Question

Calculate the height of a column of liquid glycerol \(\overline{\left(d=1.26 \mathrm{g} / \mathrm{cm}^{3}\right), \text { in meters, required to exert the }}\) same pressure as \(3.02 \mathrm{m}\) of \(\mathrm{CCl}_{4}(\mathrm{l})\left(d=1.59 \mathrm{g} / \mathrm{cm}^{3}\right)\)

Short answer

The height of glycerol column required to exert the same pressure as 3.02 m of CCl4 is approximately 3.81 m.

Step-by-step solution

Convert densities to SI units

The densities of glycerol and CCl4 are given in g/cm^3. To convert them to SI units (kg/m^3), we need to multiply the given densities by 1000. So, the densities becomes: \(d_{glycerol} = 1.26 * 1000 kg/m^3\), \(d_{CCl4} = 1.59 * 1000 kg/m^3\)

Convert height of CCl4 to meters

The height of the CCl4 column is given in metres, so no conversion is required.

Set up the pressure equation

We will use the formula for pressure in fluids: Pressure = density * gravity * height. Since the pressures exerted by both columns are equal, we can write: \(d_{glycerol} * g * h_{glycerol} = d_{CCl4} * g * h_{CCl4}\) where g is the acceleration due to gravity (9.81 m/s^2), \(h_{glycerol}\) is the height of the glycerol column, and \(h_{CCl4}\) is the height of the CCl4 column (3.02 m). The g factor can be cancelled from both sides.

Solve for height of glycerol column

Now, we can rearrange the formula and solve for \(h_{glycerol}\): \(h_{glycerol} = \frac{d_{CCl4} * h_{CCl4}}{d_{glycerol}}\). Substituting the known values: \(h_{glycerol} = \frac{1.59 * 1000 * 3.02}{1.26 * 1000} m\). To calculate the result, we multiply and divide the numbers.

Key concepts

Density Conversion

Density conversion plays a pivotal role in fluid statics problems, such as calculating the height of a glycerol column exerting equivalent pressure to a CCl4 column. Density is often given in various units, but in scientific calculations, using SI units is essential. In our scenario, we start with densities in grams per cubic centimeter (g/cm³). To convert these to the standard unit of kilograms per cubic meter (kg/m³), we multiply the given value by 1000.

So, for glycerol with a density of 1.26 g/cm³, the conversion is:

• Density of glycerol: 1.26 g/cm³ * 1000 = 1260 kg/m³

And for carbon tetrachloride (CCl4), with a density of 1.59 g/cm³:

• Density of CCl4: 1.59 g/cm³ * 1000 = 1590 kg/m³

This conversion allows us to use these densities in further calculations involving pressure that rely on SI units.

Pressure Calculation

To understand fluid statics, knowing how to calculate pressure in a fluid column is fundamental. The pressure exerted by a column of liquid depends on three factors: the density of the fluid, the gravitational acceleration, and the height of the fluid column.

The pressure can be calculated using the formula:
\[ P = \rho \cdot g \cdot h \]
Where:

• P represents pressure
• \(\rho\) is the fluid density (kg/m³)
• \(g\) is the acceleration due to gravity (9.81 m/s²)
• \(h\) is the height of the fluid column (m)

In the exercise, these elements are applied to maintain equal pressure between columns of glycerol and CCl4. Since the pressure exerted by both columns must remain equal, we set:\[\rho_{glycerol} \cdot g \cdot h_{glycerol} = \rho_{CCl4} \cdot g \cdot h_{CCl4}\]
This relationship allows us to solve for the unknown height of the glycerol column by isolating \(h_{glycerol}\) in the equation.

Fluid Statics

Fluid statics involves the study of fluids at rest and the forces acting upon them. This principle is fundamental when comparing columns of different liquids, like in our exercise where glycerol and CCl4 must exert the same pressure.

In a fluid at rest, pressure is uniform in all directions at a given depth and increases with depth due to the weight of the fluid above. Understanding this helps us set up equations as seen in the step-by-step solution:\[\rho_{glycerol} \cdot h_{glycerol} = \rho_{CCl4} \cdot h_{CCl4}\] since gravity \(g\) cancels out.

This insight leads to setting up equations to calculate unknown variables, like height in this case, emphasizing the crucial aspect of fluid behavior: maintaining equilibrium. By manipulating these principles, we achieve solutions consistent with the laws governing incompressible fluids in a gravitational field without needing intense calculations every time.