Respuesta :
Answer: [tex]\rho_{h}=\frac{m}{m_{air}} \frac{\rho_{c} T_{c}}{T_{h}}[/tex]
Explanation:
We have the followin data:
[tex]W=mg[/tex] is the combined weight of the pilot basket together with that of the balloon fabric and other equipment
[tex]m[/tex] is the combined mass of the pilot basket together with that of the balloon fabric and other equipment
[tex]g[/tex] is the acceleration due gravity
[tex]V[/tex] is the volume of the hot air inside the balloon when it is inflated
[tex]T_{h}[/tex] is the absolute temperature of the hot air at the bottom of the balloon, being [tex]T_{h}>T_{c}[/tex]
[tex]T_{c}[/tex] is the absolute temperature of the cold air outside
[tex]\rho_{c}[/tex] is the density of the cold air outside
[tex]\rho_{h}[/tex] is the density of the hot air inside
[tex]P_{in}=P_{out}[/tex] where [tex]P_{in}[/tex] is the pressure at the inside and [tex]P_{out}[/tex] is the pressure at the outside
Well, let's begin by writting the equations for the density:
Density cold air outside:
[tex]\rho_{c}=\frac{m_{air}}{V_{air}}[/tex] (1)
Where [tex]m_{air}[/tex] is the mass of air outside and [tex]V_{air}[/tex] is the volume of air outside
Isolating [tex]V_{air}[/tex] we have:
[tex]V_{air}=\frac{m_{air}}{\rho_{c}}[/tex] (2)
Density hot air inside:
[tex]\rho_{h}=\frac{m}{V}[/tex] (3)
Where [tex]m=\frac{W}{g}[/tex]
Then:
[tex]\rho_{h}=\frac{\frac{W}{g}}{V}[/tex] (4)
On the other hand, the The Ideal Gas equation is:
[tex]P.V=n.R.T[/tex] (5)
Where:
[tex]P[/tex] is the pressure of the gas
[tex]n[/tex] the number of moles of gas
[tex]R[/tex] is the gas constant
[tex]T[/tex] is the absolute temperature of the gas in Kelvin.
[tex]V[/tex] is the volume
This can be rewritten as:
[tex]P=\frac{n.R.T}{V}[/tex] (6)
Since [tex]P_{in}=P_{out}[/tex]:
[tex]\frac{n.R.T_{h}}{V}=\frac{n.R.T_{c}}{V_{air}}[/tex] (7)
Isolating [tex]V[/tex]:
[tex]V=\frac{T_{h}V_{air}}{T_{c}}[/tex] (8)
Substituting (8) in (3):
[tex]\rho_{h}=\frac{m}{\frac{T_{h}V_{air}}{T_{c}}}[/tex] (9)
Substituting (2) in (9):
[tex]\rho_{h}=\frac{m}{\frac{T_{h}\frac{m_{air}}{\rho_{c}}}{T_{c}}}[/tex] (10)
Rearranging:
[tex]\rho_{h}=\frac{m}{m_{air}} \frac{\rho_{c}T_{c}}{T_{h}}[/tex]