In: Chemistry
The severity 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 40.0 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? _____L
Ans. We have-
Pressure at the edge of Typhoon Odessa = 1.0 atm = 1013.25 mbar
Pressure at the center of Typhoon Odessa = 40 mbar lower than 1.0 atm
= 1013.25 mbar – 40.0 mbar
= 973.25 mbar = 0.96 atm
Volume of balloon at the edge = 40.0 L
Temperature: Assumed to be constant
# Now, calculate the volume of balloon at the center using Boyle’s law-
Ans. We have-
Pressure at the edge of Typhoon Odessa = 1.0 atm = 1013.25 mbar
Pressure at the center of Typhoon Odessa = 40 mbar lower than 1.0 atm
= 1013.25 mbar – 40.0 mbar
= 973.25 mbar = 0.96 atm
Volume of balloon at the edge = 40.0 L
Temperature: Assumed to be constant
# Now, calculate the volume of balloon at the center using Boyle’s law-
Boyle’s Law: For a gas at constant temperature confined in a closed vessel the product of pressure and volume is a constant.
That is, P1V1 (at the edge) = P2V2 (at the center)
Or, V2 = (P1V1 / P2) = ( 1.00 atm x 40.0 L) / 0.96 atm = 41.67 L
# Hence, the volume of balloon at the center = 41.67 L
# Hence, the volume of balloon at the center = 41.67 L