It is known that density `rho` of air decreases with height y as
`-y//y_(0)`
`rho=rho_(o)e`
where `rho_(o)=1.25kgm^(-3)` is the density at sea level and `y_(o)` 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 `1425m^(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 ? `(y_(o)=8000mandrho_(He)=0.18kgm^(-3))`.

Explanation

The rate of decrease of density `rho` of air is directly proportional to the height y. It is given as ,
`-(drho)/(dy)proprhoor-(drho)/(dy)=-krho`
where k= constant and negative sign indicates that density decreases as height increases.
From height 0 to y density becomes `rho_(o)` to `rho` . By integrating above equation,
`int_(rho_(o))^(rho)(1)/(rho)=-kint_(o)^(y)dy`
`[Inrho]_(rho_(o))^(rho)=-k[y]_(o)^(y)`
`[In rho-Inrho_(o)]=-k[y=0]`
ln `((rho)/(rho_(o)))=-ky`
`log_(e)((rho)/(rho_(o)))=-ky`
`therefore(rho)/(rho_(o))=e^(-ky)`
`thereforerho=rho_(o)^(e-ky)`
Taking constant `k=(1)/(y_(o))`
`rho=rho_(o)e^((y)/(y_(0)))`
The balloon will rise upto a height where density of air equal to the density of balloon Volume of ballon `V=1425m^(3)`
Mass of He has in balloon,
`V=1425xx0.18=256.5kg`
Total mass of balloon (with payload),
`M=400+256.5`
`=656.5kg`
Density of balloon `rho=(M)/(V)=(656.5)/(1425)`
`rho=0.46kg//m^(3)`
`y_(o)=8000m`
`rho_(o)=1.25kg//m^(3)`
`rho=0.46kg//m^(3)`
`rho=rho_(o)e^((y)/(y_(o)))`
`therefore0.46=1.25_(e)^((-y)/(8000))`
`thereforee^((y)/(8000))=(1.25)/(0.46)=2.7`
`thereforey=8000loge^(2.7)`
`=8000xx1.025`
`=8200m`
`=8.2km`