Question

Despite the destruction from Hurricane Katrina in September 2005, the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered an atmospheric pressure of 88.2 kPa on October 19, 2005, that was 2.0 kPa lower than Hurricane Katrina. What was the difference in pressure between the two hurricanes?

Answer 1

Complete Question

Despite the destruction from Hurricane Katrina in Sept 2005,the lowest pressure for a hurricane in the Atlantic Ocean was measured several weeks after Katrina. Hurricane Wilma registered at atmospheric pressure of 88.2 kPa on Oct 19, 2005,about 2 kPa lower than Hurricane Katrinia. What was the difference between the two hurricanes in:

millimeters of Hg? ___ mm Hg

atmospheres? ___ atm

millibars? ___ mb

Answer:

The difference in pressure in mm of Hg is
`Pg = 15.05 \ mm\ Hg`

The difference in pressure in atm is
`P_(atm) = 0.0198 \ atm`

The difference in pressure in millibar is
`P_(bar) = 20 \ mbar`

Step-by-step explanation:

From the question we are told that

The pressure of Hurricane Wilma is
`P = 88 kPa`

The difference is pressure is
`\Delta P =2\ k Pa`

Generally

`1\ atm = 1.01 *10^5 \ Pa`

`1 \ atm = 760\ mmHg`

The pressure difference in mm Hg is mathematically evaluated as

`Pg = \Delta P * (1000 Pa)/(1kPa) * (1\ atm)/(1.01 *10^5) * (760 mm Hg)/(1 \ atm)`

Substituting the value

`Pg = 2 kPa * (1000 Pa)/(1kPa) * (1\ atm)/(1.01 *10^5) * (760 mm Hg)/(1 \ atm)`

`Pg = 15.05 \ mm\ Hg`

The pressure difference in atm is

`P_(atm ) = 2kPa * (1000)/(1\ kPa) *(1 \ atm)/(1.01 *10^5 Pa)`

`= 0.0198 \ atm`

The pressure in millibars is

`= 2kPa * (1000Pa)/(1 kPa) * (1\ bar )/(1*10^5 Pa) * (1000 \ milli bar)/(1 bar)`

`= 20 \ m bar`