)
From equation (1) find the expression for rho and substitute in eq. (3)
 
  • #3
21
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thank rl.bhat!

but the notations are confusing.
n = (ma)/(mu)
so ma is the mass of air column ? Can you explain a bit further please ?
 
  • #4
rl.bhat
Homework Helper
4,433
7
ma is the mas of the air column on unit area, mu is the molecular weight of air.
So number of mole = n = ma/mu.
 
  • #5
21
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figured i have to derive yet another expression. This time I have to assume an adiabatic expansion and show

dp/p = {[tex]\gamma[/tex]/([tex]\gamma[/tex]-1)}{dT/T}

i.e.

dp/dT = {[tex]\gamma[/tex]/([tex]\gamma[/tex]-1)} {p/T}


My idea:

since it is an adiabatic expansion, i have
pz([tex]\gamma[/tex]-1) / T(Z)[tex]\gamma[/tex] = constant

I used it and tried to cook the needed expression, but i got lost.
 
  • #6
rl.bhat
Homework Helper
4,433
7
For adiabatic expression pV^gamma = K
You replace V by nR(T/p)
So p(nRT/p)^gamma = k or
p(T/p)^gamma = K'
i.e. T^gamma*p^(1-gamma) = K.
Take the differentiation and simplify.
 
  • #7
1
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How would you derive the hydrostatic equation in spherical polars?

Thanks
 

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