Calculating Air Density and Temperature in a Hot Air Balloon

In summary, the conversation discusses a hot-air balloon with a volume of 2000 m^3 and a lift of 2720 N. Using the equations Fnet = Fbouyant - Fgrav, PV = nRT, P=F/A, and d=m/V, the density of the air inside the balloon is found to be 1.06 kg/m^3 and the number of moles of air inside is calculated to be 75714 mol. The temperature of the air in the balloon is determined to be 259 K and the mean kinetic energy and mean velocity of the air are calculated to be 6.66 x 10^-21 J and 2.51 x 10^-12 m/s,
  • #1
toothpaste666
516
20

Homework Statement


A hot-air balloon has a volume of 2000 m^3 and generates a lift of 2720 N (assume the outside temperature is 20◦C with an air density of 1.2 kg/m^3).

A)What is the density of the air inside the balloon?
B)How many moles of air are inside of the balloon (the molecular mass of air is 28 g/mol)?
C) What is the temperature of the air in the balloon ?
D)What is the mean kinetic energy of the air in the balloon?
E) What is the mean velocity (not vrms, and not speed) of the air in the balloon?

Homework Equations


Fnet = Fbouyant - Fgrav
PV = nRT
P=F/A
d=m/V

The Attempt at a Solution


A) the lift is the net force on the balloon
Flift = Fbouyant - Fgrav

[itex] F_{lift} = m_{displaced}g - m_{balloon}g [/itex]

where mdisplaced is the mass of the displaced air and mballoon is the mass of the air in the balloon
since denisty d = m/V then m = Vd

[itex] F_{lift} = d_{displaced}Vg - d_{balloon}Vg [/itex]

[itex] \frac{F_{lift}}{Vg} = d_{displaced} - d_{balloon} [/itex]

[itex] d_{balloon} = d_{displaced} - \frac{F_{lift}}{Vg} [/itex]

plugging in the info from the problem

[itex] d_{balloon} = 1.2 kg/m^3 - \frac{2720 kgm/s^2}{(2000 m^3)(9.8 m/s^2)} [/itex]

[itex] d_{balloon} = 1.06 kg/m^3 [/itex]

B) d= m/V so m = Vd
mass of air in balloon = 1.06 kg/m^3 * 2000m^3 = 2120 kg
molar mass of air is given as 28g/mol or .028kg/mol
2120 kg (mol/.028 kg) = 75714 mol of air n the balloon

C) PV = nRT
we have moles , volume and the ideal gas constant. Assuming the pressure is 1 atm we can find T
T = PV/nR = (101300 Pa)(2000m^3)/(75714mol)(8.314 J/molK) = 322 K

D) K = (3/2)kT where k is boltzmanns constant
K = 1.5 * (1.38x10^-23J/K) (322K) = 6.66 x10 ^-21 J

E) K = .5mv^2
we know the mass of air in the balloon and K of the air in the balloon. solving for v:
v^2 = 2K/m
v = sqrt(2K/m) = sqrt((2*(6.66E-21))/2120) = 2.51E-12

can anyone provide feedback to see if i solved this question correctly?
 
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  • #2
I agree with A and B, but I don't think you should assume ambient pressure is 1atm in C. There is enough info provided to calculate it. E.g., from the answer to A you can quickly deduce C by considering the ratio of the densities and the ratio of the absolute temperatures. This gives a rather higher temperature.
 
  • #3
so I can compare the ratios because the change in density is a result of the change in temp?
dballoon/dair = Tballoon/Tair
Tballoon = Tair(dballoon/dair) = 293K (1.06/1.2) = 259 K ?
 
  • #4
toothpaste666 said:
so I can compare the ratios because the change in density is a result of the change in temp?
dballoon/dair = Tballoon/Tair
Tballoon = Tair(dballoon/dair) = 293K (1.06/1.2) = 259 K ?
Nearly right, but do you think a higher density means a higher temperature or a lower temperature?
 
  • #5
the mass increases with the number of moles. the density increases with mass and decreases with volume.
n/v = P/RT
if you increase the number of moles the density increases and T must decrease. if you increase the volume the density decreases and T must increase ( does this make sense? want to be sure I am understanding)
so it would be
dballoon/dair = Tair/Tballoon
?
 
  • #6
toothpaste666 said:
the mass increases with the number of moles. the density increases with mass and decreases with volume.
n/v = P/RT
if you increase the number of moles the density increases and T must decrease. if you increase the volume the density decreases and T must increase ( does this make sense? want to be sure I am understanding)
so it would be
dballoon/dair = Tair/Tballoon
?
Yes.
 
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Likes toothpaste666
  • #7
thank you!
 

1. What is the "hot air balloon problem"?

The "hot air balloon problem" refers to the challenge of controlling the altitude and direction of a hot air balloon. This problem has been studied by scientists and engineers in the field of aeronautics.

2. How does a hot air balloon work?

A hot air balloon works by using a burner to heat the air inside the balloon's envelope, creating a difference in air density and causing the balloon to rise. The pilot can control the altitude by adjusting the amount of heat in the envelope.

3. What are the main challenges in controlling a hot air balloon?

The main challenges in controlling a hot air balloon include wind direction and speed, temperature changes in the surrounding air, and the weight of the balloon and its passengers. These factors can affect the balloon's ability to maintain a steady flight path.

4. How do scientists and engineers address these challenges?

Scientists and engineers use various techniques and technologies to address the challenges of controlling a hot air balloon. This includes advanced weather forecasting, lightweight materials for the balloon envelope, and precise control systems for the burner and other components.

5. Are there any safety concerns with hot air balloons?

Like any form of aviation, there are safety concerns with hot air balloons. However, strict regulations and safety protocols are in place to ensure the safety of the passengers and crew. It is important for pilots and operators to regularly maintain and inspect their equipment to minimize any potential risks.

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