(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)`]

Explanation

(a) Let the pressure and density of air at the sea level be `P_0 and p_0,` respectively. Now let us consider a very thin layer of thickness dy at a height y from the sea level where the pressure is P and density of air is p. As we move dy upwards, the decrease in pressure is given by: `dP=-dypg ......(i)`
We shall learn the gas law in the later chapters:
`pV=m/M RT`
`m/V=(PM)/(RT)`
`p=((PM)/(RT))`
where m is the mass of gas sample, M is the molecular mass of the gas, R is the gas constant and T is the temperature.
Substituting p in equation (i)
`dP=-dy ((PM)/(RT))g`
`underset(P_(C))overset(P)int (dP)/(P)=-underset(0)overset(y)int ((M)/(RT)) gdy`
It is given that the tempearture T, accerleration due to gravity g are constant. M and R are also constant. Hence, `(Mg)/(RT)` can be taken out of integration,
`underset(P_(0))overset(P)int (dP)/(P)=-((Mg)/(RT)) underset(0)^(y) int dy`
`[ln P]_(P_(0))^(P) =-((Mg)/(RT)) [y]_(0)^(y)`
`rArr P=P_(0) e^(-(Mg)/(RT)y)`
Let `((RT)/(Mg))=y_(0) ("constant")`
Thus, `P=P_(0) e^(-y//y_(e))`
Now from the gas relation `p=((PM)/(RT))`
`rArr P=(pRT)/(M)`
Thus, `(pRT)/(M)=(p_(0)RT)/(M) e^(-y//y_(0))`
`rArr p=p_(0)e^(-y//y_(0))`
(b) The ballon will rise to a height where its density will become equal to density of air.
Mass of He gas in ballon `=1425 xx 0.18 =256.5 kg`
Density of ballon `=("Pay load +mass of He gas")/("Volume")`
`p=(400+256.5)/(1425)=0.46 km//m^(3)`
`p_(0)=1.25 kg//m^(3), y_(0)=8000 m`
`p=p_(0) e^(-y//y_(0))`
`0.46 =1.25 e^(-y//8000)`
`e^(y//8000)=(1.25)/(0.46) =2.72`
`(y)/(800)=log_(e) 2.72`
`y=8000 xx 1=8000 m=8km`