# Evacuated Air Pressure Calculations

## Explanation:

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level).

The surface area of the container lid is A.

Given data:

- Altitude of container in Denver, \(d_D = 6000\) ft
- Surface Area of the container lid, \(A = 0.0155m^2\)
- Air pressure in Denver, \(P_D = 79000Pa\)
- Air pressure in New Orleans, \(P_{No} = 100250Pa\)

## Calculation:

Solving for the force exerted on the container in New Orleans:

The expression for force, \[F_{No} = \triangle P \times A\]

Since the inside pressure is half the pressure at sea level, then:

\[\triangle P = P_{No} - \frac{P_{area}}{2}\]Where standard pressure \(P_{area} = 101000Pa\)

This gives: \[F_{No} = [P_{No} - \frac{P_{area}}{2}] \times A\]

Plugging in the values:

\[F_{No} = [100250 - \frac{101000}{2}] \times 0.0155\] \[F_{No} = [100250 - 50500] \times 0.0155\] \[F_{No} = 49750 \times 0.0155\] \[F_{No} = 771.125N\]Therefore, the force exerted on the container in New Orleans is 771.125N