Problem 29 - Physics

Question

A \(1.00-\mathrm{m}\) -tall container is filled to the brim, partway with mercury and the rest of the way with water. The container is open to the atmosphere. What must be the depth of the mercury so that the absolute pressure on the bottom of the container is twice the atmospheric pressure?

Step-by-Step Solution

Answer

The mercury should be approximately 0.78 meters deep.

Step 1: Understanding the Problem

We have a 1.00-meter tall container filled with mercury and water. We need to find the height of the mercury layer that causes the pressure at the bottom to be twice the atmospheric pressure.

Step 2: Setting Up the Pressure Equation

The total pressure at the bottom of the container is the sum of the atmospheric pressure, the pressure due to mercury, and the pressure due to water. This total pressure should be twice the atmospheric pressure. We express this as: \[ P_{atm} + \rho_{Hg}gh_{Hg} + \rho_{H_2O}g(1.00 - h_{Hg}) = 2P_{atm} \], where \( \rho_{Hg} \) and \( \rho_{H_2O} \) are the densities of mercury and water respectively, and \( h_{Hg} \) is the height of the mercury.

Step 3: Substituting Known Values

Standard atmospheric pressure \( P_{atm} \) is \( 101325 \ ext{Pa} \). The densities are \( \rho_{Hg} = 13600 \ ext{kg/m}^3 \) and \( \rho_{H_2O} = 1000 \ ext{kg/m}^3 \). Gravity \( g \) is approximately \( 9.81 \ ext{m/s}^2 \). Substituting these values, our equation becomes: \[ 101325 + 13600 \times 9.81 \times h_{Hg} + 1000 \times 9.81 \times (1.00 - h_{Hg}) = 202650 \].

Step 4: Solving for the Depth of Mercury

Rearranging the equation to solve for \( h_{Hg} \), we have: \[ 13600 \times 9.81 \times h_{Hg} + 1000 \times 9.81 \times (1.00 - h_{Hg}) = 101325 \]. Simplifying this, we get: \[ 13600 \times 9.81 \times h_{Hg} - 1000 \times 9.81 \times h_{Hg} = 183615 \]. Factoring out \( h_{Hg} \), we get: \[ h_{Hg} = \frac{183615}{(12600 \times 9.81)} \].

Step 5: Calculating the Final Answer

Now, calculate \( h_{Hg} \): \[ h_{Hg} = \frac{183615}{12600 \times 9.81} \approx 0.78 \ ext{m} \]. Therefore, the depth of the mercury should be approximately 0.78 meters.

Key Concepts

Pressure in Fluids

When dealing with fluids, understanding pressure is key. Pressure in fluids is the force exerted by the fluid per unit area. This force acts in all directions due to the fluid's ability to flow and adapt its shape. Fluid pressure is uniform at any given depth and is affected by both the fluid's density and depth.

In the context of our exercise, we are looking at a container filled with two different fluids — mercury and water. Each fluid contributes to the total pressure at the container's bottom. The deeper a point within the fluid, the higher the pressure, due to the weight of the fluid above it. This concept is simply expressed through the equation:

• The pressure difference in a fluid column: \( \Delta P = \rho g h \)
• Where \( \rho \) is the fluid density, \( g \) is the acceleration due to gravity, and \( h \) is the height or depth of the fluid column.

Pressure from both water and mercury in the problem must add up with atmospheric pressure to achieve the desired total pressure.

Atmospheric Pressure

Atmospheric pressure is the pressure exerted by the weight of the air in the Earth's atmosphere. It is a crucial factor in fluid dynamics and needs to be considered when calculating pressure in open containers.

In most conditions, the standard atmospheric pressure is approximately 101325 Pascals (Pa). It provides a base pressure to which any additional pressure from fluids would be added.
Atmospheric pressure can vary with altitude and weather conditions but remains consistent enough at sea level for standard calculations.
When solving the given problem, atmospheric pressure provides the starting point for determining total pressure at the container's bottom. We need this pressure plus the contributions from both mercury and water to equal twice the atmospheric pressure to solve for the desired depth of mercury.

Density of Mercury

Density is a measure of mass per unit volume of a substance and plays a significant role in calculating pressure in fluids. Mercury is a dense liquid metal, and its density is a factor that greatly influences the pressure it exerts in the container.

In our exercise, the density of mercury is 13600 kg/my, which means it is significantly heavier than many other liquids including water.
This high density is why even a relatively small depth of mercury can contribute significantly to the overall pressure at the bottom of the container.

Mercury's pressure contribution: \( P = \rho_{Hg} g h_{Hg} \)

When pumped into calculations, this density, combined with gravitational acceleration and its depth, permits us to solve the exercise by adjusting the mercury's height until the total target pressure is met.

Density of Water

Water's density is a benchmark in pressure calculations due to its commonplace nature and significant influence in many fluid-related problems.
Typically, the density of water is 1000 kg/m^3, making it much less dense than mercury. This means it exerts less pressure per unit height than mercury does. However, it's the most ubiquitous fluid on Earth, serving as a reference for many physical property comparisons.
When we combine the density of water with gravitational force and depth, water's contribution to total pressure is given by:

Pressure from water: \( P = \rho_{H_2O} g h_{H_2O} \)

In the exercise, the remaining portion of the container filled with water needs to be calculated in conjunction with the mercury to ensure the bottom pressure reaches twice the atmospheric rate, balancing the combined fluid effects to precisely meet that requirement.