Question

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

ρ = ρ₀e^-y/ye

where ρ₀ = 1.25 kg m^-3 is the density at sea level, and y₀ is a constant. This density variation is called the law of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.

(b) A large H_e 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]

Answer 1

(a) Consider an imaginary cylinder of air having an area 'Q' of cross section 'a' (of each face) and placed vertically above the surface of earth. Let the Cap I of the cylinder be at a height (y) and the Cap II be at the height of (h + dy).

Let p and (p + dp) be the pressures at the caps I and II respectively. Then, the net force acting on the cap due to gravity is the weight of the cylinder, i.e., W = mass of cylinder x g

ady ρg   ...(i)

Since the cylinder is considered to be placed in air, therefore, force due to the difference in pressure is balanced by the weight of the cylinder under the equilibrium conditions.

Therefore,

[p - (p + dp)]a = W

⇒ -dp.a = ady ρg    [Using eqn.(i)]

⇒ dp = -ρg dy    ...(ii)

Now, under isothermal conditions,

pV = constant

⇒ pm/ρ = constant = K^-1

i.e., p = K'ρ/m = Kρ   (Where K = K'/m)

Here m is the mass of the air contained in the cylinder.

Thus, dp = Kdρ    ...(iii)

From eqns. (ii) and (iii), we have

Kdρ = -ρgdy

or, dρ/ρ = -gdy/K

Integrating both sides, we get

∫dρ/ρ = -g/K ∫dy + C

Where C is the constant of integration.

⇒ In ρ = -g/K y + C

Now, using the boundary conditions, when y = 0 (on sea level), ρ = ρ₀

From eqn. (iii),

In ρ₀ = -g/K x 0 + C

⇒ C = In ρ₀

Using this value of C in eqn. (iii), we get

In ρ = -g/K x y + In ρ₀

⇒ In ρ - In ρ₀ = -g/K y

Put g/K = 1/y₀

so that In ρ - In ρ₀ = -y/y₀

⇒ In ρ/ρ₀ = -y/y₀

⇒ ρ = ρ₀e^-y/y0   ...(iv)

which is the required relation

(b) Here, vol. of balloon.

V = 1425 m³

Density of helium,

ρ_He = 0.18 kg m^-3

y₀ = 8000 m

and mass of the payload = 400 kg

Let y be the height upto which the balloon is able to rise,

Then, upthrust at the height

y = Weight of balloon + Weight of payload

⇒ ρgV = ρ_HegV + mg

Dividing both sides by gV,

ρ = ρ_He + m/V = 0.18 + 400/1425 = 0.46

Now using eqn. (iv),

ρ = ρ₀e^-y/y0

⇒ ρ₀e^-y/y0 = 0.46

i.e., 1.25 e^-y/8000 = 0.46

i.e., e^y/8000 = 1.25/0.46 2.71

Taking natural log of both sides, we have

y/8000 = In 2.71

= 2.3026 log 2.71 = 2.3026 x 0.4330 = 0.997

or, y = 0.997 x 8000 = 7975 m

i.e., the balloon shall be able to rise to 7975 m i.e., approximately 8 km, above the sea level.