one hell of a question........

er ..... chemistry forum?

How in the world do you answer this question? where do you start?

"The modern hot-air balloon is fuelled by inexpensive propane and constructed of
lightweight nylon fabric. The air in the balloon cavity is heated by a propane burner
located at the top of the passenger cage. The flames from the burner that shoot into
the balloon heat the air in the balloon cavity, raising the air temperature at top of the
balloon from 65oC to over 120oC. The air temperature is maintained at the desired
levels by periodically firing the propane burner. The buoyancy force that pushes the
balloon upward is proportional to the density of the cooler air outsider the balloon and
the volume of the balloon, and can be expressed as Fb = ρcool air gVballoon where g is the
gravitational acceleration. Consider a 20-m-diameter hot-air balloon that, together
with its cage, has a mass of 80 kg when empty. This balloon is hanging still in the air
at a location where the atmospheric pressure and temperature are 90 kPa and 15oC,
respectively, while carrying three 65 kg people. Determine the average temperature of
the air in the balloon. What would your response be if the atmospheric air temperature
were 30oC".

why, thank you kind sir!

and you're right - that's what I thought, it's more physics than anything else...

takes a deep breath....

Fb = ρ(cool air).g.V(balloon)

ok, assume that the ballloon is a sphere.
It's diameter is 20m therefore its volume 4/3pi.r^3 = 4189 m^3

when hovering, force due to gravity on the balloon = mass x g = ρ(cool air).g.V(balloon)
g cancels out

mass = ρ(cool air).V(balloon)

ρ(cool air) = [80 + (3 x 65)]/4189 = 0.065648 kg/m^3
----------------------------------------------------------------------

When the atmospheric pressure = 90kPa and T = 288k the mass of air displaced = the difference in
mass of air inside and outside the balloon.

Mass displaced = (volume x density)outside - (volume x density)inside
but volume(inside) = volume(outside)

mass displaced/volume = density(outside - inside)

275/4189 = density(outside) - density(inside)
0.071999 = density(outside) - density(inside)

therefore density(inside) = density(outside) - 0.071999

so lets work out the density outside from the conditions given...

conditions outside PV=RT (for 1 mole)
molar volume = RT/P = 26.6048 dm^3

conditions inside balloon PV=RT (for 1 mole)

average RMM of air = (28 x 0.8) + (32 x 0.2) = 28.8
so 1 mole of air = 28.8g = 0.0288kg
volume occupied by 1 mole = 26.6048dm^3 = 0.026604 m^3

density(outside) = mass/volume = 0.0288/0.026604 = 1.0825 kg/m^3

density(inside) = density(outside) - 0.071999

density(inside) = 1.0825 - 0.0720 = 1.01054

so we've got the inside density, we know the volume so we can work out the mass of air

density = mass/volume so mass = density x volume = 1.01054 x 4189 = 4233.1725kg

expressed in grams = 4233172.5g
expressed in moles = 146985.15 moles

pV = nRT

pressure must be the same as outside = 90kPa
volume is the volume of the balloon = 4189m^3 = 4189000dm^3
n= 146985.15 moles
R = 8.314

therefore the average inside air temperature = PV/nR = 308.5K = 35.5ºC

in your answer you write:

"275/4189 = density(outside) - density(inside)
0.071999 = density(outside) - density(inside)"

but 275/4189 = 0.06565 and NOT 0.071999
I am not sure if you done this differently but the answer change from 35.5degrees to 33.6

And you are right about the universal gas constant because it says "ideal gas constant of air = 0.287 kPa.m(cubed)/kg-1.K-1"
Would that still give me the same value of 8.314???