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\(\cdot 2\) A partially cvacuated airtight container has a tight-fitting lid of surface area \(77 \mathrm{~m}^{2}\) and negligible mass. If the force required to remove the lid is \(480 \mathrm{~N}\) and the atmospheric pressure is \(1.0 \times 10^{5} \mathrm{~Pa}\), what is the internal air pressure?

Short Answer

Expert verified
The internal air pressure is approximately \(99993.77\,\text{Pa}\).

Step by step solution

01

Understand the Problem

We have a lid with a known surface area, a force required to remove it, and the atmospheric pressure. We need to calculate the internal air pressure inside the container.
02

Apply the Force Equation for Pressure Difference

The force required to remove the lid is due to the difference between the atmospheric pressure (\(P_{atm} = 1.0 \times 10^{5} \mathrm{~Pa} \)) and the internal air pressure (\(P_{int}\)). This force is given by:\[ F = (P_{atm} - P_{int}) \cdot A \]where \(F = 480 \mathrm{~N}\) and \(A = 77 \mathrm{~m}^{2}\).
03

Rearrange the Equation to Solve for Internal Pressure

Rearrange the equation from Step 2 to solve for the internal pressure:\[ P_{int} = P_{atm} - \frac{F}{A} \]
04

Substitute Known Values into the Equation

Substitute the given values into the equation:\[ P_{int} = 1.0 \times 10^{5} \mathrm{~Pa} - \frac{480 \mathrm{~N}}{77 \mathrm{~m}^{2}} \]
05

Calculate the Internal Air Pressure

Perform the calculation:\[ \frac{480 \mathrm{~N}}{77 \mathrm{~m}^{2}} = 6.23 \mathrm{~Pa} \]Then,\[ P_{int} = 1.0 \times 10^{5} \mathrm{~Pa} - 6.23 \mathrm{~Pa} = 99993.77 \mathrm{~Pa} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atmospheric Pressure
Atmospheric pressure is a fundamental concept that refers to the force exerted by the weight of the air above us at any given point. It is usually measured in pascals (Pa) and is affected by altitude and weather conditions. At sea level, the standard atmospheric pressure is approximately \(1.013 \times 10^{5} \mathrm{~Pa}\).Understanding atmospheric pressure is crucial when dealing with problems involving pressure differences, such as the one with the airtight container. The atmospheric pressure acts on the lid from outside the container, creating a differential force if the internal pressure differs. This force differential is what presses the lid down onto the container, making it harder to remove when the internal pressure is less.To get a mental picture, consider the air around us as an ocean of air molecules, pressing down on us from every direction. This pressure is what we refer to as atmospheric pressure, and it can significantly influence the behavior of both natural and man-made systems.
Force Equation for Pressure
The force equation for pressure is a vital tool for understanding how pressure differences create force. In the context of removing the lid from the container, the formula isgiven by:\[F = (P_{atm} - P_{int}) \cdot A\]Where:
  • \(F\) is the force required to lift the lid.
  • \(P_{atm}\) is the atmospheric pressure outside the container.
  • \(P_{int}\) is the internal air pressure inside the container.
  • \(A\) is the surface area of the lid.
This equation helps us understand that the force required to remove the lid is the result of the pressure difference (between the outside and inside of the container) acting over the surface area.In practical terms, it's like an invisible hand pushing down with the air's weight. This is why high pressure outside compared to inside necessitates a greater force to remove the lid. Solving these equations step by step, especially substituting real-world values, can demystify the often abstract nature of physical formulas.
Surface Area Calculation
Surface area plays a critical role in the physics of pressure, as it influences the magnitude of force exerted by or into a system. In the given exercise, the surface area of the lid is provided as \(77 \mathrm{~m}^{2}\). The force experienced by an object due to pressure differences is directly proportional to its surface area, as explained by the force equation.A larger surface area means more room for the atmospheric pressure to act on, thereby increasing the force applied. Conversely, a smaller area would experience less force for the same pressure difference.Think of the lid on the container. It covers an area of \(77 \mathrm{~m}^{2}\). The larger this area, the more total force the outside air can apply to keep the lid sealed. When dealing with problems involving pressure, always pay close attention to the surface area involved, since it greatly impacts the final force and the effort needed to overcome it.

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