__Question 10.31__:*ρ*of air decreases with height y as ρ

_{0}e

^{-y/y0}

Where ρ

_{0}= 1.25 kg m

^{-3}

is the density at sea level, and

*y*

_{0}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 gremains constant.

(b) A large He balloon of volume 1425 m

^{3}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

*y*

_{0}= 8000 m and ρ

_{He}= 0.18 kg m

^{-3}]

__Solution__:**(a)**Volume of the balloon,

*V*= 1425 m

^{3}

Mass of the payload,

*m*= 400 kg

Acceleration due to gravity, g = 9.8 m/s

^{2}

y

_{0}= 8000 m

ρ

_{He}= 0.18 kg m

^{-3}

ρ

_{0}= 1.25 kg m

^{-3}

Density of the balloon =

*ρ*

Height to which the balloon rises =

*y*

Density (

*ρ*) of air decreases with height (

*y*) as:

ρ = ρ

_{0}e

^{-y/y0}

ρ / ρ

_{0 }= e

^{-y/y0}....(i)

This density variation is called the law of atmospherics.

It can be inferred from equation (

*i*) that the rate of decrease of density with height is directly proportional to

*ρ*, i.e.,

-(dρ / dy) ∝ ρ

(dρ / dy) = -kρ

(dρ / ρ) = -k dy

Where,

*k*is the constant of proportionality

Height changes from 0 to

*y*, while density changes from ρ

_{0}to ρ

Integrating the sides between these limits, we get:

y

_{0}= 1/k

k = 1/y

_{0}....(iii)

From equations (i) and (iii), we get

ρ = ρ

_{0}e

^{-y/y0}

^{}

(b) Density ρ = Mass / Volume

= (Mass of the payload + Mass of helium) / Volume

( m + Vρ

_{He}) / V

= (400 + 1425 X 0.18) / 1425

= 0.46 kg m

^{-3}

From equations (ii) and (iii), we can obtain y as:

ρ = ρ

_{0}e

^{-y/y0}

^{}

log

_{e}(ρ / ρ

_{0}) = -y / y

_{0}

∴ y = - 8000 X log

_{e}(0.46 / 1.25)

= -8000 X (-1)

= 8000 m = 8 km

Hence, the balloon will rise to a height of 8 km.

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