Question

A partially evacuated airtight container has a tight-fitting lid of surface area 79.9 cm2 and negligible mass. If the force required to remove the lid is 388 N and the atmospheric pressure is 1.00 × 105 Pa, what is the air pressure in the container before it is opened?

Answer 1

Answer:

$P_1 = 5.14 \times 10^4 Pa$

Explanation:

As we know that the air container is fixed by the lid on it and it is removed by external force

F = 388 N

now this external force must be same as the force due to pressure difference of the object

so it is given as

$F = P_2 A - P_1 A$

here we know that

$A = 79.9 cm^2 = 79.9 \times 10^{-4} m^2$

here we have

$P_2 = 1.00 \times 10^5 Pa$

now pressure is given as

$P_1 = P_2 - \frac{F}{A}$

$P_1 = 1.0 \times 10^5 - \frac{388}{79.9 \times 10^{-4}}$

$P_1 = 5.14 \times 10^4 Pa$