Question

The density of palladium at \(20^{\circ} \mathrm{C}\) is \(12.0 \mathrm{~g} / \mathrm{mL}\), and at \(1550^{\circ} \mathrm{C}\) the density is \(11.0 \mathrm{~g} / \mathrm{mL}\). What is the change in volume (in \(\mathrm{mL}\) ) of \(1.00 \mathrm{~kg}\) Pd in going from \(20^{\circ} \mathrm{C}\) to \(1550^{\circ} \mathrm{C}\) ?

Short answer

The change in volume is 7.58 mL.

Step-by-step solution

- Calculate the initial volume at 20°C

Given that the density of palladium at 20°C is 12.0 g/mL, and the mass of palladium is 1.00 kg (or 1000 g), calculate the volume using the formula: \[ V = \frac{m}{\rho} \]Substitute the given values: \[ V_{20} = \frac{1000 \text{ g}}{12.0 \text{ g/mL}} = 83.33 \text{ mL} \]

- Calculate the final volume at 1550°C

Given that the density of palladium at 1550°C is 11.0 g/mL, and the mass of palladium is still 1.00 kg (or 1000 g), calculate the volume at this temperature using the formula: \[ V_{1550} = \frac{m}{\rho} \]Substitute the given values: \[ V_{1550} = \frac{1000 \text{ g}}{11.0 \text{ g/mL}} = 90.91 \text{ mL} \]

- Calculate the change in volume

Now, to find the change in volume, subtract the initial volume at 20°C from the final volume at 1550°C: \[ \Delta V = V_{1550} - V_{20} \]\[ \Delta V = 90.91 \text{ mL} - 83.33 \text{ mL} = 7.58 \text{ mL} \]

Key concepts

volume change

Understanding the change in volume is crucial when studying density and temperature effects in substances like palladium. The volume of a substance can vary depending on its temperature due to thermal expansion or contraction. This means that as temperature increases, the volume of a solid usually increases as well (and vice versa when the temperature decreases).

In the exercise, you learned to calculate the initial and final volumes of palladium at different temperatures using its density. By knowing the initial and final densities and the corresponding temperatures, you could find the initial and final volumes using the formula: \( V = \frac{m}{\rho} \) where \(V\) is volume, \(m\) is mass, and \(\rho\) is density. Finally, the change in volume ( \[\Delta V\] ) is found by subtracting the initial volume from the final volume.

palladium

Palladium is a rare and lustrous silvery-white metal, which belongs to the Platinum Group Metals (PGMs). It is widely used in various applications, including jewelry, automotive catalytic converters, and electronics due to its desirable properties like high melting point and excellent corrosion resistance. One interesting property of palladium is its ability to absorb hydrogen gas, making it useful in certain types of hydrogen storage and filtration systems.

The exercise highlighted how density changes in palladium when it is subjected to different temperatures. This is important for industries that use palladium because they need to account for volume changes due to temperature fluctuations. Understanding this property helps in designing better products and processes that achieve high efficiency and reliability.

temperature effect on density

Temperature can significantly affect the density of materials, including metals like palladium. Generally, as temperature increases, the atoms in a substance move more vigorously, which often causes the substance to expand, leading to a decrease in density. Conversely, reducing the temperature can make the substance contract, increasing its density.

In the exercise, you saw that palladium has a density of 12.0 g/mL at 20°C and 11.0 g/mL at 1550°C. This decrease in density with increasing temperature is typical for most materials. It is crucial for students to understand this relationship because it helps explain how materials behave under different thermal conditions, which is fundamental in fields ranging from material science to engineering design.

This concept is practically applied in real-world scenarios such as adjusting the volume of fuel in engines depending on temperature, understanding the buoyancy of objects in fluids, and designing structures that must withstand various thermal environments.

density formula

The density formula, which is \[ \rho = \frac{m}{V} \], where \(\rho\) denotes density, \(m\) denotes mass, and \(V\) denotes volume, is a fundamental equation in physics and engineering. It helps to explain how tightly or loosely packed a material's atoms are. In practical terms, density determines how heavy an object feels for its size and plays a key role in various scientific calculations and applications.

For the given exercise, you rearranged this basic formula to calculate volume: \( V = \frac{m}{\rho} \). By knowing both the mass of the palladium sample and its density at different temperatures, the volume could be computed. This allowed you to see how the volume of palladium changes due to the temperature effect based on density variations.

This concept is widely used in areas such as:

• Material property analysis
• Buoyancy calculations
• Fluid mechanics
• Design and optimization of industrial processes

Understanding how to manipulate and apply the density formula is essential for solving many real-world problems and contributes significantly to technological advancements.