Adiabatic rising of air

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  • #1
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Homework Statement


The hydrostatic equation expresses the change in pressure dp due to a layer of

atmosphere of thickness dz as

constant volume.

##dp = −\rho g dz ##
Using this expression, show that the change in temperature with height for a parcel of air that rises adiabatically in the atmosphere can be expressed as
##-\frac{\gamma-1}{\gamma} \frac{mg}{K_B}##


Homework Equations




The Attempt at a Solution


So I think we're trying to find ##\frac{\partial T}{\partial z}_S ## as this seems like a reversible process

starting off with ##dU=TdS-pdV##
## \frac{\partial U}{\partial T}_z dT +\frac{\partial U}{\partial z}_T dz = TdS-pdV##
## \frac{3NK_B}{2}\frac{\partial T}{\partial z}_S +\frac{\partial U}{\partial z}_T =-p\frac{\partial V}{\partial z}_S##

The fact that i can't find ## \frac{\partial V}{\partial z}_S## makes me think I've gone wrong somewhere

Many thanks
 

Answers and Replies

  • #2
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What is the relation between T and p for an adiabatic reversible expansion? From the ideal gas law, what is the density expressed as a function of the temperature, pressure, and molecular weight?
 
  • #3
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What is the relation between T and p for an adiabatic reversible expansion? From the ideal gas law, what is the density expressed as a function of the temperature, pressure, and molecular weight?

##TV^{\gamma -1} =constant## and ##\rho=\frac{m}{V} ## ?
 
  • #7
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In the equation I wrote, The m is the molecular weight and R is the ideal gas constant. $$\rho=\frac{nm}{V}$$
Oh of course. But i don't see how to use these. Was i on the right track?
 
  • #8
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Oh of course. But i don't see how to use these. Was i on the right track?
No. Actually, you weren't on the right track. You can start by substituting ##\rho=\frac{pM}{RT}## into the equation $$dp=-\rho g dz$$. Then you substitute the relationship between p and T (in terms of ##\gamma##) for an adiabatic reversible expansion.
 
  • #9
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11
No. Actually, you weren't on the right track. You can start by substituting ##\rho=\frac{pM}{RT}## into the equation $$dp=-\rho g dz$$. Then you substitute the relationship between p and T (in terms of ##\gamma##) for an adiabatic reversible expansion.

so I get ##dp=-\frac{pmg}{K_BT}dz= -\frac{Nmg}{V}dz##

I can't see where to go from here
 
  • #11
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