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Chapter 14: Problem 15

A partially evacuated airtight container has a tight-fitting lid of surface area \(50 \mathrm{~m}^{2}\) 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 99990.4 Pa.

Step by step solution

01

Understand the Pressure Difference

Determine why the lid requires a force to be removed. The force is due to the pressure difference between the inside and outside of the container. Atmospheric pressure pushes down on the lid while the internal air pressure pushes up. The net force due to this pressure difference is what needs to be overcome to remove the lid.
02

Calculate Pressure Difference

The force required to remove the lid, given as 480 N, can be expressed as the difference in pressure across the lid multiplied by its area. This can be expressed with the formula: \[ \Delta P = \frac{F}{A} \]where \( \Delta P \) is the pressure difference, \( F = 480 \textrm{ N} \) is the force, and \( A = 50 \textrm{ m}^2 \) is the area. Substitute these values into the equation:\[ \Delta P = \frac{480}{50} \]
03

Solve for Pressure Difference

Substitute the given values into the pressure difference equation and solve:\[ \Delta P = \frac{480}{50} = 9.6 \textrm{ Pa} \]This is the difference between the atmospheric pressure and the internal pressure.
04

Solve for Internal Air Pressure

Since the atmospheric pressure is \( 1.0 \times 10^5 \textrm{ Pa} \) and we know the pressure difference is 9.6 Pa, we can set up the following equation to find the internal pressure:\[ P_{\text{atm}} - P_{\text{internal}} = 9.6 \]\[ P_{\text{internal}} = P_{\text{atm}} - 9.6 \]Substitute \( P_{\text{atm}} = 1.0 \times 10^5 \textrm{ Pa} \):\[ P_{\text{internal}} = 1.0 \times 10^5 - 9.6 \]
05

Calculate the Internal Air Pressure

Substitute and calculate the internal pressure:\[ P_{\text{internal}} = 100000 - 9.6 = 99990.4 \textrm{ Pa} \]Thus, the internal air pressure is 99990.4 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 the force imposed on a surface by the weight of the air above it. It normally decreases with increasing altitude since there is less air above the surface, thus it weighs less.
At sea level, standard atmospheric pressure is approximately \[ 1.0 \times 10^5 \text{ Pa} \], or 101.3 kPa. This corresponds to the standard conditions against which other pressures are measured.
  • The pressure is caused by air molecules colliding with surfaces.
  • It is exerted equally in all directions鈥攄ownward, upward, and sideways.
In this exercise, the atmospheric pressure plays a crucial role in determining the force needed to lift the lid from the container. It pushes down on the lid, adding to the force necessary to overcome the reduced pressure inside the container.
Pressure Difference
The pressure difference is what's responsible for the need to exert force to remove the lid in this exercise. When atmospheric pressure exceeds the internal air pressure, it exerts a net force on the lid.
This difference in pressure, \( \Delta P \), is the critical quantity to understand:
  • Calculated using the formula \( \Delta P = \frac{F}{A} \), where \( F \) is the force applied, and \( A \) is the area.
  • It measures how much more atmospheric pressure the external air exerts compared to the internal pressure.
In the example, \( \Delta P \) was found to be 9.6 Pa, indicating a slight but significant difference due to the partial vacuum inside the container. A thorough grasp of pressure differences is fundamental to solving similar physics problems.
Force Calculation
Calculating force involves understanding the relation between force, area, and pressure difference. The formula used, \[ \Delta P = \frac{F}{A}, \] allows us to derive the required pressure difference across the lid.
This relationship is derived from the fundamental formula of pressure \( P = \frac{F}{A} \), where pressure is the force distributed over an area.
  • For this exercise, we rearranged that formula to solve for the difference in pressure \( \Delta P \).
  • Given: Force \( F = 480 \text{ N} \) and area \( A = 50 \text{ m}^2 \).
  • Resulting pressure difference \( \Delta P = 9.6 \text{ Pa}. \)
This part of the problem eloquently demonstrates the utility of such basic equations in elucidating real-world physics scenarios.
Physics Education
Understanding physics concepts like air pressure and force is made simple with step-by-step exercises as presented here. These exercises break down larger problems into manageable steps, encouraging deeper comprehension.
Key takeaways for effective physics education include:
  • Engage with practical scenarios that let you apply theoretical knowledge to tangible problems.
  • Utilize clear and straightforward language to describe complex phenomena, reducing cognitive load.
  • Reinforce learning through applied math formulas, fostering mathematical literacy alongside physics understanding.
By consistently practicing these problems and exploring real-world applications, physics learning becomes more intuitive and enjoyable, laying a solid foundation for more advanced studies.

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Most popular questions from this chapter

An inflatable device with a density of \(1.20 \mathrm{~g} / \mathrm{cm}^{3}\) is fully submerged in fresh water and then inflated enough for the device's density to match that of the water. What then is the ratio of the expanded volume to the full volume of the device?

A cylindrical tank with a large diameter is filled with water to a depth \(D=0.30 \mathrm{~m}\). A hole of cross-sectional area \(A=6.2 \mathrm{~cm}^{2}\) in the bottom of the tank allows water to drain out. (a) What is the drainage rate in cubic meters per second? (b) At what distance below the bottom of the tank is the cross-sectional area of the stream equal to one-half the area of the hole?

The maximum depth \(d_{\max }\) that a diver can snorkel is set by the density of the water and the fact that human lungs can function against a maximum pressure difference (between inside and outside the chest cavity) of \(0.050 \mathrm{~atm}\). What is the difference in \(d_{\max }\) for fresh water and the water of the Dead Sea (the saltiest natural water in the world, with a density of \(1.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) )?

Water stands at depth \(D=30.0 \mathrm{~m}\) behind the vertical upstream face of a dam of width \(W=250 \mathrm{~m}\). Find (a) the net horizontal force on the dam from the gauge pressure of the water and (b) the net torque due to that force about a horizontal line through \(O\) parallel to the (long) width of the dam. This torque tends to rotate the dam around that line, which would cause the dam to fail. (c) Find the moment arm of the torque.

Models of torpedoes are sometimes tested in a horizontal pipe of flowing water, much as a wind tunnel is used to test model airplanes. Consider a circular pipe of internal diameter \(25.0 \mathrm{~cm}\) and a torpedo model aligned along the long axis of the pipe. The model has a \(6.00 \mathrm{~cm}\) diameter and is to be tested with water flowing past it at \(2.00 \mathrm{~m} / \mathrm{s}\). (a) With what speed must the water flow in the part of the pipe that is unconstricted by the model? (b) What will the pressure difference be between the constricted and unconstricted parts of the pipe?
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