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An adiabatic process is one that involves no heat transfer. What is the relationship between work and the change in internal energy in an adiabatic process?

Short Answer

Expert verified
In an adiabatic process, \(\triangle U = -W\).

Step by step solution

01

Understand the First Law of Thermodynamics

The First Law of Thermodynamics states that the change in internal energy (\(\triangle U\) in a system is equal to the heat added to the system (Q) minus the work done by the system (W): \[\triangle U = Q - W\].
02

Define an Adiabatic Process

In an adiabatic process, there is no heat transfer, which means Q = 0. Hence, the equation becomes: \[\triangle U = -W\].
03

Establish the Relationship

Since Q = 0 in an adiabatic process, the First Law of Thermodynamics simplifies to: \[\triangle U = -W\]. This indicates that the change in internal energy is equal to the negative of the work done by the system.

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle that provides an understanding of energy in a system. It is crucial for grasping how energy is conserved and transferred in various physical processes. The law states:
i.e., \[ \triangle U = Q - W \] In simpler terms, it says that the change in internal energy (\triangle U) in a system is equal to the heat added to the system (Q) minus the work done by the system (W). This highlights that the energy within a closed system remains constant, accounting for energy added or removed through heat and work. Understanding this relationship is very important when we analyze processes like adiabatic processes where heat transfer is absent.

Internal Energy
Internal energy refers to the total energy contained within a system. It includes both the kinetic energy of all moving particles and the potential energy due to the interactions between them. When discussing thermodynamic processes, internal energy is a core concept. It helps in describing how energy within the system changes, especially during processes where heat and work are involved.
In an adiabatic process, the heat transfer (Q) is zero (Q = 0). This means any change in internal energy is directly related to the work done (W) by or on the system. Thus, for an adiabatic process the First Law of Thermodynamics simplifies to: \[ \triangle U = -W \]The negative sign indicates that if the system does work (W > 0), its internal energy decreases, whereas if work is done on the system (W < 0), the internal energy increases. Understanding this helps in analyzing how energy shifts within a system in response to an absence of heat transfer.
Work-Energy Relationship
In thermodynamics, work refers to the energy transferred when a force is applied over a distance. When we discuss the work-energy relationship, it is directly connected to how energy changes within the system in response to work done. An adiabatic process is a clear example of this relationship where the system's change in internal energy is resultant of work done, not heat transfer. In an adiabatic process: \[ \triangle U = -W \]This helps us understand that work done by the system reduces its internal energy, and work done on the system increases it. This inverse relationship is essential for understanding energy dynamics when analyzing systems undergoing adiabatic changes. When studying the first law, internal energy changes, and work-energy dynamics, remember these key points:
  • Energy conservation is fundamental.
  • No heat transfer in adiabatic processes means internal energy change matches the work done.
  • Adiabatic processes exemplify the direct work-internal energy relationship.
Regularly reinforcing these relationships provides a robust understanding of thermodynamic processes and principles.

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Most popular questions from this chapter

Calculate \(\Delta H\) for $$\mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{CaCO}_{3}(s)$$given the following reactions: $$\begin{aligned}\mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & & \Delta H=-635.1 \mathrm{~kJ} \\\\\mathrm{CaCO}_{3}(s) & \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) & & \Delta H=178.3 \mathrm{~kJ}\end{aligned}$$

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