Problem 2 - Fundamentals of Physics

Question

A partially evacuated 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 \(1.0 \times 10^{5}\) Pa, what is the internal air pressure?

Step-by-Step Solution

Answer

The internal air pressure is approximately 99993.77 Pa.

Step 1: Understand the problem

We have an airtight container with a lid that can be removed with a specific force. We need to find the internal air pressure of the container using the given force, surface area of the lid, and atmospheric pressure.

Step 2: Recall the relevant formula

The force required to remove the lid is related to the pressure difference between the atmospheric pressure and the internal pressure of the container. This can be expressed as \( F = (P_{ ext{atm}} - P_{ ext{internal}}) imes A \) where \( F \) is the force applied, \( P_{ ext{atm}} \) is the atmospheric pressure, \( P_{ ext{internal}} \) is the internal pressure, and \( A \) is the area of the lid.

Step 3: Substitute known values

Given \( F = 480 \, \text{N} \), \( A = 77 \, \text{m}^2 \), and \( P_{ ext{atm}} = 1.0 \times 10^5 \, \text{Pa} \), substitute into the equation: \[ 480 = (1.0 \times 10^5 - P_{ ext{internal}}) \times 77 \] Solve for \( P_{ ext{internal}} \).

Step 4: Solve for internal pressure

Rearrange the equation to find \( P_{ ext{internal}} \): \[ 480 = (1.0 \times 10^5 - P_{ ext{internal}}) \times 77 \] Divide both sides by 77: \[ \frac{480}{77} = 1.0 \times 10^5 - P_{ ext{internal}} \] Calculate \( \frac{480}{77} \approx 6.23 \).

Step 5: Calculate the final value of internal pressure

Now solve for \( P_{ ext{internal}} \): \[ P_{ ext{internal}} = 1.0 \times 10^5 - 6.23 \approx 99993.77 \] Therefore, the internal air pressure is approximately 99993.77 Pa.

Key Concepts

Atmospheric Pressure

When considering problems related to pressure, atmospheric pressure is often a key factor. It is the force exerted by the weight of the atmosphere above us. This pressure is uniform from all directions and affects everything on our planet.
Atmospheric pressure is typically measured using a unit called pascals (Pa), where one pascal is equivalent to one newton per square meter.
In our exercise, the atmospheric pressure is given as \(1.0 \times 10^5\) Pa. This means that every square meter of the lid's surface is being pushed down with a force of 100,000 N due to the atmosphere. Consequently, understanding how atmospheric pressure interacts with other pressures, such as internal pressure, is crucial for solving the problem.

Internal Pressure

Internal pressure refers to the pressure exerted from within a container, in this case, the partially evacuated airtight container. It is simply the force per unit area inside the container.
The internal air pressure is often less than the atmospheric pressure when a container is evacuated or has air removed from it. This creates a pressure differential, pushing forces against the lid.
In our problem, internal pressure is calculated by considering the difference between atmospheric pressure and the force required to remove the lid. By using the equation \( F = (P_{\text{atm}} - P_{\text{internal}}) \times A \), we rearrange for internal pressure and discover that it is just slightly less than the atmospheric pressure due to the force needed to overcome this pressure differential.

Physics Problem Solving

Physics problems related to pressure require a methodical approach. It's important to:

• Understand the problem statement, identifying known and unknown variables.
• Recall and apply relevant formulas, as with the pressure formula used in our exercise.
• Substitute the given values into the formula and solve for the unknowns.
• Double-check calculations to ensure there are no errors.

In our exercise, we first identified the given values: the force, surface area of the lid, and atmospheric pressure. Then, we used them to set up the equation relating force and pressure. This helped us determine the internal pressure with confidence. Solving pressure problems requires attention to detail and clear understanding of physics concepts.