Question

A \(25.0-\mathrm{cm}\) long cylindrical glass tube, sealed at one end, is filled with ethanol. The mass of ethanol needed to fill the tube is found to be \(45.23 \mathrm{~g}\). The density of ethanol is \(0.789 \mathrm{~g} / \mathrm{mL}\). Calculate the inner diameter of the tube in centimeters.

Short answer

The inner diameter of the tube is approximately \(2.16\,\mathrm{cm}\).

Step-by-step solution

Find the volume of ethanol using mass and density relation

We know the mass (m) and density (ρ) of ethanol. We have to find the volume (V) of the ethanol-filled in the tube. We use the formula: \(ρ = \frac {m}{V}\) Now make V the subject in the formula: \(V = \frac {m}{ρ}\) Given data: Mass of ethanol, \(m = 45.23\,\mathrm{g}\) Density of ethanol, \(ρ = 0.789\,\mathrm{g/mL}\) Now substituting these values in the formula above, we can calculate the volume of ethanol.

Calculate the volume of ethanol

Plug in the given values for mass and density: \(V = \frac {45.23\,\mathrm{g}}{0.789\,\mathrm {g/mL}}\) Now, calculate the volume: \(V = 57.324\,\mathrm{mL}\) So the volume of the ethanol inside the tube is \(57.324\,\mathrm{mL}\).

Find the cross-sectional area of the tube using the volume of a cylinder

Now we have the volume and length of the tube filled with ethanol. We can use the formula for the volume of a cylinder to find the cross-sectional area (A) of the tube: Volume of a cylinder, \(V = A \times L\) Where L is the length of the tube. To find the area A, we will rearrange this equation: \(A = \frac{V}{L}\) Given data: Volume of ethanol, \(V = 57.324\,\mathrm{mL}\) Length of the tube, \(L = 25.0\,\mathrm{cm}\) (since 1cm = 10 mL in unit conversion, L = 2.5 mL) Now substituting these values in the formula above, we can calculate the cross-sectional area, A

Calculate the cross-sectional area of the tube

Plug in the given values for volume and length: \(A = \frac {57.324\,\mathrm{mL}}{2.5\,\mathrm {mL}}\) Now calculate the area: \(A = 22.9296\,\mathrm {{cm^2}}\) So, the cross-sectional area of the tube is \(22.9296\,\mathrm {{cm^2}}\).

Calculate the inner diameter of the tube using the cross-sectional area and circular area formula

As the tube has a cylindrical shape, the cross-sectional area will be in the shape of a circle. So by using the area of a circle formula, we can find the diameter of the circular cross-section: Area of a circle, \(A = \pi\, (\frac{d}{2})^2\) where \(d\) is the diameter. Rearranging the equation, \(d = \sqrt{\frac{4A}{\pi}}\) Now substitute the calculated area value \(A = 22.9296\,\mathrm{{cm^2}}\)

Calculate the inner diameter of the tube

Plug in the given value for the area: \(d = \sqrt{\frac{4 \times 22.9296\,\mathrm{{cm^2}}}{\pi}}\) Now calculate the diameter (rounded to two decimal places): \(d = 2.16\,\mathrm{cm}\) So, the inner diameter of the tube is approximately \(2.16\,\mathrm{cm}\).

Key concepts

Volume

In scientific terms, the volume of a substance is the amount of space that it occupies. It is usually measured in cubic centimeters (\(\mathrm{cm^3}\)) or milliliters (\(\mathrm{mL}\)), especially when dealing with liquids. Understanding volume is crucial when calculating the amounts needed for a specific container size or when conducting experiments. In our problem, the volume of ethanol is determined from its mass and density. This relationship is defined by the formula:\[\rho = \frac{m}{V}\]Where \(\rho\) is density, \(m\) is mass, and \(V\) is volume. By solving this formula for \(V\), we find it equals:\[V = \frac{m}{\rho}\]Here, the mass of ethanol is \(45.23\,\mathrm{g}\), and the density is \(0.789\,\mathrm{g/mL}\). By plugging these into the formula, we find the volume of ethanol to be \(57.324\,\mathrm{mL}\). This calculation is pivotal as it lays the foundation for further calculations needed to determine the tube's dimensions.

Cylinder

In geometry, a cylinder is a solid with straight parallel sides and a circular cross-section. It's a common shape in everyday objects, like pipes or cans. Understanding the properties of cylinders is important for calculating volume, area, and more.Our task involves a cylindrical glass tube. The volume of the cylinder can be calculated using:\[V = A \times L\]In this formula, \(V\) is volume, \(A\) is the cross-sectional area, and \(L\) is the length of the cylinder. Knowing two of these will allow you to find the third. For the glass tube filled with ethanol, we use the volume and length to find the cross-sectional area. Cylinders are symmetrical and have uniform cross-sections deeply influencing calculations related to liquid fill, pressures, and capacities of such containers.

Cross-Sectional Area

The cross-sectional area of a cylindrical object represents the size of a slice cut perpendicular through its length. In our exercise, this is the circular area one would see looking directly into the tube.To find this area, we use the relationship known as:\[A = \frac{V}{L}\]Here \(A\) is the cross-sectional area, \(V\) is the volume (which we've calculated as \(57.324\,\mathrm{mL}\)), and \(L\) is the length of the tube (\(25.0\,\mathrm{cm}\)). Substituting these values, we compute the area as approximately \(22.9296\,\mathrm{{cm^2}}\).This understanding of cross-section contributes to determining other measures like the diameter, which offers insight into the internal shape and size of a cylinder within many engineering and scientific contexts.

Diameter

In circular geometry, the diameter is the longest straight-line distance across the circle, passing through its center. Knowing the diameter is crucial for understanding the breadth of circular objects like pipes, tubes, or wheels.For a cylinder, the diameter is directly related to its cross-sectional area. Once the area (\(A = 22.9296\,\mathrm{cm^2}\)) is known, we find the diameter using the formula for the area of a circle:\[d = \sqrt{\frac{4A}{\pi}}\]This formula is derived by rearranging the basic circle area formula \(A = \pi (\frac{d}{2})^2\).Solving gives us an approximate diameter of \(2.16\,\mathrm{cm}\). This diameter not only helps visualize the internal width of the tube but is key in engineering for designing systems where liquid flow or air passage size matters.