Question

(a) It is known that density of air decreases with height y as
$\rho ={\rho }_0{e}^{-y/y_0}$
where ${\rho }_0$ = 1.25 kg m${}^{-3}$ is the density at sea level, and y${}_0$ is constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of the atmosphere remains a constant (isothermal conditions). Also, assume that the value of g remains constant.
(b) A large He balloon of volume 1425 m${}^3$ 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${}_0$ = 8000 m and ${\rho }_{He}$ = 0.18 kg m${}^{-3}$].

Answer 1

(a)

Consider an atmospheric layer of thickness dy at height y and cross section area A at static equilibrium.

$\text{Mass of the layer = density}\times \text{volume=Number of atoms per unit volume}\times \text{volume}\times \text{mass of an atom}$

$M=\rho Ady=mANdy$

$⟹\rho =mN.......\left(i\right)$

For equilibrium, upward force = downward force + weight

$PA-(P+dP)A=mNAdyg$

where $m:\text{mass of an atom},N:\text{Number of atoms per unit volume}$

$dP=-mANgdy$

Substituting from (i),
$dP=-A\rho gdy..........\left(ii\right)$

From gas law,

$P=NkT$

$P=\rho kT/m$

$dP=\frac{kT}{m}d\rho ..........\left(iii\right)$

From (ii) and (iii),

$\frac{kT}{m}d\rho =-A\rho gdy$

$\frac{d\rho }{\rho }=-\alpha dywhere\alpha \text{is a constant}$

${\int }_{{\rho }_o}^{\rho }\frac{d\rho }{\rho }={\int }_0^y-\alpha dy$

$ln\rho -ln{\rho }_o=-\alpha y$

$ln\frac{\rho }{{\rho }_o}=-\alpha y$

$\rho ={\rho }_o{e}^{-\alpha y}...........\left(iv\right)$

Putting $y=y_o,\rho ={\rho }_o$

$\alpha =1/y_o............\left(v\right)$

From (iv) and (v),

$\rho ={\rho }_o{e}^{-y/y_o}$

Hence proved.

(b) Density $\rho =$ Mass / Volume

$\rho =$ (Mass of the payload + Mass of helium) / Volume
$=(m+V{\rho }_{He})/V$
$=(400+1425\times 0.18)/1425$

$=0.46kg{m}^{-3}$

From part (a)

$\rho ={\rho }_0{e}^{-y/y_0}$

$lo{g}_e\left(\rho /{\rho }_0\right)=-y/y_0$
$\therefore y=-8000\times lo{g}_e\left(0.46/1.25\right)$
$=-8000\times (-1)$
$=8000m=8km$
Hence, the balloon will rise to a height of 8 km.