Bill and Bob have been testing a new aluminised PET (Polyethylene Terephthalate) film and they now use the material to make helium filled balloons. Bill’s balloon is less inflated than Bob’s, and the latter jokes about it. However, Bill insists he has to be careful because if he were to take the balloon on hills and mountains, the atmospheric pressure would decrease and the balloon would expand and burst. Bill then makes a bet that he can walk, drive or climb somewhere high enough for his balloon to burst...
The balloons are partially filled with helium and they unfold as helium expands into them. The balloons burst once they reach a volume of 22.0 dm3; the pressure and temperature of the helium inside the balloons are always equal to the surrounding atmospheric pressure and temperature. Bill and Bob added respectively 8600 cm3 and 18000 cm3 of helium in their balloons, by the sea side where the altitude is 0 meters, the temperature is 25.0oC and the atmospheric pressure is 760mm Hg.
The conditions on Mount Everest during climbing season are typically: P = 0.330atm and
T = -20.0oC.
(a) Calculate the volume that would be occupied by the helium in Bill’s balloon on Mount Everest (3 sig. fig.). Can Bill ever hope to win his bet? [3 marks]
(b) The dependence of the atmospheric pressure with altitude can be approximated by the following formula:
Patm = -7.57x10-2 h + 1
where h is the altitude in km, Patm is the atmospheric pressure in atm.
Calculate the altitude at which Bob’s balloon would burst, assuming that the temperature remains constant and equal to 250C (3 sig. fig).
To calculate the volume that would be occupied by the helium in Bill's balloon on Mount Everest, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, let's calculate the number of moles of helium in Bill's balloon:
Volume in sea level = 8600 cm^3 = 8.6 dm^3
Pressure in sea level = 760 mmHg
Temperature in sea level = 25.0 °C = 298 K
To convert pressure from mmHg to atm, we divide by 760:
Pressure in sea level = 760 mmHg / 760 = 1 atm
Rearranging the Ideal Gas Law equation to solve for n:
n = PV / RT
n = (1 atm) * (8.6 dm^3) / [(0.0821 L*atm/mol*K) * (298 K)]
n = 0.35 mol (rounded to two decimal places)
Now, let's calculate the volume that would be occupied by the helium in Bill's balloon on Mount Everest:
Pressure on Mount Everest = 0.330 atm
Temperature on Mount Everest = -20.0 °C = 253 K
Rearranging the Ideal Gas Law equation to solve for V:
V = nRT / P
V = (0.35 mol) * (0.0821 L*atm/mol*K) * (253 K) / (0.330 atm)
V = 65.03 L (rounded to three significant figures)
Therefore, the volume that would be occupied by the helium in Bill's balloon on Mount Everest is 65.0 L. Since this is smaller than the burst volume of 22.0 dm^3 (or 22.0 L), Bill's balloon will not burst on Mount Everest.
Now, let's move on to part (b) and calculate the altitude at which Bob's balloon would burst. We can use the formula provided:
Patm = -7.57x10^-2 * h + 1
where Patm is the atmospheric pressure in atm and h is the altitude in km.
Given that the temperature remains constant at 25.0 °C = 298 K, we can substitute the pressure and solve for h:
0.330 atm = -7.57x10^-2 * h + 1
Rearranging the equation:
0.330 - 1 = -7.57x10^-2 * h
-0.670 = -7.57x10^-2 * h
Dividing both sides by -7.57x10^-2 :
h = -0.670 / (-7.57x10^-2)
h = 8.85 km (rounded to three significant figures)
Therefore, Bob's balloon would burst at an altitude of 8.85 km.