$(a)$ It is known that the density $\rho$ 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 a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of the atmosphere remains 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 a constant radius as it rises. How high does it rise?
[Take $y_{0} = 8000 \; m$ and $\rho_{He} = 0.18 \; kg \, m^{-3}$]

(N/A) Let $P$ be the pressure at height $y$. The change in pressure $dP$ for a small change in height $dy$ is $dP = -\rho g dy$.
Since the atmosphere is isothermal,$PV = nRT$,which implies $P = \frac{\rho RT}{M}$,where $M$ is the molar mass of air.
Thus,$\rho = \frac{PM}{RT}$. Substituting this into the pressure equation: $dP = -\frac{PM}{RT} g dy$,or $\frac{dP}{P} = -\frac{Mg}{RT} dy$.
Integrating from $y=0$ (where $P=P_{0}$) to $y$ (where $P=P$): $\ln(\frac{P}{P_{0}}) = -\frac{Mg}{RT} y$.
So,$P = P_{0} e^{-y/y_{0}}$,where $y_{0} = \frac{RT}{Mg}$. Since $\rho \propto P$ at constant temperature,$\rho = \rho_{0} e^{-y/y_{0}}$.
$(b)$ The balloon rises until its density $\rho$ equals the density of the surrounding air.
Total mass of the balloon system $M_{total} = m_{payload} + m_{He} = 400 + (1425 \times 0.18) = 400 + 256.5 = 656.5 \; kg$.
Density of the balloon $\rho = \frac{M_{total}}{V} = \frac{656.5}{1425} \approx 0.4607 \; kg \, m^{-3}$.
Using the law of atmospheres: $\rho = \rho_{0} e^{-y/y_{0}} \implies 0.4607 = 1.25 e^{-y/8000}$.
$\ln(\frac{0.4607}{1.25}) = -\frac{y}{8000} \implies \ln(0.36856) = -\frac{y}{8000}$.
$-0.998 \approx -\frac{y}{8000} \implies y \approx 8000 \; m = 8 \; km$.