Question

(a) How high in meters must a column of water be to exert a pressure equal to that of a \(760-\mathrm{mm}\) column of mercury? The density of water is \(1.0 \mathrm{~g} / \mathrm{mL}\), whereas that of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\). (b) What is the pressure in atmospheres on the body of a diver if he is \(39 \mathrm{ft}\) below the surface of the water when atmospheric pressure at the surface is \(0.97\) atm?

Short answer

The height of the water column needed to exert the same pressure as the $760-\mathrm{mm}$ mercury column is approximately \(10.4 \ \mathrm{m}\). The pressure exerted on the diver who is $39 \mathrm{ft}$ below the surface of the water when atmospheric pressure at the surface is $0.97$ atm is approximately \(2.121 \ \mathrm{atm}\).

Step-by-step solution

Calculate pressure of the mercury column

The formula for pressure is Pressure = Density * Height * Gravity. We are given the height of the mercury column and the density of mercury. We will use the standard value of gravity (9.8 m/s^2). To find the pressure of the mercury column, we first need to convert the height of the mercury column from mm to m and the density from g/mL to kg/m^3.\[ h_{Hg} = 760 \ \mathrm{mm} \times \frac{1 \ \mathrm{m}}{1000 \ \mathrm{mm}} = 0.76 \ \mathrm{m} \]\[ ρ_{Hg} = 13.6 \frac{\mathrm{g}}{\mathrm{mL}} \times \frac{1000 \ \mathrm{mL}}{1 \ \mathrm{L}} \times \frac{1 \ \mathrm{L}}{0.001 \ \mathrm{m^{3}}} \times \frac{1 \ \mathrm{kg}}{1000 \ \mathrm{g}} = 13600 \ \mathrm{kg/m^{3}} \]Now, we can plug in the values into the formula for pressure:\[ P_{Hg} = ρ_{Hg} \times h_{Hg} \times g = 13600 \ \mathrm{kg/m^3} \times 0.76 \ \mathrm{m} \times 9.8 \ \mathrm{m/s^2} \approx 101874.4 \ \mathrm{Pa} \]

Calculate the height of the water column

We are given the density of water, and we have found the pressure exerted by the mercury column in step 1. Using the formula for pressure, we can solve for the height of the water column. First, convert the density of water from g/mL to kg/m^3.\[ ρ_{H_2O} = 1.0 \frac{\mathrm{g}}{\mathrm{mL}} \times \frac{1000 \ \mathrm{mL}}{1 \ \mathrm{L}} \times \frac{1 \ \mathrm{L}}{0.001 \ \mathrm{m^{3}}} \times \frac{1 \ \mathrm{kg}}{1000 \ \mathrm{g}} = 1000 \ \mathrm{kg/m^{3}} \]Now, we can rearrange the pressure formula and plug in the values to find the height of the water column.\[ h_{H_2O} = \frac{P_{Hg}}{ρ_{H_2O} \times g} = \frac{101874.4 \ \mathrm{Pa}}{1000 \ \mathrm{kg/m^3} \times 9.8 \ \mathrm{m/s^2}} \approx 10.4 \ \mathrm{m} \]So, the height of the water column needed to exert the same pressure as the mercury column is approximately 10.4 meters. #b: Finding the pressure on the diver#

Convert depth to meters and calculate pressure

First, we need to convert the depth of the diver from feet to meters.\[ D = 39 \ \mathrm{ft} \times \frac{1 \ \mathrm{m}}{3.281 \ \mathrm{ft}} \approx 11.9 \ \mathrm{m} \]Now, we can calculate the pressure exerted by the water column at this depth, using the water density and height values from part (a).\[ P_{H_2O} = ρ_{H_2O} \times D \times g = 1000 \ \mathrm{kg/m^3} \times 11.9 \ \mathrm{m} \times 9.8 \ \mathrm{m/s^2} \approx 116620 \ \mathrm{Pa} \]

Calculate the pressure on the diver

To find the pressure exerted on the diver, we need to add the pressure due to the water column (at their depth) to the atmospheric pressure at the surface. The total pressure on the diver will be measured in atmospheres, so we need to first convert pressure values to atmospheres.\[ P_{H_2O(atm)} = \frac{P_{H_2O}}{101325 \ \mathrm{Pa/atm}} \approx 1.151 \ \mathrm{atm} \]Now, we can find the total pressure on the diver by adding the atmospheric pressure at the surface to the pressure due to the water column.\[ P_{total} = P_{surface} + P_{H_2O(atm)} = 0.97 \ \mathrm{atm} + 1.151 \ \mathrm{atm} \approx 2.121 \ \mathrm{atm} \]The pressure exerted on the diver is approximately 2.121 atmospheres.

