How much does pressue drop when you move from 8km to 10km at key west?

To calculate the pressure drop when moving from 8 km to 10 km at Key West, we can use the barometric formula:

$$ P = P_0 * e^{(-gh/RT)}$$

where:

- P is the pressure at the given altitude (in Pascals)

- P0 is the pressure at sea level (in Pascals)

- g is the acceleration due to gravity (approximately 9.81 m/s²)

- h is the altitude above sea level (in meters)

- R is the gas constant for air (approximately 287 J/kg-K)

- T is the air temperature (in Kelvin)

Let's assume the following values for Key West:

- P0 = 101325 Pa (standard atmospheric pressure)

- h1 = 8000 m (altitude at 8 km)

- h2 = 10000 m (altitude at 10 km)

- T = 298 K (approximately 25°C)

Plugging these values into the formula, we can calculate the pressure at 8 km:

$$ P_1 = 101325 * e^{(-9.81 * 8000 / (287 * 298)}$$

$$ P_1 = 50247 Pa $$

Now, we can calculate the pressure at 10 km:

$$ P_2 = 101325 * e^{(-9.81 * 10000 / (287 * 298)}$$

$$ P_2 = 34865 Pa$$

Finally, we can calculate the pressure drop by subtracting the pressure at 10 km from the pressure at 8 km:

$$ \Delta P = P_1 - P_2 $$

$$ \Delta P = 50247 - 34865 = 15382 Pa $$

Therefore, the pressure drop when moving from 8 km to 10 km at Key West is approximately 15382 Pa or 15.382 kPa.