Question

Hot air balloons float in the air because of the difference in density between cold and hot air. In this problem, you will estimate the minimum temperature the gas inside the balloon needs to be, for it to take off. To do this, use the following variables and make these assumptions: The combined weight of the pilot basket together with that of the balloon fabric and other equipment is W. The volume of the hot air inside the balloon when it is inflated is V. The absolute temperature of the hot air at the bottom of the balloon is Th (where Th>Tc). The absolute temperature of the cold air outside the balloon is Tc and its density is ?c. The balloon is open at the bottom, so that the pressure inside and outside the balloon is the same. As always, treat air as an ideal gas. Use g for the magnitude of the acceleration due to gravity. Here is the actual question!!!! What is the density ?h of hot air inside the balloon? Assume that this density is uniform throughout the balloon. Express the density in terms of Th, Tc, and ?c.

Answer 1

Answer: $\rho_{h}=\frac{m}{m_{air}} \frac{\rho_{c} T_{c}}{T_{h}}$

Explanation:

We have the followin data:

$W=mg$ is the combined weight of the pilot basket together with that of the balloon fabric and other equipment

$m$ is the combined mass of the pilot basket together with that of the balloon fabric and other equipment

$g$ is the acceleration due gravity

$V$ is the volume of the hot air inside the balloon when it is inflated

$T_{h}$ is the absolute temperature of the hot air at the bottom of the balloon, being $T_{h}>T_{c}$

$T_{c}$ is the absolute temperature of the cold air outside

$\rho_{c}$ is the density of the cold air outside

$\rho_{h}$ is the density of the hot air inside

$P_{in}=P_{out}$ where $P_{in}$ is the pressure at the inside and $P_{out}$ is the pressure at the outside

Well, let's begin by writting the equations for the density:

Density cold air outside:

$\rho_{c}=\frac{m_{air}}{V_{air}}$ (1)

Where $m_{air}$ is the mass of air outside and $V_{air}$ is the volume of air outside

Isolating $V_{air}$ we have:

$V_{air}=\frac{m_{air}}{\rho_{c}}$ (2)

Density hot air inside:

$\rho_{h}=\frac{m}{V}$ (3)

Where $m=\frac{W}{g}$

Then:

$\rho_{h}=\frac{\frac{W}{g}}{V}$ (4)

On the other hand, the The Ideal Gas equation is:  

$P.V=n.R.T$  (5)

Where:  

$P$ is the pressure of the gas  

$n$ the number of moles of gas  

$R$ is the gas constant  

$T$ is the absolute temperature of the gas in Kelvin.

$V$ is the volume

This can be rewritten as:

$P=\frac{n.R.T}{V}$  (6)

Since $P_{in}=P_{out}$:

$\frac{n.R.T_{h}}{V}=\frac{n.R.T_{c}}{V_{air}}$  (7)

Isolating $V$:

$V=\frac{T_{h}V_{air}}{T_{c}}$  (8)

Substituting (8) in (3):

$\rho_{h}=\frac{m}{\frac{T_{h}V_{air}}{T_{c}}}$ (9)

Substituting (2) in (9):

$\rho_{h}=\frac{m}{\frac{T_{h}\frac{m_{air}}{\rho_{c}}}{T_{c}}}$ (10)

Rearranging:

$\rho_{h}=\frac{m}{m_{air}} \frac{\rho_{c}T_{c}}{T_{h}}$