# Homework Help: I really need help and no one can seem to help

1. May 2, 2005

### 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.

2. May 2, 2005

### OlderDan

Same problem- different numbers

3. May 2, 2005

### xanthym

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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Last edited: May 2, 2005
4. May 2, 2005

### Andrew Mason

Correction

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

Last edited: May 2, 2005