Question

(a) It is known that density r of air decreases with height y as ρ = ρₒ e ¯ _ y/yₒ where ρₒ = 1.25 kg m¯³is the density at sea level, and yₒ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of 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 yₒ = 8000 m and ρ _He = 0.18 kg m¯³].

Answer 1

The rate of decrease of density of the air is directly proportional to the height y.

dρ dy =− ρ y 0

Where y is constant of proportionality and negative sign shows the decrement with increase in height.

Integrate the above equation.

∫ dρ ρ =− ∫ dy y 0 [ lnρ ] ρ 0 ρ =− [ y y 0 ] 0 y ln ρ ρ 0 =− y y 0 ρ= ρ 0 e − y y 0

(b)

Given:

Volume of the balloon =V=1425  m 3

Mass of payload m=400 kg

y 0 =8000 m

Density of He gas is, ρ He =0.18  kgm -3

Mean density of the balloon is,

ρ= m+V ρ He V

Substitute the values.

ρ= 400+1425×0.18 1425 ρ=0.46  kgm -3

It is given that the density of the air at sea level is ρ 0 =1.25  kgm -3 . The balloon will rise up to a height of y where density of air is same as the density of the balloon.

ρ= ρ 0 e − y y 0 ln( ρ ρ 0 )=− y y 0 ln( ρ 0 ρ )= y 0 y y= y 0 ln( ρ 0 ρ )

Substitute the values.

y= 8000 ln( 1.25 0.46 ) y=8002 m y≃8 km