It is known that density ρ of air decreases with height y as O^e(^(y/y0)) Where ρ0 = 1.25 kg m^-3 is the density

Question

a) It is known that density ρ of air decreases with height y as O^{e^(y/y₀)} 

Where ρ₀ = 1.25 kg m^-3  is the density at sea level, and y₀ 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 y₀= 8000 m and ρ_He= 0.18 kg m^–3 ].

Answer 1

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.,

Where, k is the constant of proportionality

Height changes from 0 to y, while density changes from P₀ to ρ.

Integrating the sides between these limits, we get:

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