Change in atmopheric pressure with height (Thermodynamics)

[Slepton]

Homework Statement

Think of Earth's atmosphere as an ideal gas of molecular weight $$\mu$$ 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($$\left($$-$$\mu$$g/RT)

where,
dp = change in pressure
dz = change in height
$$\mu$$ = 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

 

[rl.bhat]

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)

 

[Slepton]

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 ?

 

[rl.bhat]

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.

 

[Slepton]

figured i have to derive yet another expression. This time I have to assume an adiabatic expansion and show

dp/p = {$$\gamma$$/($$\gamma$$-1)}{dT/T}

i.e.

dp/dT = {$$\gamma$$/($$\gamma$$-1)} {p/T}


My idea:

since it is an adiabatic expansion, i have
pz^{($$\gamma$$-1)} / T(Z)^$$\gamma$$ = constant

I used it and tried to cook the needed expression, but i got lost.

 

[rl.bhat]

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.

 

[Aroo]

How would you derive the hydrostatic equation in spherical polars?

Thanks