Question

An airtight box, having a lid of area 120 cm2, is partially evacuated. Atmospheric pressure is 1.01 x 105 Pa. A force of 120 lb is required to pull the lid off the box. What was the pressure in the box (in Pa)? Round your answer to the nearest hundred.

Answer 1

Required force to pull the lid off, F = 120 lb

Convert this in Newton.

F = 120 x 4.448 N = 533.76 N

Area of the lid, A = 120 cm^2 = 120 x 10^-4 m^2

Pressure required to open P1 = F/A

= 533.76 / (120 x 10^-4) = 4.45 x 10^4 Pa

Now, let us assume the pressure inside the box is P2 Pa,

So -
=>P1+P2 = P(atmospheric)
=>P2 = P-P1
=>P2 = 1.01 x 10^5 - 4.45 x 10^4 = 1.01 x 10^5 - 0.445 x 10^5
=>P2 = 0.565 x 10^5 Pa = 5.65 x 10^4 Pa

Round the answer to the rearest hundred.

P2 = 6.00 x 10^4 Pa (Answer)