What is the air pressure in the container before it is opened?

[tigers4]

Homework Statement

A partially evacuated airtight container has a tight-fitting lid of surface area 78 cm2 and negligible mass. If the force required to remove the lid is 480 N and the outside atmospheric pressure is 1.01*10^5 Pa, what is the air pressure in the container before it is opened?

Homework Equations

p=F/A

The Attempt at a Solution

(1.01*10^5)*(78cm^2)+480N=Force needed to open lid
/area of lid=pressure needed to open lid...

I tried doing this but cannot get the write answer, please any help would be greatly appreciated

 

[gneill]

The force required to open the lid is equal to the force due to the pressure difference between outside and inside the container. So first you should calculate the pressure that would result in a 480N force on the lid (assume that only one pressure is operating).

 

[tigers4]

so p=480N/78cm^2?

 

[gneill]

so p=480N/78cm^2?

Fine. That's the pressure difference that has to exist between outside and inside. Now find the inside pressure.

 

[rwisz]

!

Based on the given information, we can use the formula p=F/A to calculate the pressure inside the container before it is opened. We know that the force required to remove the lid is 480 N and the surface area of the lid is 78 cm^2.

Using the formula, we can rearrange it to solve for pressure: p=F/A. Plugging in the values, we get p=480N/78cm^2 = 6.15 N/cm^2.

However, this is not the final answer because the units for pressure should be in Pascals (Pa). To convert N/cm^2 to Pa, we can use the conversion factor 1 N/cm^2 = 1*10^4 Pa.

Therefore, the pressure inside the container before it is opened is 6.15 N/cm^2 * 1*10^4 Pa/N/cm^2 = 6.15*10^4 Pa.

This means that the air pressure inside the container before it is opened is 6.15*10^4 Pa.