Key concepts

Density and Pressure Relationship

Understanding the relationship between density and pressure is fundamental in fluid mechanics and applications like hydrostatic pressure calculation. In hydrostatics, pressure is the force exerted by a fluid due to gravity and can be calculated using the formula:

\[\begin{equation} Pressure = Density \times Height \times Gravity\text{,}\tag{1}\text{where}\begin{itemize} Density \text{ is a measure of how much mass of a substance is contained in a given volume, typically expressed in } \text{kg/m}^3. It is an intrinsic property of the material and does not change with size or shape. <\end{itemize} \text{Height } \text{is the depth or the vertical distance over which the fluid's weight acts, typically given in meters.}\end{equation}\]Gravity, often denoted by the symbol g, is the acceleration due to Earth's gravity and is approximately 9.8 m/s². All these factors are directly proportional to pressure i.e., as density or depth increases, the pressure also increases, assuming gravity remains constant. In the exercise above, by substituting the density and height values of mercury and water into equation (1), we can compare their pressures and thus determine the equivalent heights.While explaining concepts like these, always ensure you highlight that density is central to the calculation of pressure and that pressure increases with depth because of the weight of the fluid above the point where the measurement is taken. This is why divers experience more pressure as they go deeper underwater. By understanding how density and pressure are related, students can better grasp the implications of diving depths, the construction of dams, and the behavior of atmospheric layers.

Pressure Unit Conversion

In physics and engineering, the pressure is often measured in different units, and the ability to convert between these units is a practical necessity. The Pascal (Pa) is the SI unit of pressure, which is equal to a force of one Newton per square meter. However, other units such as atmospheres (atm), bar, mmHg (millimeters of mercury), and psi (pounds per square inch) are also widely used depending on the context.To understand the conversion process, consider the equation used previously where the pressure was first found in Pascals (101874.4 Pa for mercury's pressure) and then needed to be converted to atmospheres to solve the diver's pressure problem. The conversion factor between Pascals and atmospheres is: \[\begin{equation} 1 \text{ atm} = 101325 \text{ Pa}\end{equation}\]In the provided solution, we see that to convert the underwater pressure from Pascals to atmospheres, we simply divide the pressure in Pascals by 101325 Pa/atm. Therefore, the underwater pressure on the diver originally calculated in Pascals is transformed to 1.151 atm. Understanding and being able to perform such conversions are essential, as it aids in interpreting and relating various pressure values across different fields such as meteorology, diving, and many engineering disciplines.Furthermore, emphasizing the correct application of conversion factors will help students avoid mistakes and understand the reasons for using different units. This added familiarity with unit conversions will greatly benefit students in problem-solving across various scientific contexts.

Underwater Pressure Effects

The effects of underwater pressure are a topic of keen interest to divers, marine biologists, and engineers. As one descends below the water surface, the pressure increases significantly. This can be attributed to the weight of the water above, which adds to the pressure exerted by the atmosphere at the surface. The pressure experienced by an object underwater is known as hydrostatic pressure, and it increases by approximately 1 atm for every 10 meters of depth, due to the increasing mass of water above that point.In the exercise, we calculate the total pressure on a diver at a certain depth. This requires adding the hydrostatic pressure caused by the water column to the atmospheric pressure at the water surface. The effects of such pressures on the human body can be profound; the increased pressure can lead to conditions such as decompression sickness (the bends) if a diver ascends too quickly, due to dissolved gases coming out of solution in the blood.Unsurprisingly, these effects are not only a concern for humans but also a critical factor in the design of underwater structures, submersible vehicles, and equipment. High-pressure environments demand materials and structures that can withstand significant forces, highlighting engineering considerations such as material strength, buoyancy, and pressure regulation. Including examples like divers and submarines makes the concept more relatable and emphasizes the vital role pressure calculations play in safety and design within the underwater environment.