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 or she is \(39 \mathrm{ft}\) below the surface of the water when atmospheric pressure at the surface is \(0.97\) atm?

Short answer

(a) The height of a water column that exerts the same pressure as a \(760\,\text{mm}\) column of mercury is \(10.33 \, \text{m}\). (b) The pressure on the diver is \(2.12 \, \text{atm}\).

Step-by-step solution

Part (a) - Calculate the height of a column of water

We will use the pressure formula for fluids, which is: \[P = \rho gh\] where \(P\) is the pressure, \(\rho\) is the density, \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)), and \(h\) is the height of the fluid column. First, we will calculate the pressure exerted by the given column of mercury (height \(760\,\text{mm}\) and density \(13.6\,\text{g/mL}\)). We will use the height in meters and convert the density to \(\text{kg/m}^3\). Height of mercury column in meters: \(0.76 \, \text{m}\) Density of mercury in \(\text{kg/m}^3\): \(13.6 \,\text{g/mL} * 1000\,\text{kg/g} * \frac{1}{1000}\,\frac{\text{m}^3}{\text{L}} = 13{,}600\,\text{kg/m}^3\) Pressure exerted by mercury column: \[P_{\text{mercury}} = \rho_\text{mercury} g h_\text{mercury} = (13,600\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(0.76\,\text{m}) = 101{,}104.96\,\text{Pa}\] Now, we will find the height of a water column that exerts the same pressure \(P_{\text{mercury}}\). We are given the density of water as \(1.0\,\text{g/mL}\). Let's convert it to \(\text{kg/m}^3\). Density of water in \(\text{kg/m}^3\): \(1.0\,\text{g/mL} * 1000\,\text{kg/g} * \frac{1}{1000}\,\frac{\text{m}^3}{\text{L}} = 1000\,\text{kg/m}^3\) Now, we have: \[101{,}104.96\,\text{Pa} = (1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(h_\text{water})\] To find the height of the water column (\(h_\text{water}\)), we will rearrange the equation and solve for \(h_\text{water}\): \(h_\text{water} = \frac{101{,}104.96\,\text{Pa}}{(1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)} = 10.33\,\text{m}\) So, the height of a water column that exerts the same pressure as a \(760\,\text{mm}\) column of mercury is \(10.33 \, \text{m}\).

Part (b) - Calculate the pressure on a diver

Given the depth of the diver is \(39\,\text{ft}\), we need to convert this depth to meters to work with SI units. We will use the conversion factor: \(1 \, \text{m} = 3.281 \, \text{ft}\). Depth of diver in meters: \(39\,\text{ft} * \frac{1}{3.281}\,\frac{\text{m}}{\text{ft}} = 11.89\,\text{m}\) Now, we can calculate the pressure on the diver due to the water column above them, using the formula and the given density of water: \[P_{\text{water}} = \rho_\text{water} g h_\text{diver} = (1000\,\text{kg/m}^3)(9.81\,\text{m/s}^2)(11.89\,\text{m}) = 116{,}767.9\,\text{Pa}\] Now, we need to add this pressure to the atmospheric pressure at the surface, which is given as \(0.97 \, \text{atm}\). We will first convert this pressure to Pascal using the conversion factor: \(1 \, \text{atm} = 101{,}325 \, \text{Pa}\). Atmospheric pressure at the surface in \(\text{Pa}\): \(0.97\,\text{atm} * 101{,}325 \, \frac{\text{Pa}}{\text{atm}} = 98{,}284.25\,\text{Pa}\) Total pressure on the diver: \[P_\text{diver} = P_{\text{water}} + P_{\text{surface}} = 116{,}767.9\,\text{Pa} + 98{,}284.25\,\text{Pa} = 215{,}052.15\,\text{Pa}\] Finally, we will convert this pressure back to atmospheres: Pressure on the diver in atmospheres: \[\frac{215{,}052.15\,\text{Pa}}{101{,}325\,\frac{\text{Pa}}{\text{atm}}} = 2.12\, \text{atm}\] So, the pressure on the diver is \(2.12 \, \text{atm}\).

Key concepts

Density of Liquids

Density plays a crucial role in understanding fluid pressure. In the context of this exercise, we work with the densities of water and mercury.
Density is defined as the mass of a substance per unit volume, typically expressed in units of grams per milliliter (g/mL) or kilograms per cubic meter (kg/m³).

• Water has a density of 1.0 g/mL, which converts to 1000 kg/m³.
• Mercury has a significantly higher density of 13.6 g/mL, equivalent to 13,600 kg/m³.

This difference in density means that mercury can exert more pressure at a shorter column height compared to water. The density of a liquid directly affects how much pressure it can exert at a given height. This forms the basis for pressure calculations in fluids.

Pressure Calculations

Calculating fluid pressure involves understanding how density, gravity, and height interact. In this exercise, we use the formula:\[P = \rho gh\]where:

• \( P \) is the pressure exerted by the fluid (in Pascal, Pa),
• \( \rho \) is the density of the fluid (in kg/m³),
• \( g \) is the gravitational acceleration (approximately 9.81 m/s²), and
• \( h \) is the height of the fluid column (in meters).

For example, to find the pressure exerted by a 0.76-meter column of mercury with a density of 13,600 kg/m³, you calculate it as:\[P_{\text{mercury}} = 13,600 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.76 \, \text{m} = 101,104.96 \, \text{Pa}\]To find how high a column of water needs to exert the same pressure:\[h_{\text{water}} = \frac{101,104.96 \, \text{Pa}}{1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2} \approx 10.33 \, \text{m}\]This illustrates how liquids with different densities exert the same pressure at different heights.

Unit Conversion

Unit conversion is crucial for coherent calculations, especially in science and engineering. The exercise requires converting units to maintain consistency in pressure calculations.

• Convert lengths from feet to meters. For instance, 39 feet is converted to 11.89 meters using the conversion factor 1 meter = 3.281 feet.
• Convert pressure from atmospheres to Pascals, since 1 atmosphere is equal to 101,325 Pascals. Thus, 0.97 atm becomes 98,284.25 Pa.
• When working with density, ensure to use the appropriate conversion to kg/m³, as seen with densities of liquids in the problem.

These conversions mean calculations are performed accurately within the same unit system, leading to correct results.

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of the air above us.
In our exercise, understanding atmospheric pressure is essential for calculating the total pressure on a diver. At sea level, the standard atmospheric pressure is around 1 atm, equivalent to 101,325 Pa.
Atmospheric pressure decreases with altitude and can vary due to weather conditions. In this problem, the atmospheric pressure is given as 0.97 atm at the water's surface, equal to 98,284.25 Pa.

• This means a diver 39 feet underwater faces pressure from both the weight of the water and the atmosphere.
• The total pressure experienced underwater is calculated by adding the water pressure to the atmospheric pressure.

Therefore, understanding atmospheric pressure is vital for determining the overall pressure in different environments.