What volume of copper (density 8.96 g/cm³) would be needed to balance a 2.57 cm³ sample of lead (density 11.4 g/cm³) on a two-pan laboratory balance? cm³
Kinetic Theory of Gas
The Kinetic Theory of gases is a classical model of gases, according to which gases are composed of molecules/particles that are in random motion. While undergoing this random motion, kinetic energy in molecules can assume random velocity across all directions. It also says that the constituent particles/molecules undergo elastic collision, which means that the total kinetic energy remains constant before and after the collision. The average kinetic energy of the particles also determines the pressure of the gas.
P-V Diagram
A P-V diagram is a very important tool of the branch of physics known as thermodynamics, which is used to analyze the working and hence the efficiency of thermodynamic engines. As the name suggests, it is used to measure the changes in pressure (P) and volume (V) corresponding to the thermodynamic system under study. The P-V diagram is used as an indicator diagram to control the given thermodynamic system.
![**Physics Problem - Balancing Lead and Copper on a Laboratory Balance**
**Problem Statement:**
What volume of copper (density 8.96 g/cm³) would be needed to balance a 2.57 cm³ sample of lead (density 11.4 g/cm³) on a two-pan laboratory balance?
**Given Data:**
- Density of copper: 8.96 g/cm³
- Volume of lead sample: 2.57 cm³
- Density of lead: 11.4 g/cm³
**Solution:**
To balance the two substances on the two-pan laboratory balance, their masses must be equal.
1. **Calculate the mass of the lead sample:**
\[
\text{Mass of lead} = \text{Density of lead} \times \text{Volume of lead}
\]
\[
\text{Mass of lead} = 11.4 \text{ g/cm³} \times 2.57 \text{ cm³} = 29.298 \text{ g}
\]
2. **Determine the volume of copper needed to have the same mass:**
\[
\text{Mass of copper} = \text{Mass of lead} = 29.298 \text{ g}
\]
\[
\text{Volume of copper} = \frac{\text{Mass of copper}}{\text{Density of copper}}
\]
\[
\text{Volume of copper} = \frac{29.298 \text{ g}}{8.96 \text{ g/cm³}} = 3.27 \text{ cm³}
\]
**Final Answer:**
The volume of copper needed to balance the lead sample is **3.27 cm³**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d55ae7d-5a0b-457d-9765-fafe019f255e%2Fa20e523a-3d90-4e2e-958b-d8d11f2f65f4%2Fex9fjuj_processed.jpeg&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
