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Homework Statement
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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