# State functions/Ideal gas

1. Mar 7, 2015

### Incand

1. The problem statement, all variables and given/known data
Let $h$ be the height above sea level. Show that the reduction in air pressure $dp$ as a function of $dh$ is given by $\frac{dp}{p} = -\frac{Mg}{RT}dh$. Where $M$ is the molar mass of air and $T$ the temperature at height $h$.
2. Solution
This problem got a solution provided by our professor, however i how a question about the solution since i don't fully understand it.
Consider the attached figure. We can write $dV = Adh$ (A is the area under our segment of air) and the ideal gas law gives us
$dn = \frac{pdV}{RT}$
The mass segment $dm$ is given by
$dm = MdV = \frac{MpAdh}{RT}$
The forces on the segment balance out
$gdm + (p+dp)A = pA \Longrightarrow \frac{dp}{p} = -\frac{Mg}{RT}dh$

So what I'm wondering about is the second step where we express
$dm = MdV = \frac{MpAdh}{RT}$
My problem being that the pressure and temperature isn't constant in the airsegment so in actuality we would have to express $dn$ as
$pdV + Vdp = RTdn + nRdT \Longleftrightarrow dn = \frac{pdV+Vdp - nRdT}{RT}$
I asked our professor about this and he said that the other differentials doesn't affect the result since they're negligible compared too $pdV$ but I fail to see why.

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Last edited: Mar 7, 2015
2. Mar 7, 2015

### Quantum Defect

For the tiniest slab, the temperature difference is negligible (remember that there all sorts of currents/winds that will tend to stir things up over short differences). The temperature is constant over short distances. You go on to integrate the equation to get the barometric formula. If you wanted to take into account the actual variation of temperature as a function of height, you could do the integral where you use T=T(h) on the right side integral.

There is actually very interesting temperature structure in the atmosphere. Look at a book on atmospheric chemistry/physics to see why this is the case.

3. Mar 7, 2015

### Incand

Thanks for responding. I'm afraid I'm just as confused as before. I guess I can accept that the temperature difference doesn't matter compared to the difference in volume and pressure but we still have a change of pressure in the volume as well (in fact we are trying to calculate that!). Actually the following questions is actually about estimating the temperature difference from the formula but for a much simpler case, then the general one, when we assume the pressure difference is from an adiabatic expand.

But back to the question maybe something a long the line of using $dn = \frac{pdV +Vdp}{RT}$ and then the question is what is $V$ really. We can choose that to be identical to $dV=Adh$ i guess and we get
$\frac{dp}{p} = \frac{gMPdh}{RT(1-dh)}$
I guess this doesn't make a whole lot of sense, maybe its just beyond the course (first year intro thermodynamics) and i should just move on and accept it.

4. Mar 7, 2015

### Quantum Defect

All that you are calculating when you calculate the change dP between the top and bottom of the thin slab is the weight of the thin slab (a force) divided by the area, A.

Mass of thin slab = n x M
Weight of thin slab = (nxM)g
We can calculate n using the ideal gas law n = PV/RT
The volume is the volume of the thin slab:V=A x dh

So... Weight of thin slab = PAdh/RT x M x g. You get pressure by dividing by A.

So the tiny pressure drop is: dP = -(PAMgdh/RT)/A. The negative is because the pressure drops. There is a smaller mass of air above.

We need to get all of the P terms on the left, so this becomes:

dP/P = -Mg/RT x dh

Integrating both sides from P0 to P and h=0 to h gives the barometric formula.

You are thinking that you get the differential from the ideal gas law, but you are not. Think about it in terms of there being less mass of air pushing down on you as you go up. The same thing happens as you go down into the ocean, but water is much less compressible than air.

5. Mar 7, 2015

### Incand

Thank you, that makes a lot more sense to me!