1.A hot-air balloon achieves its buoyant lift by heating theair inside the ballo
ID: 1678196 • Letter: 1
Question
1.A hot-air balloon achieves its buoyant lift by heating theair inside the balloon, which makes it less dense than the airoutside. Suppose the volume of a balloon is 1600 m3 and the required lift is2200 N (rough estimate of the weightof the equipment and passenger). Calculate the temperature of theair inside the balloon which will produce the required lift. Assumethat the outside air temperature is 0°C and that air is anideal gas under these conditions.1°C
What factors limit the maximum altitude attainable by this methodfor a given load? (Neglect variables like wind.)
2.An ordinary glass is filled to the brim with 359.0 mL of water at 100°C. If the temperaturedecreased to 22°C, how much watercould be added to the glass?
1 mL
1.A hot-air balloon achieves its buoyant lift by heating theair inside the balloon, which makes it less dense than the airoutside. Suppose the volume of a balloon is 1600 m3 and the required lift is2200 N (rough estimate of the weightof the equipment and passenger). Calculate the temperature of theair inside the balloon which will produce the required lift. Assumethat the outside air temperature is 0°C and that air is anideal gas under these conditions.
1°C
What factors limit the maximum altitude attainable by this methodfor a given load? (Neglect variables like wind.)
2.An ordinary glass is filled to the brim with 359.0 mL of water at 100°C. If the temperaturedecreased to 22°C, how much watercould be added to the glass?
1 mL
Explanation / Answer
in the given problem there are three forcesto consider
the upward bouytant force (FB)
the downward force which is the weight of the hot airballoon (Wh)
the downward force which is the weight of thepassenger and his equipment (Wp)
accoeding to equilibrium when the blloon is to beraised at constant speed the bouyant force
must equal to the two weigts
FB = Wh + Wp
if mh is the mass of the hot air balloon and gis acceoleration due to gravity then
FB = mh + 2200 N
let c be the density of the cod andh is the density of the hot then
as we know that
density = mass m / volume V
according to newtons second law we can write
V c g = V h g + 2200N
the ideal gas law is given by
PV = nRT
if the gas law is written in terms of the gas density and the molecular mass M then the ideal gas
law changes to
PV = nRT
= (m / M)RT
(PM / R) = (m / V) T
= T
here we must recollect that
the gas inside and outside the balloon is the same soM will be same and moreover the pressure
is also same so the above equation reduces to
T = C (some constant)
= (T)c
= (T)h
so finally we get
V c g = V h g +2200 N
= V c (Tc / Th) g + 2200N
temperature of the air inside the balloon will be
Th = V cTc g / V cg - 2200 N
= ...........K
the factor that will be limiting is, the densityof the air decreases and the temperature required
will be increasing as a result the air becomestoo hot and the balloon fabric becomes burnt
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