Answer to Question #99886 in General Chemistry for Beverlie

Question

The severity of a tropical storm is related to the depressed atmospheric pressure at its center. In August 1985, Typhoon Odessa in the Pacific Ocean featured maximum winds of about 90 mi/hr and pressure that was 40.0 mbar lower at the center than normal atmospheric pressure. In contrast, the central pressure of Hurricane Andrew (pictured) 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 58.0 L at a pressure of 1.00 atmosphere was deployed at the edge of Typhoon Odessa, what was the volume of the balloon when it reached the center? answer in L

Answer 1

If the normal pressure is taken as 1 atm, then at the edge of the typhoon the pressure is 40 mbar (0.04 atm) less, in the center - another 90 mbar less in the center (0.09 atm), as a result, the pressure in the center of the typhoon

$$p_{center} = 1 - 0.04 - 0.09 = 0.87 atm$$

Then, according to the Boyle-Marriott law

$$V_{center} = {p_{center}*V \over p}$$

$$V_{center} = {0.87*58.00 \over 1.00} = 5.46L$$

Answer: 5.46 L