I really and no one can seem to help

• Jayhawk1
In summary: ENDMENT: I am not sure how to get rid of the "n's" in this problem but it would be something like "Assuming that the hot air is all dry air, the n's cancel and I get: T_{inside} = \frac{T_{outside}}{1-\frac{Lift}{Vg\rho_{cold}}} So (in K) T_{inside} = 273.1/ (1-(3139/(1794*9.81*1.29))) = 356.1 So (in C) T_{inside} = 356.1 - 273.1 = 83.0 C This is approximately 182 F or 356.1
Jayhawk1
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.

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

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

{Bouyant 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}

Bouyant 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).

~~

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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:

$$\frac{T_{inside}}{T_{outside}} = \frac{n_{cold}}{n_{hot}} = \frac{\rho_{cold}}{\rho_{hot}}$$

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)$$Lift = (\rho_{cold} - \rho_{hot})Vg$$

$$\rho_{hot} = \frac{\rho_{cold}T_{outside}}{T_{inside}}$$

So (1) becomes:

$$Lift = (\rho_{cold} - \frac{\rho_{cold}T_{outside}}{T_{inside}})Vg$$

$$\frac{Lift}{Vg\rho_{cold}} = (1 - \frac{T_{outside}}{T_{inside}})$$

Solve for the inside temp.

AM

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