Question

(a) How high in meters must a column of glycerol be to exert a pressure equal to that of a \(760-\mathrm{mm}\) column of mercury? The density of glycerol is 1.26 \(\mathrm{g} / \mathrm{mL}\) , whereas that of mercury is 13.6 \(\mathrm{g} / \mathrm{mL}\) . (b) What pressure, in atmospheres, is exerted on the body of a diver if she is 15 ft below the surface of the water when the atmospheric pressure is 750 torr? Assume that the density of the water is \(1.00 \mathrm{g} / \mathrm{cm}^{3}=1.00 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) The gravitational constant is \(9.81 \mathrm{m} / \mathrm{s}^{2},\) and \(1 \mathrm{Pa}=1 \mathrm{kg} / \mathrm{m}-\mathrm{s}^{2} .\)

Short answer

(a) To find the height of the glycerol column exerting the same pressure as a 760-mm column of mercury, we use the formula \( h_{glycerol} = \frac{Pressure_{Hg}}{Density_{glycerol} \times g} \). After calculation, we get \( h_{glycerol} \approx 8.17 \mathrm{m} \). (b) The total pressure exerted on the diver is the sum of the atmospheric pressure and the pressure due to the water column. Calculating the total pressure, we get \(Total Pressure_{atm} \approx 1.62 \) atmospheres.

Step-by-step solution

(a) Calculate the pressure of a 760-mm column of mercury

To find the pressure of a 760-mm column of mercury, we can use the formula, \( Pressure = Density \times g \times h \), where h is the height of the column in meters. Convert 760 mm to meters: \( h_{Hg} = 760 \frac{mm}{1} \times \frac{1m}{1000mm} = 0.76 \mathrm{m} \). So the pressure of the mercury column can be calculated as \( Pressure_{Hg} = 13.6 \frac{g}{mL} \times \frac{1000g}{1kg} \times 9.81 \frac{m}{s^{2}} \times 0.76 m \).

(a) Calculate the height of the glycerol column

We need to set the pressure of the glycerol column equal to the pressure of the mercury column. The equation becomes, \(Pressure_{Hg} = Density_{glycerol} \times g \times h_{glycerol}\). Rearrange the equation to solve for \(h_{glycerol}\): \( h_{glycerol} = \frac{Pressure_{Hg}}{Density_{glycerol} \times g} \). Substitute the known values and solve for \(h_{glycerol}\).

(b) Convert 15 ft to meters

To find the pressure exerted on the body of the diver, we first need to convert the given height from feet to meters. \(height_{water} = 15 \frac{ft}{1} \times \frac{0.3048m}{1ft} = 4.572 \mathrm{m}\).

(b) Convert 750 torr to Pascals

To add the atmospheric pressure to the pressure due to the water column, we need to convert 750 torr into Pascals. \(Pressure_{atm} = 750 \frac{torr}{1} \times \frac{101325 Pa}{760 torr} \).

(b) Calculate the pressure due to the water column

Using the formula for pressure, \( Pressure_{water} = Density_{water} \times g \times height_{water} \), we can calculate the pressure exerted by the water column. \( Pressure_{water} = 1000 \frac{kg}{m^3} \times 9.81 \frac{m}{s^2} \times 4.572 m \).

(b) Find the total pressure on the diver

To find the total pressure exerted on the diver, we need to add the atmospheric pressure to the pressure due to the water column: \(Total Pressure = Pressure_{atm} + Pressure_{water}\).

(b) Convert total pressure to atmospheres

Finally, we want the total pressure in atmospheres. Convert the total pressure from Pascals to atmospheres using the conversion factor: \( Total Pressure_{atm} = \frac{Total Pressure}{101325 Pa} \).

Key concepts

Density

Density is a fundamental concept used to describe how much mass is contained in a specific volume of a substance. It is typically expressed in units like grams per milliliter (g/mL) or kilograms per cubic meter (kg/m³). The formula used to calculate density is simple:

• Density (\(d\)) = Mass (\(m\)) / Volume (\(V\)).

This characteristic property is crucial when calculating pressures exerted by different materials, like liquids. As seen in the exercise, different substances, such as glycerol and mercury, have varying densities that directly affect pressure. For instance, mercury is much denser than glycerol, which means a shorter column of mercury will exert the same pressure as a much taller column of glycerol.
Understanding density can help in comprehending other phenomena, such as buoyancy, where objects with varying densities will float or sink when placed in a fluid.

Pressure Conversion

Pressure is a force applied over a surface area and is often expressed in various units, such as atm (atmospheres), torr, or Pa (Pascals). Thus, understanding pressure conversion is essential, especially when solving problems involving different scales like in the given exercise.

• 1 atm = 760 torr = 101325 Pa.

In the exercise, we converted 750 torr into Pascals to combine atmospheric pressure with the liquid pressure conveniently. This resulted in the formula:

• \[Pressure_{atm} = 750 \frac{torr}{1} \times \frac{101325 Pa}{760 torr} \]

Proper conversion enables accurate calculations and understanding, especially when different systems or scientific fields require specific units. Always remember the key conversion factors!

Pressure Exerted by Liquids

The pressure exerted by a liquid in a column depends on its density, the gravitational acceleration, and the height of the liquid column. The formula to determine this pressure is:

• Pressure = Density \( \times \) Gravitational constant \( \times \) Height.

In the exercise, this formula helped us find pressures for mercury and glycerol columns. Liquids exert pressure equally in all directions, increasing with depth. This principle is why the diver in the exercise experiences additional pressure from the water column above them in addition to atmospheric pressure.
Such calculations are vital for engineering, fluid mechanics, and atmospheric sciences, providing insight into how different factors affect the pressure a specific liquid exerts.

Gravitational Constant

The gravitational constant (usually denoted by \(g\)) on Earth's surface is approximately \(9.81 \ m/s^2\). It plays a crucial role in fluid pressure calculations since gravity influences how much force a liquid exerts.

• In the pressure formula ( \(Pressure = Density \times g \times Height\)), \(g\) gives the acceleration due to gravity.

In the problem, this constant allows you to understand how gravity pulls the fluid downward, causing it to exert pressure.
Without considering the gravitational constant, one could overlook how gravity affects the system's dynamics. This constant is essential for many calculations involving weight, force, and potential energy.