When a person got onto an airplane, she brought with her a half-full 1.0000 L aluminum water bottle that was closed inside the airport at standard pressure and a temperature of 26.000 C. After the plane takes off and climbs to 36,000. feet, the cabin pressure is at 0.82000 atm and the cabin temperature is 26.000 C. Assume her water bottle and its contents always take the temperature of its surroundings.
a) What would happen if she opened the water bottle for a drink?
P1V1 = P2V2
(1) (.05) = (.82) (V2) = .609 L
I am assuming there would be slightly more water?
b) What if the cabin pressure was at 0.82000 atm and the cabin temperature was at -28 C? (Ignore the expansion of frozen water.)
V1/T1 = V2/T2
0.5/299 K = V2/245 K
The volume would slightly decrease?
I am not really sure if this is what I am supposed to get out of these questions. They seem a little odd, but based off of P1V1 = P2V2 and V1/T1 = V2/T2. That seems to be correct.
(a) but when she opens the bottle the air inside is still at 1 atm so must escape out the open top--> whoosh !
a) When the person opens the water bottle for a drink, the pressure inside the bottle will equalize with the cabin pressure, which is 0.82000 atm. Using the equation P1V1 = P2V2, we can calculate the new volume of the water in the bottle.
Given:
P1 = 1.0000 atm (standard pressure inside the airport)
V1 = 0.5000 L (half-full water bottle)
P2 = 0.82000 atm (cabin pressure)
V2 =?
Using the equation P1V1 = P2V2:
(1.0000 atm) (0.5000 L) = (0.82000 atm) (V2)
V2 = (1.0000 atm)(0.5000 L) / (0.82000 atm)
V2 ≈ 0.609 L
So, if the person opens the water bottle for a drink, the volume of water in the bottle would slightly increase to approximately 0.609 L, due to the lower cabin pressure.
b) If the cabin pressure is still at 0.82000 atm, but the cabin temperature is at -28 °C, we can use the equation V1/T1 = V2/T2 to calculate the new volume of the water in the bottle.
Given:
V1 = 0.5000 L (initial volume of the water bottle)
T1 = 26.000 °C (initial temperature)
V2 =?
T2 = -28 °C (new temperature)
Converting the temperatures to Kelvin:
T1 = 26.000 °C + 273.15 = 299.15 K
T2 = -28 °C + 273.15 = 245.15 K
Using the equation V1/T1 = V2/T2:
(0.5000 L) / (299.15 K) = V2 / (245.15 K)
V2 = (0.5000 L)(245.15 K) / (299.15 K)
V2 ≈ 0.409 L
So, if the cabin temperature is -28 °C, the volume of water in the bottle would slightly decrease to approximately 0.409 L, assuming the water does not freeze and expand.