Q. 21 (a) Pressure decreases as one ascends the atmosphere. If the density of air is ρ, what is the change in pressure dp over a differential height dh?

(b) Considering the pressure p to be proportional to the density, find the p at a height h if the pressure on the surface of the earth is po.

(c) If po=1.03×103Nm-2,ρo=1.29kgm-3andg=9.8m/s2, at what height will be pressure drop to (1/10) the value at the surface of the earth?

(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

Hint: The upthrust of the underlying air balances the weight of the upper air.
Step 1: Find the pressure difference at height h.
(a) Consider a horizontal parcel of air with cross-section A and height dh.
Let the pressure on the top surface and bottom surface be P and P+dP. If the parcel is in equilibrium, then the net upward force must be balanced by the weight.
i.e. (P+dp)A - PA =—ρgAdh (.: Weight = Density x Volume x g)
dp=-ρgdh(ρ=densityofair)
The negative sign shows that the pressure decreases with height.
 
Step 2: Find the equation of pressure at height h.
 
(b) Let ρ0 be the density of air on the surface of the earth.
As per the question, pressure  density
PPo=ρρo
ρ=ρoPoP
dp=-ρogPoPdh[dp=-pgdh]
dpP=-ρogPodh
PoPdpP=-ρogPo0hdhath=0,P=Poandath=h,P=P
InPPo=-ρogPoh
By removing log, P=Poe-ρoghPo
Step 3: Find the height h for the required pressure using the equation of pressure at height h.
(c) As P = Poe-ρoghPo,
ln110PoPo=-ρogPoh
ln110=-ρogPoo
h=-Poρogln110=-PoρogIn(10)-1=Poρogln10
=Poρog×2.303[In(x)=2.303log10x]
=1.013×1051.22×9.8×2.303=0.16×105m
=16×103m
(d) We know that Pp (When T=constant i.e. isothermal pressure)
Temperature (T) remains constant only near the surface of the earth, not at the greater heights.