The seventy of a tropical storm is related to the depressed atmospheric pressure
ID: 1069188 • Letter: T
Question
The seventy of a tropical storm is related to the depressed atmospheric pressure at its center. Shown below is a photograph of Typhoon Odessa taken from the space shuttle Discovery in August 1985, when the maximum winds of the storm were about 90 mi/hr and the pressure was 40.0 mbar lower at the center than normal atmospheric pressure. In contrast, the central pressure of Hurricane Andrew was 90.0 mbar lower than its surroundings when it hit south Florida with winds as high as 165 mi/hr. If a small weather balloon with a volume of 52.0 L at a pressure of 1.00 atmosphere was deployed at the edge of Typhoon Odessa (in picture), what was the volume of the balloon when it reached the center?Explanation / Answer
This is a case of boyle-mariotte law which says that the relationship between Pressure and volume is constant:
P*V = k, p is pressure, v is volume and k is a constant value
another way to write it is
P1 * V1 = P2 * V2, this means that if you have a system and you change the pressure then the volume will change in order to keep the relationship so for our particular case
initial data:
at the beginning you have an atmospheric pressure equals to 1 atm(from the statement) the value of 1 atm can be expressed in many ways: 760 mmHg, 760 torr, 101325 Pa, etc...
for our particular case we will use the next equivalence
1 atm = 1013.25 mbar for initial pressure
Volume is 52 Liters
final state
the statement says that pressure is 40 mbar lower at the center compared to the pressure on the edge so:
P2 = 1013.25 - 40 = 973.25 mbar
V2 = ?, this is the volume we want to know
P1 * V1 = P2 * V2
1013.25 * 52 = 973.25 * V2
V2 = 1013.25 * 52 / 973.25 = 54.137 or 54.14 Liters
Volume of the weather ballon will be 54.14 liters
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