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Question

(a) It is known that density of air decreases with height y as
ρ=ρ0ey/y0
where ρ0 = 1.25 kg m3 is the density at sea level, and y0 is constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of the atmosphere remains a constant (isothermal conditions). Also, assume that the value of g remains constant.
(b) A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?
[Take y0 = 8000 m and ρHe = 0.18 kg m3].

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Solution

(a)
Consider an atmospheric layer of thickness dy at height y and cross section area A at static equilibrium.

Mass of the layer = density×volume=Number of atoms per unit volume×volume×mass of an atom
M=ρAdy=mANdy
ρ=mN.......(i)

For equilibrium, upward force = downward force + weight
PA(P+dP)A=mNAdy g
where m:mass of an atom,N:Number of atoms per unit volume
dP=mANgdy
Substituting from (i),
dP=Aρgdy..........(ii)

From gas law,
P=NkT
P=ρkT/m
dP=kTmdρ..........(iii)

From (ii) and (iii),
kTmdρ=Aρgdy
dρρ=αdy where α is a constant
ρρodρρ=y0αdy
lnρlnρo=αy
lnρρo=αy
ρ=ρoeαy...........(iv)

Putting y=yo,ρ=ρo
α=1/yo............(v)

From (iv) and (v),
ρ=ρoey/yo
Hence proved.

(b) Density ρ= Mass / Volume
ρ= (Mass of the payload + Mass of helium) / Volume
=(m+VρHe)/V
=(400+1425×0.18)/1425
=0.46kgm3
From part (a)
ρ=ρ0ey/y0
loge(ρ/ρ0)=y/y0
y=8000×loge(0.46/1.25)
=8000×(1)
=8000m=8km
Hence, the balloon will rise to a height of 8 km.

381990_419498_ans_69c0207d6b9b4a79a7f64ea0c664cc44.png

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