# Question about calculating the minimum temperature in hot air balloons

• ianc1339
ianc1339
Homework Statement
A hot air balloon has a volume of 2500 m^3 and the fabric and the passengers have mass of 800 kg. It is taking-off when the air around the balloon has a temperature of 10 degrees Celsius and a density of 1.29 kgm^-3. Calculate the minimum temperature of the air inside the balloon for take off. Note that the pressure inside and outside the balloon must be equal, since the balloon is open at the bottom.
Relevant Equations
pV=nRT
pV=1/3Nm<c^2>
<E_k>=1/2m<c^2>=3/2kT
p = 1/3 x density x <c^2>
First, I tried using the Archimedes principle and calculated the weight of the surrounding air displaced when taking off.

##W = 2500\times 1.29\times 9.81 = 31637.25 N##

But then, I got stuck and do not know how to proceed from here on.
I don't want the full solution yet but can I get some guidance/hints as to how to approach this question?

Thanks

ianc1339
What is the buoyant force on the balloon, in terms of its variable volume ##V##?

Is it ##\rho V g##?

Apologies, I misinterpreted the question. The volume is most definitely not variable!

You have the balloon at that given volume, and you know the buoyant force and the weight of the material. What must be the weight of the air inside for equilibrium? How might we then relate density and temperature?

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• Lnewqban and ianc1339
ianc1339
Apologies, I misinterpreted the question. The volume is most definitely not variable!

You have the balloon at that given volume, and you know the buoyant force and the weight of the material. What must be the weight of the air inside for equilibrium? How might we then relate density and temperature?

So the weight of the material is ##W=mg=800\times 9.81 = 7848N##. So the weight of the air inside must be ##31637.25 - 7848 = 23789.25N##.

Since volume is proportional to temperature, as temperature increases, density decreases. So temperature is inversely proportional to temperature?

So temperature is inversely proportional to temperature?

Careful , though I think you meant to say density, which is indeed correct. You can show it from the good old ideal gas law ##PV = nRT \implies P = \frac{n}{V} RT##. Scale ##n## by the molar mass and you get a different constant of proportionality, but essentially ##\rho \propto \frac{1}{T}##.

You're already given one pair of values for density and temperature, so can you work out the new temperature? You could write out the proportionality and find the constant, or you could equate some 'conserved' quantity.

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• Lnewqban and ianc1339
ianc1339
So the relation becomes ##\frac{1}{T}=\frac{R}{p}\times\rho## where ##\frac{R}{p}## is a constant.
##\frac{R}{p} = \frac{1}{T\rho} = 2.7377\times 10^{-3}##.

Since the weight of the air inside the balloon before take off is ##23789.25N##, the density of the air inside the balloon is ##\rho = \frac{m}{V}=\frac{W/9.81}{V} = \frac{23789.25/9.81}{2500} = 0.97kgm^{-3}##

So the temperature inside the balloon just before take off is ##\frac{1}{T}=2.7377\times 10^{-3}\times 0.97kgm^{-3}\\ \therefore T = 376.6K##

And yes, this was the answer (The answer stated 380K rounded to 2 s.f.).

Thank you so much for your help!

• Lnewqban and etotheipi
So the relation becomes ##\frac{1}{T}=\frac{R}{p}\times\rho## where ##\frac{R}{p}## is a constant.
##\frac{R}{p} = \frac{1}{T\rho} = 2.7377\times 10^{-3}##.

This is essentially correct, just note that the ##R## in that equation is the so-called "specific" gas constant, i.e. just divided by the molar mass: ##R_{\text{specific}} = \frac{R}{M_{r}}##.

• ianc1339 and Lnewqban