1. When filling a hot air balloon, the air inside the balloon is heated from 20°
ID: 1043732 • Letter: 1
Question
1. When filling a hot air balloon, the air inside the balloon is heated from 20°C to 99°C at constant atmospheric pressure (1.00 atm). The hot air expands from its initial volume at 20.°C to fill the entire volume of the balloon at 99°C. Since the volume that the air occupies has increased (while the mass of the air stays the same), the density of the air inside the balloon is less than the density of the surrounding air which causes the balloon to float. The mass of air in the balloon is approximately 2700 kg and the specific heat of air is 1.003 J/g. C. a. If the density of air at 20°C is 1.204 kg/m3, what volume (in L) does the air occupy at 20°C? b. What volume (in L) does the air occupy at 99°C? C. Calculate q, w, ??, and ?? for this process.Explanation / Answer
1. assuming ideal gas behavior
mass of air in the ballon is 2700 kg
use the density value to calculate the volume
density = mass / volume
1.204 = 2700 / volume
volume = 2700 / 1.204 = 2242.525 m3
multiply by 1000 to get liters
volume = 2 242 525 Liters
2. use the relationship
V1 / T1 = V2 / T2
V is volume and T is temperature, T must be in kelvin
2242.5 / (20 + 273.15) = V2 / (99 + 273.15)
solve this to get
V2 = 2846.855 m3 or 2 846 855 Liters (multiply by 1000)
3. 1 atm is equal to 101 325 pascals so:
work is defined as - pressure * (V2 - V1), volume in m3 to get joules
w = -101 325 * ( 2846.855 - 2242.525) = -61 233 781 Joules or -61 233.8 kJ
q = mass * heat capacity * (T2 - T1)
q = 2700 000 * 1.003 * (372.15 - 293.15) = 213 939 900 Joules, divide this by 1000 to get
q = 213 939.9 kJ
q = delta H
E = q + w
E = 213 939.9 kJ + (-61 233.8 kJ) = 152 706.118 kJ
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