Question

Define adiabatic lapse rate and show that
dT/dh= {(–γ–1)/γ} (Mg/R)
Plot a graph between atmospheric temperature with altitude.

Answer 1

The air parcel undergoes only adiabatic transformations $dQ=0$, and the atmosphere is in hydrostatic equilibrium.

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

$\frac{dp}{\rho}=-g\cdot dh$ (1)

$-dA=dU\to -pdV=mc_VdT$

$d(pV)=pdV+Vdp$

$Vdp-d(pV)=mc_VdT$

From the ideal gas low $pV=nRT$

$Vdp-d(nRT)=Vdp-nRdT=mc_VdT$

$\frac{1}{\rho}dp-\frac{n}{nM}RdT=c_VdT\to \frac{1}{\rho}dp=(\frac{R}{M}+c_V)dT\to$

$\frac {dp}{\rho}=c_pdT$ (2)

$c_pdT=-g\cdot dh$

Combining (1) with the equation (2), yields

$\frac{dT}{dh}=-\frac{g}{c_p}$ or $\frac{dT}{dh}=-\frac{g}{\frac{\gamma R}{(\gamma-1)M}}\to \frac{dT}{dh}=-\frac{\gamma-1}{\gamma}\frac{Mg}{R}$