Problem 154
Question
The internal energy change when a system goes from state \(\mathrm{A}\) to \(\mathrm{B}\) is \(40 \mathrm{~kJ} / \mathrm{mol}\). If the system goes from \(\mathrm{A}\) to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy? (a) \(40 \mathrm{~kJ}\) (b) \(>40 \mathrm{~kJ}\) (c) \(<40 \mathrm{~kJ}\) (d) zero
Step-by-Step Solution
Verified Answer
(d) zero
1Step 1: Understanding Internal Energy Change
Internal energy is a state function, meaning it depends only on the initial and final states of the system, not on the path taken to get there. Therefore, if a system starts at state A, moves to state B, and then returns to state A, the net change in internal energy is determined by the state function property.
2Step 2: Calculating Net Internal Energy Change
When the system goes from state A to state B, the change in internal energy is given as 40 kJ/mol. Since internal energy is a state function, returning from state B to state A, regardless of the reversibility of the path, will result in the energy change between state A and state B being zero over the entire cycle. Therefore, the net internal energy change is zero.
Key Concepts
Internal EnergyState FunctionsReversible and Irreversible Processes
Internal Energy
Internal energy is a key concept in thermodynamics. It represents the total energy contained within a system. This energy consists of both kinetic energy, due to the movement of molecules, and potential energy, arising from the interactions between molecules.
There are a few important characteristics of internal energy that are crucial for understanding problems like the one presented:
There are a few important characteristics of internal energy that are crucial for understanding problems like the one presented:
- Internal energy is an extensive property, meaning it depends on the amount of substance present in the system.
- It is a state function, which implies that the change in internal energy depends only on the initial and final states, not the path taken to move between these states.
- In a closed system, changes in internal energy can occur through heat transfer or performance of work.
State Functions
State functions are properties of a system that depend only on the current state of the system, not how it reached there. This is an essential concept in thermodynamics because it allows simplification of complex problems.
Some critical aspects include:
Some critical aspects include:
- Examples of state functions include internal energy, enthalpy, entropy, and Gibbs free energy.
- State functions enable us to evaluate the change in properties between two states simply by subtracting the initial state value from the final state value.
- Since they do not depend on the path, knowing initial and final states is sufficient to understand the change in a state function.
Reversible and Irreversible Processes
Understanding reversible and irreversible processes is vital in the study of thermodynamics.
Reversible processes are idealized processes that occur infinitely slowly, allowing the system to adjust and remain in equilibrium with its surroundings at each step.
Irreversible processes are more real-world processes that occur quickly, creating some level of irreversibility due to factors like friction or rapid expansion.
Irreversible processes are more real-world processes that occur quickly, creating some level of irreversibility due to factors like friction or rapid expansion.
- In reversible processes, you can restore both the system and the surroundings to their original states without any extra energy.
- In irreversible processes, achieving the original states requires external intervention and often expends additional energy.
- Irreversible processes are more prevalent in natural phenomena because they happen faster and are influenced by uncontrolled external conditions.
Other exercises in this chapter
Problem 150
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant \(K_{c}\) is (a) \(\Delta \mathrm{G}=\mathrm{RT}\)
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