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A sealed container is fitted with a massless lid of area 8 cm² and is partially evacuated. A force of 48 N is needed to pull the lid off. If atmospheric pressure is 1.0 x 10⁵ Pa, what was the pressure of the gas inside the container before opening?
- 0.6 atm
- 0.5 atm
- 0.4 atm
- 0.2 atm
Correct answer: 0.4 atm
Solution
Net inward pressure on the lid is (P_atm - P_inside). To pull the lid off, F = (P_atm - P_inside)*A. So P_atm - P_inside = F/A = 48/(8x10⁻⁴) = 6x10⁴ Pa. Thus P_inside = 1.0x10⁵ - 0.6x10⁵ = 0.4x10⁵ Pa = 0.4 atm.
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