Problem 107 - General Chemistry

Question

Obtain the difference in volume between two spheres, one of radius \(5.61 \mathrm{~cm},\) the other of radius \(5.85 \mathrm{~cm} .\) The volume \(V\) of a sphere is \((4 / 3) \pi r^{3},\) where \(r\) is the radius. Give the result to the correct number of significant figures.

Step-by-Step Solution

Answer

The difference in volume is approximately 184 cm³.

Step 1: Write the Formula for the Volume of a Sphere

The formula to calculate the volume of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \] where \(r\) is the radius of the sphere.

Step 2: Calculate the Volume of the First Sphere

Substitute \(r = 5.61\) cm into the formula. \[ V_1 = \frac{4}{3} \pi (5.61)^3 \] Simplify to find \(V_1\): \[ V_1 \approx \frac{4}{3} \pi (176.837781) \approx 741.351 \pi \approx 2330.96 \text{ cm}^3\] when multiplying by \(\pi\).

Step 3: Calculate the Volume of the Second Sphere

Substitute \(r = 5.85\) cm into the formula. \[ V_2 = \frac{4}{3} \pi (5.85)^3 \] Simplify to find \(V_2\): \[ V_2 \approx \frac{4}{3} \pi (200.037625) \approx 800.1505 \pi \approx 2514.87 \text{ cm}^3\] when multiplying by \(\pi\).

Step 4: Calculate the Difference in Volumes

Subtract the volume of the smaller sphere (calculated in Step 2) from the volume of the larger sphere (calculated in Step 3): \[ \Delta V = V_2 - V_1 \approx 2514.87 - 2330.96 \approx 183.91 \text{ cm}^3 \]

Step 5: Ensure the Correct Number of Significant Figures

Both original measurements (5.61 cm and 5.85 cm) have three significant figures, so the final answer should also have three significant figures: \[ \Delta V \approx 184 \text{ cm}^3 \]

Key Concepts

Understanding Significant Figures

In mathematics and science, significant figures play a crucial role. They help determine the accuracy and precision of a calculated result. When performing calculations, it's vital to consider the number of digits that are meaningful in a given measurement. This prevents us from overestimating the precision of our results.

• Significant figures include all non-zero numbers, any zeros between significant digits, and trailing zeros in the decimal place.
• Leading zeros are not considered significant. For example, in 0.0456, only the 456 part is significant.
• It's important to carry the correct precision through all steps in a calculation.

When you report your final result, ensure that it reflects the precision of the initial data. In this exercise, since both radii had three significant figures, our final answer should have the same number. Thus, we express the volume difference as 184 cm³.

Definition of Radius of a Sphere

The radius of a sphere is a fundamental concept in geometry. It’s the distance from the center of the sphere to any point on its surface. Understanding this measurement is key to calculating various properties of a sphere, like its volume or surface area.

• The radius is a consistent distance, always the same from the center to any boundary point of a sphere.
• In problems involving spheres, the radius is typically the most crucial measurement.
• It determines not just the size but also the volume and shape of the whole sphere.

In the calculation example, we have two different radii, 5.61 cm, and 5.85 cm. Using these measurements in the volume formula gives us insights into how small changes in the radius affect the sphere's overall volume.

Calculating Volume of a Sphere

The volume of a sphere is calculated using a specific formula, which is derived from geometry principles. The formula you need to use is:
\[ V = \frac{4}{3} \pi r^3 \]
Where \( V \) is the volume, \( r \) is the radius, and \( \pi \) is approximately 3.14159.

• The cube of the radius (\( r^3 \)) means that even small changes in the size of the radius will result in large differences in volume.
• The constant fraction \( \frac{4}{3} \) and the presence of \( \pi \) make this calculation specific to spheres, different from other geometric shapes.
• When using the radius to calculate volume, ensure units are consistent to avoid errors.

Using this formula, we've calculated the volumes for both spheres in our exercise. The difference in these volumes, when adjusted to the correct significant figures, gives a deeper understanding of how these mathematical principles apply to real-world measurements.