Question

A balloon is to be used to carry meteorological instruments to an elevation of 18000 feet. The balloon is filled with helium at the ground level temperature and pressure ( 70 degrees Fahrenheit, 1 atm). The balloon material weighs .001 lb per square foot. If the instruments weigh 12 lbs, what should the diameter be of the spherical balloon ? Assume the temperature to drop 10 degrees Fahrenheit for every 4000 feet height.

Answer 1

Volume of ballon at ground level can be calculated from

The ballon material weighs 0.001lb per /ft2 and the instrument weighs 12 lbs

Surface area of ballon =12/0.001 = 12000 ft2

Surface area =4PiR2 = 12000 ft2

R is radius of ballon = R2= 12000/(4*3.14) R =30.91ft

volume of ballon =(4/3) PIR3= 1.33* 3.42* 30.91 ³= 123751.6 ft3 =123751.6* 28.32 liters=3504645 liters

T in deg. K

Deg.c =(70-32)/1.8 =21.11 deg.c =21.11+ 273.15= 294.26

from n= PV/RT = 1* 3504645/( 0.08206* 294.26)=145138 moles

Moels remain the same

at 18000 ft the temperature is ( 10deg. F drop for every 4000 ft

drop in temperature = (18000/4000)*10=45 Deg.F

Temperature at 18000ft = 70-45 =25 Deg.F =-(25-32)*5/9 deg.c = -3.9 deg.c =-3.9+273.15=269.25 K

The moles of Helium remains constant

at 1800 ft, the pressure of air =0.5 atms

PV= nRT

V=nRT/P = 145138*0.08206* 269.25/0.5= 6413548 liters = 6413548*0.0353 ft3= 226398.2 ft3

Froom 4/3PIR3= 226398

R3=2263998/ (1.33*3.14),

R =37.84 ft