Question

(a). It is known that density `rho` of air decreases with height y as
`rho=rho_(0)e^(-y//y(_o))` ltb rgt where `p_(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?
[take `y_(o)=8000m and rho_(He)=0.18kgm^(-3)`]

Answer 1

(a). We know that rate of decrease of density `p` of air is directly proportional to the height y. it is given as `drho//dy=-rho//y_(0)`
where y is constant of proportionally and -ve sign signifies that density is decreasing with increase in height. On integration, we get
`underset(rho_(0))overset(rho)int(drho)/(rho)=-int_(0)^(y)(1)/(y_(0))dy`
`implies[logrho]_(rho_(0))^(rho)=-[(y)/(y_(0))]_(0)^(y),` where `rho_(0)=` density of air at sea level i.e., `y=0`
or `log_(2)(rho)/(rho_(0))=-(y)/(y_(0))` or `rho=rho_(0)e^(-(y)/(y_(0)))`
here dimensions and units of constant `y_(0)` are same as of y.
(b). Here volume of He balloon, `V=1425m^(3)`, mass of payload, `m=400kg`
`y_(0)=8000m,` density of He `rho_(He)=0.18kgm^(-3)`
Mean density of balloon `rho=("Total mass of balloon")/("Volume")=(m+V*rho_(He))/(V)Pa`
`=(400+1425xx0.18)/(1425)=0.4608=0.46kgm^(-3)`
As density of air at sea level `rho_(0)=1.25kgm^(-3)` the balloon will rise up to a height y
where density of air `=`density of balloon `rho=0.46kgm^(-3)`
As `rho=rho_(0)e^(-(y)/(y_(0)))` or `(rho_(0))/(rho)=e^((y_(0))/(y))`
`thereforelog_(e)((rho_(0))/(rho))=(y_(0))/(y)` or `y=(y_(0))/(log_(e)((rho_(0))/(rho)))=(8000)/(log_(e)((1.25)/(0.46)))`
`=8002m` or `8.0km`