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If a closed system would undergo an internally reversible process and an irreversible process between the same end states, how would the changes in entropy for the two processes compare? How would the amounts of entropy produced compare?

Short Answer

Expert verified
The changes in entropy are equal for both processes, but the entropy produced is greater in the irreversible process.

Step by step solution

01

Identify Given Information

Determine what is given in the problem: A closed system undergoing both an internally reversible process and an irreversible process between the same end states.
02

Understand Entropy Change in Reversible Process

Recall that in a reversible process, the change in entropy \(\triangle S_{\text{rev}}\) depends only on the initial and final states of the system. Since the system is closed, \(\triangle S_{\text{rev}} = S_2 - S_1\), where \(S_1\) is the entropy at the initial state and \(S_2\) is the entropy at the final state.
03

Understand Entropy Change in Irreversible Process

For an irreversible process, the change in entropy \(\triangle S_{\text{irrev}}\) is also determined by the initial and final states. Hence, \(\triangle S_{\text{irrev}} = S_2 - S_1\).
04

Compare Entropy Changes

Since both processes start and end at the same states, the entropy change for both the internally reversible process and the irreversible process will be the same: \(\triangle S_{\text{rev}} = \triangle S_{\text{irrev}}\).
05

Entropy Produced in Reversible vs Irreversible Processes

Note that in internally reversible processes, there is no increase in entropy due to irreversibilities. Therefore, the entropy produced \(\triangle S_{\text{gen}}\) in a reversible process is zero. In contrast, for an irreversible process, there will always be positive entropy generation, making \(\triangle S_{\text{gen, irrev}} > 0\).
06

Summarize Findings

The changes in entropy for both the internally reversible and irreversible processes are equal, but the amount of entropy produced is greater in the irreversible process.

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Key Concepts

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

Closed System
A closed system in thermodynamics is one where no mass is allowed to enter or exit. However, energy can be transferred into or out of the system in the form of heat or work. This is different from an open system, which can exchange both energy and mass with its surroundings. In our problem, the closed system undergoes a process without changing its mass. Only its entropy, a measure of disorder, changes.
When we talk about changes in a closed system, we only consider the changes within the system's boundaries. This simplifies the calculations and focuses on the intrinsic properties, like entropy, temperature, and pressure.
Reversible Process
A reversible process is an idealized, hypothetical concept in thermodynamics. It occurs infinitely slowly, so the system remains very close to equilibrium at all times. No real process is truly reversible, but this concept helps us understand the upper limits of efficiency.
In a reversible process, the entropy change \[ \triangle S_{\text{rev}} \] is determined only by the initial and final states. Since our system is closed, the entropy change is expressed as \[ \triangle S_{\text{rev}} = S_2 - S_1 \].
This means that the system's entropy only depends on where it starts and where it ends, not on the path it takes.
Irreversible Process
An irreversible process, on the other hand, happens spontaneously and often quickly. Real-life processes are typically irreversible. They include factors like friction, unrestrained expansion, rapid mixing, and heat transfer through a finite temperature difference, all of which generate entropy.
Just like the reversible process, the entropy change \[ \triangle S_{\text{irrev}} \] is given by the initial and final states: \[ \triangle S_{\text{irrev}} = S_2 - S_1 \].
However, unlike a reversible process, irreversible processes result in an increase in entropy due to irreversibilities within the system.
Entropy Production
Entropy production is a key concept in determining the difference between reversible and irreversible processes. In an internally reversible process, there is no entropy produced; hence \[ \triangle S_{\text{gen}} = 0 \]. This is an ideal scenario where no additional entropy is generated within the system.
For an irreversible process, entropy is always produced. No real process can occur without some entropy production. Hence, \[ \triangle S_{\text{gen, irrev}} > 0 \].
This means that while the total change in entropy between initial and final states remains the same for both reversible and irreversible processes, the entropy produced within the system is greater for the irreversible process.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and forms of energy. It helps us understand how energy is transferred and transformed in systems. One of the key principles in thermodynamics involves entropy and the second law, which states that entropy in an isolated system always increases.
By examining reversible and irreversible processes within a closed system, we can better understand how energy efficiency and entropy production are intrinsically linked. The concepts of entropy change and production are vital for deciphering the differences in system behavior between idealized and real-world processes.

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Most popular questions from this chapter

Two insulated tanks are connected by a valve. One tank initially contains \(0.45 \mathrm{~kg}\) of air at \(93^{\circ} \mathrm{C}, 1\) bar, and the other contains \(0.9 \mathrm{~kg}\) of air at \(38^{\circ} \mathrm{C}, 2\) bar. The valve is opened and the two quantities of air are allowed to mix until equilibrium is attained. Employing the ideal gas model with \(c_{v}=0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the final temperature, in \({ }^{\circ} \mathrm{C}\). (b) the final pressure, in bar. (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

A system initially consists of a rivet at \(982^{\circ} \mathrm{C}\) whose mass is \(0.2 \mathrm{~kg}\) and a two-phase solid-liquid mixture of water at \(0^{\circ} \mathrm{C}, 1\) bar in which the mass of ice is \(1.2 \mathrm{~kg}\) and the mass of liquid is \(2.27 \mathrm{~kg}\). The specific heat of the rivet is \(0.5 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). The system attains an equilibrium state while pressure remains constant. If heat transfer with the surroundings is negligible, determine (a) the final temperature, in \({ }^{\circ} \mathrm{C}\). (b) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\). For water, the specific enthalpy change for a phase change from solid to liquid at 1 bar is \(335 \mathrm{~kJ} / \mathrm{kg}\).

An electric water heater having a 200 liter capacity employs an electric resistor to heat water from 23 to \(55^{\circ} \mathrm{C}\). The outer surface of the resistor remains at an average temperature of \(80^{\circ} \mathrm{C}\). Heat transfer from the outside of the water heater is negligible and the states of the resistor and the tank holding the water do not change significantly. Modeling the water as incompressible, determine the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\), for (a) the water as the system. (b) the overall water heater including the resistor. Compare the results of parts (a) and (b), and discuss.

An electric motor operating at steady state draws a current of \(10 \mathrm{amp}\) with a voltage of \(220 \mathrm{~V}\). The output shaft rotates at 1000 RPM with a torque of \(16 \mathrm{~N} \cdot \mathrm{m}\) applied to an external load. The rate of heat transfer from the motor to its surroundings is related to the surface temperature \(T_{\mathrm{b}}\). and the ambient temperature \(T_{0}\) by \(\mathrm{hA}\left(T_{\mathrm{b}}-T_{0}\right)\), where \(\mathrm{h}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, \mathrm{A}=0.195 \mathrm{~m}^{2}\), and \(T_{0}=293 \mathrm{~K} .\) Energy transfers are considered positive in the directions indicated by the arrows on Fig. P6.51. (a) Determine the temperature \(T_{\mathrm{b}}\), in \(\mathrm{K}\). (b) For the motor as the system, determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\). (c) If the system boundary is located to take in enough of the nearby surroundings for heat transfer to take place at temperature \(T_{0}\), determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for the enlarged system.

Air enters an insulated diffuser operating at steady state at 1 bar, \(7^{\circ} \mathrm{C}\), and \(200 \mathrm{~m} / \mathrm{s}\) and exits with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). Employing the ideal gas model and ignoring potential energy, determine (a) the temperature of the air at the exit, in \({ }^{\circ} \mathrm{C}\). (b) the maximum attainable exit pressure, in bar.

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