I really and no one can seem to help

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Homework Help Overview

The discussion revolves around a physics problem related to buoyancy and hot-air balloons. The original poster presents a scenario where a hot-air balloon needs to achieve a specific lift by heating the air inside it, which affects its density compared to the outside air. The problem provides specific values for the balloon's volume, required lift, and outside air temperature, asking for the temperature of the air inside the balloon to achieve the desired lift.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between temperature, pressure, and density of gases, referencing the ideal gas law. Some express frustration with the problem's complexity and seek clarification on the assumptions involved, such as the pressure conditions inside the balloon.

Discussion Status

The discussion is ongoing, with participants sharing various approaches and hints related to the problem. Some have attempted to break down the equations involved, while others express confusion and seek further assistance. There is no clear consensus yet, but several lines of reasoning are being explored.

Contextual Notes

Participants note the challenge posed by the problem's requirements and the need for a deeper understanding of buoyancy and gas behavior under different conditions. The original poster's frustration and the mention of similar problems indicate a broader context of difficulty with this topic.

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I really need help... and no one can seem to help!

Here's the question: A hot-air balloon achieves its buoyant lift by heating the air inside the balloon, which makes it less dense than the air outside. Suppose the volume of a balloon is 1794 m3 and the required lift is 3139 N (weight of equipment and passenger). Calculate the temperature of the air inside the balloon which will produce the desired lift. Assume that the outside air temperature is 0oC and that air is an ideal gas under these conditions. Express your answer in oC. The density of air at STP is 1.29 kg/m3.

Someone else even posted this on... I think... it is too hard. I do not understand and no one can help me at all. I am really frustrated and all of my friends have given up, but I really need to get it right. Please help me.
 
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Jayhawk1 said:
Here's the question: A hot-air balloon achieves its buoyant lift by heating the air inside the balloon, which makes it less dense than the air outside. Suppose the volume of a balloon is 1794 m3 and the required lift is 3139 N (weight of equipment and passenger). Calculate the temperature of the air inside the balloon which will produce the desired lift. Assume that the outside air temperature is 0oC and that air is an ideal gas under these conditions. Express your answer in oC. The density of air at STP is 1.29 kg/m3.

Someone else even posted this on... I think... it is too hard. I do not understand and no one can help me at all. I am really frustrated and all of my friends have given up, but I really need to get it right. Please help me.

Same problem- different numbers

https://www.physicsforums.com/showthread.php?p=552427&posted=1#post552427
 
Jayhawk1 said:
Here's the question: A hot-air balloon achieves its buoyant lift by heating the air inside the balloon, which makes it less dense than the air outside. Suppose the volume of a balloon is 1794 m3 and the required lift is 3139 N (weight of equipment and passenger). Calculate the temperature of the air inside the balloon which will produce the desired lift. Assume that the outside air temperature is 0oC and that air is an ideal gas under these conditions. Express your answer in oC. The density of air at STP is 1.29 kg/m3.

Someone else even posted this on... I think... it is too hard. I do not understand and no one can help me at all. I am really frustrated and all of my friends have given up, but I really need to get it right. Please help me.
SOLUTION HINTS:
{Density} = ρ ∝ P/(R*T) ::: ⇒ ρ12 = T2/T1
{Density @ STP (0 degC)=(273.1 degK)} = ρ(273.1) = (1.29 kg/m^3)
{Density @ SP & Temp=(T degK)} = ρ(T) = (1.29 kg/m^3)*(273.1 degK)/T

{buoyant Force} = {Volume Displaced}*(g)*{ρ(273.1) - ρ(T)} =
= {Volume Displaced}*(g)*{(1.29 kg/m^3) - (1.29 kg/m^3)*(273.1 degK)/T} =
= {Volume Displaced}*(g)*(1.29 kg/m^3)*{1 - (273.1 degK)/T} =
= (1794 m^3)*(9.81 m/sec^2)*(1.29 kg/m^3)*{1 - (273.1 degK)/T} =
= (22703)*{1 - (273.1 degK)/T}

buoyant Force of (3139 N) required:
(3139 N) = (22703)*{1 - (273.1 degK)/T}
::: ⇒ {1 - (273.1 degK)/T} = (0.13826)
::: ⇒ (273.1 degK)/T = 1 - (0.13826)

Solve last equation for "T" in (degK). Convert "T" to (degC) by subtracting (273.1).


~~
 
Last edited:
Correction

Jayhawk1 said:
Here's the question: A hot-air balloon achieves its buoyant lift by heating the air inside the balloon, which makes it less dense than the air outside. Suppose the volume of a balloon is 1794 m3 and the required lift is 3139 N (weight of equipment and passenger). Calculate the temperature of the air inside the balloon which will produce the desired lift. Assume that the outside air temperature is 0oC and that air is an ideal gas under these conditions. Express your answer in oC. The density of air at STP is 1.29 kg/m3.

Someone else even posted this on... I think... it is too hard. I do not understand and no one can help me at all. I am really frustrated and all of my friends have given up, but I really need to get it right. Please help me.
Since the hot air balloon is open, the pressure inside is the same as the pressure outside. Since PV=nRT or T = PV/nR:

[tex]\frac{T_{inside}}{T_{outside}} = \frac{n_{cold}}{n_{hot}} = \frac{\rho_{cold}}{\rho_{hot}}[/tex]

Now to generate buoyant lift, the density has to be such that the (mass of the air inside less the mass of same volume at 0 deg) x g = the lift.

(1)[tex]Lift = (\rho_{cold} - \rho_{hot})Vg[/tex]

[tex]\rho_{hot} = \frac{\rho_{cold}T_{outside}}{T_{inside}}[/tex]

So (1) becomes:

[tex]Lift = (\rho_{cold} - \frac{\rho_{cold}T_{outside}}{T_{inside}})Vg[/tex]

[tex]\frac{Lift}{Vg\rho_{cold}} = (1 - \frac{T_{outside}}{T_{inside}})[/tex]

Solve for the inside temp.

AM
 
Last edited:

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