What are the two main components of the internal energy of a substance? On what are they based?

Short Answer

Expert verified
The two main components are kinetic energy (based on particle motion) and potential energy (based on particle positions and interactions).

Step by step solution

01

Identify the Main Components

The internal energy of a substance can be divided into two main components: kinetic energy and potential energy.
02

Understand Kinetic Energy

Kinetic energy is based on the motion of the particles within the substance. This includes translational, rotational, and vibrational movements of the molecules.
03

Understand Potential Energy

Potential energy is based on the position and interactions between the particles within the substance. This includes energies due to chemical bonds and intermolecular forces.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
The concept of kinetic energy is crucial for understanding the internal energy of a substance. Kinetic energy is the energy that particles have due to their motion. It’s based on the speed and mass of the particles. In simpler terms, consider how fast the particles move within a substance.
This energy can be observed in different types of motion:
  • Translational motion: Where particles move from one place to another. Picture them sliding across a surface.
  • Rotational motion: Where particles spin around an axis. Think of a toy top spinning.
  • Vibrational motion: Where particles move back and forth rapidly in place, similar to how a guitar string vibrates.
These motions contribute to the overall kinetic energy of the substance. Higher temperature usually means the particles are moving more quickly, increasing the kinetic energy.
To put it mathematically, the kinetic energy of a particle can be calculated using the formula:
\( KE = \frac{1}{2}mv^2 \)
Where
  • m = mass of the particle
  • v = velocity of the particle
Potential Energy
Another key component of internal energy is potential energy. Potential energy arises from the position and interactions between particles within a substance. Think of it as the stored energy due to the arrangement of particles. There are different forms of potential energy based on various interactions:
  • Chemical bonds: These bonds between atoms store energy. Breaking these bonds (like when a chemical reaction occurs) releases this energy.
  • Intermolecular forces: The energy between molecules due to attractions and repulsions. For example, in water, hydrogen bonds between molecules are a form of potential energy.
The potential energy contributes significantly to the internal energy of substances, especially those with strong chemical or intermolecular bonds.
One fundamental way to express potential energy is through the formula:
\( U = mgh \)
Where
  • U = potential energy
  • m = mass of the object
  • g = acceleration due to gravity
  • h = height or position of the object
Note that the use of this formula depends on the specific context, like gravitational potential energy.
Particle Motion
The movement of particles plays a vital role in the internal energy of substances. Particle motion includes various ways particles within a substance can move and interact, leading to different types of energy.
Here are some key types of particle motion:
  • Translational motion: Where particles travel in straight paths, colliding with each other and the walls of their container. This motion mainly contributes to the kinetic energy of gases.
  • Rotational motion: Particles spin around their center of mass. This type of motion is prevalent in gases and liquids, contributing to their overall kinetic energy.
  • Vibrational motion: Particles oscillate around fixed positions, typically found in solid structures. This motion not only contributes to kinetic energy but also to potential energy as particles are influenced by intermolecular forces.
Understanding particle motion helps us grasp how energy changes within a substance. For example, heating a substance will increase the speed of its particles, thereby increasing both the kinetic energy and overall internal energy. Conversely, cooling slows down particle motion, reducing kinetic energy.
Key takeaway here is that internal energy is a combination of all these motion and position-based energies. Recognizing this aids in comprehending thermal dynamics and energy transformations in everyday chemicals and materials.

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Most popular questions from this chapter

A system receives \(425 \mathrm{~J}\) of heat from and delivers \(425 \mathrm{~J}\) of work to its surroundings. What is the change in internal energy of the system (in \(J\) )?

A chemist places \(1.750 \mathrm{~g}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O},\) in a bomb calorimeter with a heat capacity of \(12.05 \mathrm{~kJ} / \mathrm{K}\). The sample is burned and the temperature of the calorimeter increases by \(4.287^{\circ} \mathrm{C}\). Calculate \(\Delta E\) for the combustion of ethanol in \(\mathrm{kJ} / \mathrm{mol} .\)

Stearic acid \(\left(\mathrm{C}_{18} \mathrm{H}_{36} \mathrm{O}_{2}\right)\) is a fatty acid, a molecule with a long hydrocarbon chain and an organic acid group (COOH) at the end. It is used to make cosmetics, ointments, soaps, and candles and is found in animal tissue as part of many saturated fats. In fact, when you eat meat, you are ingesting some fats containing stearic acid. (a) Write a balanced equation for the combustion of stearic acid to gaseous products. (b) Calculate \(\Delta H_{\mathrm{rm}}^{\circ}\) for this combustion \(\left(\Delta H_{\mathrm{i}}^{\circ}\right.\) of \(\mathrm{C}_{18} \mathrm{H}_{36} \mathrm{O}_{2}=\) \(-948 \mathrm{~kJ} / \mathrm{mol})\) (c) Calculate the heat \((q)\) released in \(\mathrm{kJ}\) and kcal when \(1.00 \mathrm{~g}\) of stearic acid is burned completely. (d) A candy bar contains \(11.0 \mathrm{~g}\) of fat and provides \(100 .\) Cal from fat; is this consistent with your answer for part (c)?

Consider the following balanced thermochemical equation for the decomposition of the mineral magnesite: $$\mathrm{MgCO}_{3}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{CO}_{2}(g) \quad \Delta H=117.3 \mathrm{~kJ}$$ (a) Is heat absorbed or released in the reaction? (b) What is \(\Delta H\) for the reverse reaction? (c) What is \(\Delta H\) when \(5.35 \mathrm{~mol}\) of \(\mathrm{CO}_{2}\) reacts with excess \(\mathrm{MgO} ?\) (d) What is \(\Delta H\) when \(35.5 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) reacts with excess \(\mathrm{MgO} ?\)

Diamond and graphite are two crystalline forms of carbon. At 1 atm and \(25^{\circ} \mathrm{C},\) diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the list below, determine \(\Delta H\) for C(diamond) \(\longrightarrow\) C (graphite) (1) C(diamond) \(+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-395.4 \mathrm{~kJ}\) (2) \(2 \mathrm{CO}_{2}(g) \longrightarrow 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) \(\Delta H=566.0 \mathrm{~kJ}\) (3) C(graphite) \(+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-393.5 \mathrm{~kJ}\) (4) \(2 \mathrm{CO}(g) \longrightarrow \mathrm{C}\) ( graphite) \(+\mathrm{CO}_{2}(g) \quad \Delta H=-172.5 \mathrm{~kJ}\)

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