(a) It is known that density \[\mathbf{\rho }\] of air decreases with height \[y\] as \[{{\mathbf{\rho }}_{0}}{{e}^{y/{{y}_{0}}}}\]. Where \[{{\mathbf{\rho }}_{0}}=1.25kg\,{{m}^{-3}}\]the density at sea is 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 a constant (isothermal conditions). Also assume that the value of \[g\] remains constant.
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Ans: The given volume of the balloon is, \[V\text{ }=\text{ }1425\text{ }{{m}^{3}}\]
Mass of the payload, \[m=400kg\]
And the acceleration due to gravity is, \[g=9.8m/{{s}^{2}}\]
\[{{y}_{0}}=8000m\]
The density of helium is\[{{\rho }_{He}}=0.18kg\,{{m}^{-3}}\] , and that of air at sea level is \[{{\rho }_{0}}=1.25kg\,{{m}^{-3}}\]
Let the density of the balloon be ρ and the height to which the balloon can rise be y.
The air density (\[\rho \]) of decreases with increasing height (\[y\]) as:
\[\rho /{{\rho }_{0}}={{e}^{-y/{{y}_{0}}}}\]
This relation is also called the law of atmospherics.
It can be inferred from the above law-equation that the rate of decrease of density with height is directly proportional to \[\rho \],
\[-\frac{d\rho }{dy}\propto \rho \]
\[\Rightarrow -\frac{d\rho }{dy}=k\rho \]
\[\Rightarrow -\frac{d\rho }{\rho }=kdy\]
Here, k is a proportionality constant.
We set the limits such that the height changes from \[0\] to \[y\], and the density varies from \[{{\rho }_{0}}\]to \[\rho \].
\[\int\limits_{{{\rho }_{0}}}^{\rho }{\frac{d\rho }{\rho }=}\int\limits_{0}^{y}{kdy}\]
Next, integrating the sides between these limits, we obtain:
\[\left[ \ln \,\rho \right]_{{{\rho }_{0}}}^{\rho }=-ky\]
\[\Rightarrow \rho /{{\rho }_{0}}={{e}^{-ky}}\]
Comparing the above equation with the law of atmospherics, we get
\[\Rightarrow k=\frac{1}{{{y}_{0}}}\]
In conclusion, we have been able to derive the expression for the law of atmospherics.
(b) A large He balloon of volume \[1425{{m}^{3}}\] is used to lift a payload of\[\mathbf{400}\text{ }\mathbf{kg}\]. Assume that the balloon maintains a constant radius as it rises. How high does it rise?
(Take \[{{y}_{0}}=8000m\]and \[{{\mathbf{\rho }}_{He}}=0.18kg\,{{m}^{-3}}\] )
Ans: The generic expression for mass density is:
\[\rho =\frac{M}{V}\]
In the current case, it becomes the total mass (including the payload and the helium) per unit volume of the balloon:
\[\rho =\frac{({{M}_{payload}}+{{M}_{He}})}{V}=\frac{m+V{{\rho }_{He}}}{V}\]
\[\Rightarrow \rho =0.46kg/{{m}^{3}}\]
Using the law of atmospherics, we can obtain the value of \[y\]:
\[\rho /{{\rho }_{0}}={{e}^{-y/{{y}_{0}}}}\]
\[\Rightarrow \ln \,\rho /{{\rho }_{0}}=-y/{{y}_{0}}\]
\[\Rightarrow y=-8000\times ln(0.46/1.25)=8000m=8km\]
Therefore, the balloon can rise to a height of 8 km.