Question

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 4.0 L at a pressure of 1.00 atmospheres was deployed at the edge of Typhoon odessa what was the volume of the balloon when it reached the center?

Answer 1

To solve this problem we will apply the concepts given in the ideal gas equations for which the product between the pressure and the volume in state one must be equal to the product of the pressure and the volume in state two, that is,

$P_1 V_1 = P_2V_2$

At our case the input and the output are each state,

$P_{in}V_{in} = P_{out}V_{out}$

The value of the pressure at the center of the Odessa storm is

$P_{in} = P_{atm}-40mbar$

$P_{in} = 1.01325bar - 40*10^{-3}bar$

$P_{in} = 0.97325bar$

Rearranging to find the value of the volume we have,

$V_{in} = \frac{P_{out}V_{out}}{P_{in}}$

$V_{in} = \frac{1.01325bar*4L}{0.97325bar}$

$V_{in} = 4.164L$

Therefore the volume of the balloon when it reached the center is 4.164L