Question

An airtight box, having a lid of area 85 cm2, is partially evacuated. Atmospheric pressure is 1.01 x 105 Pa. A force of 143 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

As we know that;

= F= 143 lb

= F= 639.1 N

= F=∆P.A

Theh, 2

Answer 2

SOLUTION :

Atmospheric pressure on the lid downward from outside = 1.01*10^5 (Pa)

Let pressure inside the container and also on the lid upward = p (Pa)

So net pressure on the lid downward = 1.00*10^5 - p (Pa)

Minimum force required to open the lid = Net pressure on lid * Area in m^2

=> (143 *0.45392) * 9.8 = (1.00*10^5 - p) * 85*10^(-4) 

=> p =1.01*10^5 - (143*0.45392*9.8 / (85*10^(-4))

=> p = 26161.94 Pa (ANSWER)