Question

# On a sunny morning the atmospheric pressure is 10^5 N/m^2 and the temperature is 279 K...

On a sunny morning the atmospheric pressure is 10^5 N/m^2 and the temperature is 279 K and you are at a family reunion. A balloon has been filled with 1000m^3 of preheated air at atmospheric pressure and a temperature of 300 K. The mass of the empty balloon and its load is 240 Kg. Under these conditions, the balloon rises above its burden, but the airship remains grounded. To begin the ascent, the air within the balloon is brought to a higher temperature. It expands but remains at the pressure of 10 5 N/m2 . No air enters or leaves the balloon. How much heat must be added to the air in the balloon to initiate the fight?

P = ρRT

Where:

P is the absolute pressure of the gas, in Pa

ρ is the density of the gas, in kg/m3

R is the gas constant, in Joules/kg.K

T is the absolute temperature of the gas, in Kelvins (K)

The normal air outside has a density 0f 1.2 kg/m3

Since there will be buoyancy force coming into action due to difference in densities

the buoyancy force

Fb = ( 1.2 - ρ )*volume

Fb = (1.2 - ρ)*V2

This should be equal to 240 kg of load.

ALso the new volume V2 = (T2*V1) / T1

V2 = (T2*1000) / 300

From P = ρ*R*T

10^5 = ρ*287*T2

=> ρ = (10^5) / (287*T2)

substituting these ρ and V2 values in the force balance equation. we get:

240 = [ 1`.2 - { (10^5) / (287*T2) }]*[ (T2*1000) / 300 ]

Solving this we get T2 = 350.36K

Heat = m*Cp*(T2-T1)

Cp = 1.005 KJ/kg*K

Cp = 1005 J/kg*k

m = density * volume

m = 0.9944 *1167.87

m = 1161.32 kg.

So Heat = 1161.32*1005*(350.36-300)

Heat = 58776830.29 J

= 58776.83 KJ.

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