Question

Chapter 14, Problem 002 Your answer is partially correct. Try again. A partially evacuated airtight container has a tight-fitting lid of surface area 78.0 cm2 and negligible mass. If the force required to remove the lid is 314 N and the atmospheric pressure is 1.00 x 105 Pa, what is the air pressure in the container before it is opened? 1-1 Numbe140256 UnitsT N/m 2 or Pa

Answer 1

The re is net paessuse ading ơn tine Containeh- that is uhy314NatC = 0.0078 Mr 100000-X): 314 0.0073 X- 59孙3.589 R2

Answer 2

SOLUTION :

Atmospheric pressure on the lid downward from outside = 1.00*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

=> 314 = (1.00*10^5 - p) * 78*10^(-4) 

=> p =1.00*10^5 - (314/(78*10^(-4)) 

=> p = 59743.59 Pa (ANSWER)