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A partially evacuated airtight container has a tight-fitting lid of surface area77m2and negligible mass. If the force required removing the lid is 480 N and the atmospheric pressure is1.0×105Pa, what is the internal air pressure?

Short Answer

Expert verified

The Internal air pressure of container is9.99×104Pa

Step by step solution

01

The given data

  1. Surface area isA=77m2
  2. Applied force isF=480N
  3. Pressure outside the container = atmospheric pressure isρ0=1.0×105Pa
02

Understanding the concept of pressure

A fluid is a substance that can flow; it conforms to the boundaries of its container because it cannot withstand shearing stress. It can, however, exert a force perpendicular to its surface. If theforce is uniform over a flat area, then the force is described in terms of pressure p as,

p=FA

In the given problem, the magnitude of the force(F)required to open the lid of the container is equal to the difference in the outer pressure(P0)and inner pressure(Pi)multiplied by an area(A)of the lid.

Using the relation between internal pressure, external pressure, and force, we can write an expression internal pressure.Then, we can find the external pressure from the applied force and the given surface area.

Formulae:

The pressure at a point in a fluid at static equilibrium,

F=p0-piA …(i)

03

Calculation of the internal air pressure inside the container

Using equations (i), we get

po-pi=FApi=p0-FA=1.0×105Pa-480N77m2.........................Substitutingthegivenvalues=9.99×104Pa

Therefore, the internal air pressure of container is9.99×104Pa .

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