Adiabatic Temperature Change in Rising Air

[Physgeek64]

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

 

[Chestermiller]

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?

 

[Physgeek64]

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 $\rho=\frac{m}{V}$ ?

 

[Chestermiller]

$TV^{\gamma -1} =constant$ and $\rho=\frac{m}{V}$ ?

I asked for T vs P, not V., $$\rho=\frac{Pm}{RT}$$

 

[Physgeek64]

I asked for T vs P, not V., $$\rho=\frac{Pm}{RT}$$

would there not be a factor of n in there?

 

[Chestermiller]

would there not be a factor of n in there?

In the equation I wrote, The m is the molecular weight and R is the ideal gas constant. $$\rho=\frac{nm}{V}$$

 

[Physgeek64]

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?

 

[Chestermiller]

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.

 

[Physgeek64]

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

 

[Chestermiller]

$$pT^{\frac{\gamma}{(1-\gamma)}}=C$$

 

[Physgeek64]

$$pT^{\frac{\gamma}{(1-\gamma)}}=C$$

Thank you, I have done it now :)