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Where 
= 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 gremains 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 
= 0.18 kg m–3].
Asked by Pralhad Last Modified
Answer 
Tasneem
(a) Volume of the balloon, V = 1425 m3
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s2
Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as:
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 
to ρ.
Integrating the sides between these limits, we get:
(b)
Hence, the balloon will rise to a height of 8 km.
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