Answer to Question #96722 in Physics for Abhishek

Question

Derive an expression for adiabatic lapse
rate.

Answer 1

We will now derive an expression for the rate of change of temperature with height of a parcel of dry air as it moves about in the Earth’s atmosphere. Since the air parcel undergoes only adiabatic transformations

$dQ=0$

and the atmosphere is in hydrostatic equilibrium.

The gravity side

$dp=-g\cdot\rho\cdot dz$

$\frac{dp}{\rho}=-g\cdot dz$

The thermodynamics side

$-dA=dU \to-pdV=mc_{V}dT$

$Vdp-d(pV)=mc_{V}dT$

$Vdp-d(nRT)=Vdp-nRdT=mc_{V}dT$

$\frac{V}{m}dp-\frac{n}{m}RdT=c_{V}dT$

$\frac{1}{\rho}dp-\frac{n}{nM}RdT=c_{V}dT$

$\frac{1}{\rho}dp=(\frac{R}{M}+c_{V})dT$

$\frac{dp}{\rho}=c_{p}dT$

$c_{p}dT=-g\cdot dz$

And finally

$\frac{dT}{dz}=-\frac{g}{c_{p}}$.