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Please show work so I can follow along, thanks! 2. A hot-air balloon achieves it

ID: 1838894 • Letter: P

Question

Please show work so I can follow along, thanks!

2. 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 1900 m and the required lift is 2500 N (rough estimate of the weight of the equipment and passenger). Assume that the outside air temperature is 0°C and that air is an ideal gas under these conditions. Air is a mixture of about 20% oxygen and 80% nitrogen so assume that 1 mol of air has a mass of about 29 g. A) Express the density of the air in the balloon with variables, pressure, P, Temperature of the air balloon T and a constant R for the dry air balloons B) Calculate the temperature of the air inside the balloon which will produce the required lift.

Explanation / Answer

(A) P V = n R T

P / R T = n / V


multiplying it with molar mass M,


M P / R T = n M / V


n M / V = density where M = molar mass = 0.029 kg / mol


hence density of air = 0.029 P / R T

(B) lift force = (rho_outside air - rho_inside air) * V * g

=


2500 = [ (0.029 P / RTout) - (0.029 P / R Tin)] * 1900 * 9.8


(0.029 P / RTout) - (0.029 P / R Tin) = 0.1343

(P / R) (1 / T273 - 1 /Tin) = 0.1343 / 0.029 = 4.63

1/273 - 1 / Tin = 3.80 x 10^-4


1/Tin = 3.283 x 10^-3


Tin = 304.6 K


T (in deg ) = 31.6 deg C