An air tight container having a lid with negligible mass and an area of $8 {cm}^{2}$ is partially evacuated. If a 48N force is required to pull the lid off the container and the atmospheric pressure is $1.0 \times 10^{5}$ Ps, the pressure in the container before it is opened must be

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0.4 atm

0.5 atm

0.2 atm

0.6 atm

answer is C.

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Detailed Solution

P = \frac{F}{A}

$= \frac{48}{8 \times 10^{4}} = \frac{48 \times 10^{4}}{8} = 6 \times 10^{4}$

$= 0.6 \times 10^{5} Pa$

$P = 0 .6 atm$

$0.6 = 1 - P_{2}$

$P_{2} = 1 - 0.6 = 0.4 atm$

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