Ncert Exercises & Solutions

10.31 (a) It is known that density ρ of air decreases with height y as:

where ρ₀ = 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 a constant (isothermal conditions). Also, assume that the value of g remains constant.

(b) A large He balloon of volume 1425 m³ 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 y₀ = 8000 m and $ρ_{He}$ = 0.18 kg m^–3].

(a)

Consider a layer of thickness dy at the height y and cross-section area A.

Mass of the layer = Mass of the atoms in the layer

$M = ρAdy = No . of atoms per unit volume \times volume \times mass of an atom$
$\Rightarrow ρAdy = mNAdy$
$\Rightarrow ρ = mN$
$Pressure force = weight of the layer$
$PA - (P + dP) A = mNAdyg$
$dP = - mNdyg = - ρdyg$
$From ideal gas equation : P = NkT = \frac{ρkT}{m}$
$dP = \frac{kT}{m} dρ$
$So, \frac{kT}{m} dρ = - ρgdy$
$\frac{dρ}{ρ} = - \frac{mg}{kT} dy = - cdy$
$Integrating it :$
$\int_{ρ_{o}}^{ρ} \frac{dρ}{ρ} = - \frac{mg}{kT} dy = - \int_{0}^{y} cdy$
$\ln (ρ) - \ln (ρ_{0}) = - cy$
$\ln (\frac{ρ}{ρ_{0}}) = - cy$
$\frac{ρ}{ρ_{0}} = e^{- cy}$
$When y = y_{0} \Rightarrow ρ = ρ_{0} \Rightarrow c = \frac{1}{y_{0}}$
$\Rightarrow \frac{ρ}{ρ_{0}} = e^{- \frac{y}{y_{0}}} \Rightarrow ρ = ρ_{0} e^{- \frac{y}{y_{0}}}$

(b)

$Density, ρ = \frac{Mass}{Volume}$
$= \frac{Mass of the payload + Mass of helium}{Volume}$
$= \frac{m + {Vρ}_{He}}{V}$
$= \frac{400 + 1425 \times 0.18}{1425}$
$ρ = ρ_{0} e^{- y / y_{0}}$
$\log_{e} \frac{ρ}{ρ_{0}} = - \frac{y}{y_{0}}$
$∴ y = - 8000 \times \log_{e} \frac{0.46}{1.25}$
$= - 8000 \times - 1$
$= 8000 m = 8 km$

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