Question

It is known that density ρ of air decreases with height y as `0^(e^(y/y_0))` Where`rho_0` = 1.25 kg m^–3 is the density at sea level, and y₀ 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 gremains constant

Answer 1

Volume of the balloon, V = 1425 m³

Mass of the payload, m = 400 kg

Acceleration due to gravity, g = 9.8 m/s²

^{`y_0 = 8000m`}

`rho_"He" = 0.18 kg m^(-3)`

`rho_0 = 1.25 "kg/m"^3`

Density of the balloon = ρ

Height to which the balloon rises = y

Density (ρ) of air decreases with height (y) as:

`rho = rho_0^(e^(-y/y_0))`

`rho/rho_0 = e^(-y/y_0)` ...(i)

This density variation is called the law of atmospherics.

It can be inferred from equation (i) that the rate of decrease of density with height is directly proportional to ρ, i.e.,

`-(drho)/dy prop rho`

`(drho)/dy = -krho`

`(drho)/rho = -kdy`

Where, k is the constant of proportionality

Height changes from 0 to y, while density changes from `rho_0` to ρ.

Integrating the sides between these limits, we get:

`int_(rho_0)^rho (drho)/rho = -int_0^y kdy`

`[log_e rho]_(rho_0)^rho = -ky`

`rho/rho_0 = e^(-ky)`  ... (ii)

Comparing equation i and ii we get

`y_0 = 1/k`

`k= 1/y_0`  ...(iii)

Fromequation i and iii we get

`rho = rho_0^(e^(-y/y_0))`