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A system goes from state 1 to state 2 and back to state 1 . (a) What is the relationship between the value of \(\Delta E\) for going from state 1 to state 2 to that for going from state 2 back to state \(1 ?\) (b) Without further information, can you conclude anything about the amount of heat transferred to the system as it goes from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?\) (c) Suppose the changes in state are reversible processes. Can you conclude anything about the work done by the system upon going from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?\)

Short Answer

Expert verified
(a) The relationship between the changes in internal energy for going from state 1 to state 2 and from state 2 back to state 1 is: \(\Delta E_{21} = - \Delta E_{12}\). (b) Without further information, we cannot conclude anything about the amount of heat transferred during the process. (c) If the changes in state are reversible processes, the work done by the system upon going from state 1 to state 2 is equal in magnitude and opposite in sign to the work done upon going from state 2 back to state 1: \(W_{21} = -W_{12}\).

Step by step solution

01

Part (a) - Relationship between changes in internal energy

Let's think about the changes in internal energy for these transitions. When the system goes from state 1 to state 2, the internal energy changes by some value \(\Delta E\). When the system returns from state 2 to state 1, the internal energy has to return to its initial value. Therefore, the internal energy change for the transition from state 2 back to state 1 must be the inverse of the internal energy change from state 1 to state 2, which means \(\Delta E_{21} = - \Delta E_{12}\).
02

Part (b) - Heat transfer comparison

Without any additional information about the process, we cannot conclude anything about the amount of heat transferred (\(Q\)) during the process, as it depends on the specific paths taken between states 1 and 2 for both transitions. Keep in mind that the First Law of Thermodynamics states that \(\Delta E = Q - W\), where \(W\) denotes the work done by the system.
03

Part (c) - Comparing work done during reversed processes

Given that the processes from state 1 to state 2 and from state 2 back to state 1 are reversible, we can now analyze the work done by the system during these processes. For reversible processes, the work done by the system depends only on the initial and final states. Since both processes start and end in the same states, the work done by the system during the transition from state 1 to state 2 must be equal in magnitude and opposite in sign to the work done during the transition from state 2 back to state 1. Therefore, we can conclude that \(W_{21} = -W_{12}\).

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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, often termed the law of energy conservation, sets the foundation for understanding how energy is transformed in a physical system. At its essence, the First Law states that energy can neither be created nor destroyed; it can only be transferred or converted from one form to another.

The mathematical expression associated with this law is \(\Delta E = Q - W\), where \(\Delta E\) signifies the change in internal energy of the system, \(Q\) is the heat transferred to the system, and \(W\) is the work done by the system. If we interpret this in a practical manner, it suggests that the increase in internal energy of a closed system must be equal to the heat energy supplied to the system minus the work done by the system on its surroundings.

Understanding this law is crucial because it governs the fundamental principles of heat engines, refrigerators, and many other thermodynamic processes that involve energy transformation, making it a pivotal concept in areas like physics, chemistry, and engineering.
Internal Energy Change
Internal energy, symbolized by \(U\), represents the total energy contained within a system, arising from the kinetic and potential energies of the molecules within the system. The change in internal energy, \(\Delta E\) or \(\Delta U\), is a critical concept in thermodynamics because it helps to quantify how a system's energy changes in response to heat transfer and work done.

When a system undergoes a thermodynamic process, its internal energy may increase or decrease. For example, in our textbook problem, when the system moves from state 1 to state 2 and then returns to state 1, the net change in internal energy over the complete cycle is zero. That is to say, the system expends energy only to have it restored by the end of the cycle, which is a fundamental characteristic of thermodynamic processes in a closed system.
Reversible Processes
In thermodynamics, a reversible process is an idealized concept where the system changes states in such a manner that the system and surroundings can be restored to their original conditions without any net change. In real-world applications, truly reversible processes are not possible due to natural inefficiencies such as friction and entropy production. However, they provide a useful theoretical limit to compare with actual processes.

Diving deeper into reversible processes, one of their essential properties is that the work done by the system is maximized when the process is reversible. This implies that for any given change in state, no other process can do more work than a reversible one. It follows that if our system follows a reversible path from state 1 to state 2 and back to state 1, then the work done on each leg of the journey is precisely equal in magnitude, but of opposite sign. This reinforces the concept that reversible processes are an ideal that serves as a standard for efficiency in thermodynamic systems.
Heat Transfer
Heat transfer is a fundamental concept that describes the movement of thermal energy between a system and its environment as a result of a temperature difference. There are three primary modes of heat transfer: conduction, which occurs through direct contact; convection, which is fluid movement either naturally or forced; and radiation, which involves the transfer of energy by electromagnetic waves.

In a thermodynamic cycle, such as the one described in our exercise, the heat transfer into the system during the transition from one state to another may not be the same as the heat transfer out of the system on return to the original state. This is because heat transfer is path-dependent, meaning it is influenced by the route the system takes during a process, and this complication is what makes it challenging to find specific heat values without more information about the process itself.
Work Done by the System
The concept of work in thermodynamics typically refers to the energy transferred when a force is applied over a distance. In the context of a thermodynamic system, work can be done by the system on its surroundings, such as when gas expands against a piston, or by the surroundings on the system, such as when the piston compresses the gas.

Work is a pathway-dependent function, which means that the amount of work done by a system can vary depending on the process it undergoes. However, in the case of reversible processes, as mentioned before, the work done in each direction is identical in magnitude but opposite in direction, as highlighted in the step-by-step solution to the textbook problem. This illustrates the importance of understanding the nature of the processes at play in order to accurately calculate the work involved in a thermodynamic cycle.

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Most popular questions from this chapter

Aceticacid can be manufactured by combining methanol with carbon monoxide, an example of a carbonylation reaction: $$ \mathrm{CH}_{3} \mathrm{OH}(l)+\mathrm{CO}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}(l) $$ (a) Calculate the equilibrium constant for the reaction at \(25^{\circ} \mathrm{C}\). (b) Industrially, this reaction is run at temperatures above \(25^{\circ} \mathrm{C}\). Will an increase in temperature produce an increase or decrease in the mole fraction of acetic acid at equilibrium? Why are elevated temperatures used? (c) At what temperature will this reaction have an equilibrium constant equal to \(1 ?\) (You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, and you may ignore any phase changes that might occur.)

How does the entropy of the system change when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes?

The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a} .\) (b) By using the value of \(K_{a}\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\), \(\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} M\), and \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M?}\)

Consider the following equilibrium: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Thermodynamic data on these gases are given in Appendix C. You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (c) At what temperature will an equilibrium mixture of \(10 \mathrm{~atm}\) total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (d) Rationalize the results from parts (b) and (c) by using Le Châtelier's principle. \(\infty\) (Section 15.7)

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and discuss briefly how these factors determine the magnitude of \(K .\) (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>0 .\) (c) In each case indicate whether \(\underline{K}\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)\) (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\) (iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\) (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)

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