Problem 18 - University Physics with Modern Physics

Question

A tall cylinder with a cross-sectional area 12.0 \(\mathrm{cm}^{2}\) is partially filled with mercury; the surface of the mercury is 5.00 \(\mathrm{cm}\) above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure at the bottom of the cylinder?

Step-by-Step Solution

Answer

816 cm³ of water must be added.

Step 1: Understand the Problem

We need to find out how much water to add to double the gauge pressure at the bottom of a cylinder containing mercury.

Step 2: Calculate Initial Gauge Pressure

The initial gauge pressure due to mercury at the bottom is given by the formula: \\( P_{mercury} = \rho_{mercury} \cdot g \cdot h_{mercury} \) \where \( \rho_{mercury} = 13,600 \, \mathrm{kg/m^3} \), \( g = 9.81 \, \mathrm{m/s^2} \), and \( h_{mercury} = 0.05 \, \mathrm{m} \) (converted from 5.00 cm). This gives: \\[ P_{mercury} = 13,600 \cdot 9.81 \cdot 0.05 = 6,673.2 \, \mathrm{Pa} \]

Step 3: Set New Pressure Requirement

To double the gauge pressure, the new total gauge pressure \( P_{total} \) should be twice the initial pressure: \\[ P_{total} = 2 \times 6,673.2 = 13,346.4 \, \mathrm{Pa} \]

Step 4: Calculate Water Pressure Contribution

The gauge pressure due to the added water is given by \( P_{water} = \rho_{water} \cdot g \cdot h_{water} \) where \( \rho_{water} = 1,000 \, \mathrm{kg/m^3} \). To find \( h_{water} \), set \( P_{mercury} + P_{water} = P_{total} \), thus: \\[ 6,673.2 + 1,000 \cdot 9.81 \cdot h_{water} = 13,346.4 \]

Step 5: Solve for Water Height

Solving for \( h_{water} \) gives: \\[ 1,000 \cdot 9.81 \cdot h_{water} = 6,673.2 \] \\[ h_{water} = \frac{6,673.2}{9,810} = 0.68 \, \mathrm{m} \]

Step 6: Calculate Volume of Water

The volume of water needed is \( V = A \cdot h_{water} \), where \( A = 0.0012 \, \mathrm{m^2} \) (converted from 12.0 cm²). So: \\[ V = 0.0012 \cdot 0.68 = 0.000816 \, \mathrm{m^3} \] \which is \( 816 \, \mathrm{cm^3} \).

Key Concepts

Gauge Pressure

Gauge pressure measures the pressure in a fluid relative to atmospheric pressure. Unlike absolute pressure, which includes atmospheric pressure in its measurement, gauge pressure only measures the pressure above or below atmospheric pressure levels. This is important in fluid mechanics as it provides a practical measure of the pressure exerted by fluids in real-world applications, such as in the cylinder in this problem.
In this exercise, the initial gauge pressure was due to the mercury. We calculated it with the formula:

• \( P_{mercury} = \rho_{mercury} \cdot g \cdot h_{mercury} \)

where \( \rho_{mercury} = 13,600 \, \mathrm{kg/m^3} \), \( g = 9.81 \, \mathrm{m/s^2} \), and \( h_{mercury} = 0.05 \, \mathrm{m} \). The outcome was a pressure of \( 6,673.2 \, \mathrm{Pa} \). By doubling the gauge pressure, we could determine the extra contribution needed from the water that is added. So, the total gauge pressure was set to be twice the initial mercury pressure.

Density of Mercury

The density of mercury plays a crucial role in determining the pressure it exerts. With its density being quite high at \( 13,600 \, \mathrm{kg/m^3} \), mercury can exert significant pressure over a small height. This physical property makes mercury ideal for barometers and some pressure measurement devices. When using the formula for pressure, \( P = \rho \cdot g \cdot h \), where \( \rho \) is density, \( g \) is the acceleration due to gravity, and \( h \) is height, the large value of \( \rho_{mercury} \) shows why the pressure at a mere 5 cm height was so substantial. This density factor was evidently useful in doubling the gauge pressure when adding water in the exercise.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the gravitational force. It increases with the depth of the fluid column. The key formula used to calculate hydrostatic pressure is:

• \( P = \rho \cdot g \cdot h \)

where \( \rho \) is the density, \( g \) is gravitational acceleration, and \( h \) is the height of the fluid column. In this exercise, hydrostatic pressure was calculated initially from mercury, and later from water when added to double the very gauge pressure. Notably, despite water having a lower density (\( 1,000 \, \mathrm{kg/m^3} \)), it contributed significantly to the hydrostatic pressure given the larger height needed.

Volume Calculation

The calculation of volume is necessary whenever dealing with varying fluid levels in enclosed spaces. It integrates the concepts of cross-sectional area and fluid height. In this particular exercise, after finding the height of the water column needed (0.68 m) to double the gauge pressure, the volume of water was determined.
The formula used was:

• \( V = A \cdot h_{water} \)

where \( A \) is the cross-sectional area (\( 0.0012 \, \mathrm{m^2} \)), and \( h_{water} \) is the newly derived water height. Calculating it gave \( V = 0.000816 \, \mathrm{m^3} \), which converts to \( 816 \, \mathrm{cm^3} \). Volume calculation is critical in many fields that require precise measurements of fluid quantities.