Question

On a sunny morning in a field, the atmospheric pressure is 105N/m2 and the temperature is 279 K and you are at a family reunion. Uncle Louie has filled a balloon with 1000m3 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 105N/m2 . No air enters or leaves the balloon. How much heat must be added to the air in the balloon to initiate the flight? ( please show all the steps and try to be as clear as possible.thank you!)

Answer #1

P @ atmospheric =
10^{5} N/m^{2} : Temp of atmosphere = T_{1}
= 279 K : Volume of air filled in ballon = 1000 m^{3}

Temp of preheated
air filled in ballon = 300 K= T_{2}

mass of empty ballon = 240 Kg

**Consider
system as ballon and system boundry inside ballon. Consider air as
ideal gas**

Ein - Eout = Change of energy of system (As per first law of thermodynamics)

Q _{in} = M
_{total} C_{p} (T_{2}- T_{1})

M total = M ballon + M air = 240 + M air

M air = Density of air x volume of air occuipied

Density of air = P/
RT = 10^{5}/ (8.31 x 300) = 40.11 Kg/m^{3}_{( R
is universal gas constant with value 8.31 KJ/Kmol. K)}

M of air = 40.11 x 1000 = 40112.3 Kg

Therefore, Total Mass M total = 240 + 40112.3 = 40352.3 kg

Total heat added to
the air in the ballon to initiate the flight Q = M_{total}
C_{air} (T_{2}-T_{1})

C_{air} =
Specific heat of air at constant pressure = 1.005 KJ/kg. K

Q = 40352.3 X 1.005X (300-279) =
**851635.26** KJ

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