Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A system changes from state \(X\) to \(Y\) with a change in internal energy measuring to \(25 \mathrm{k}] \mathrm{mol}^{-1}\), by a reversible path and returns from \(Y\) to \(X\) by an irreversible path. What will be the net change in internal energy? (a) \(25 \mathrm{~kJ}\) (b) \(>25 \mathrm{~kJ}\) \((c)<25 \mathrm{k}]\) (d) zero

Short Answer

Expert verified
The net change in internal energy will be zero.

Step by step solution

01

Identify the state function property

Internal energy is a state function, which means its change is independent of the path taken between two states. It only depends on the initial and final states of the system.
02

Analyze the reversible and irreversible paths

Since the internal energy is a state function, the change in internal energy for a process that goes from state X to state Y and then back to state X will be zero, regardless of whether the path from Y to X is reversible or irreversible.
03

Conclusion

The net change in internal energy for the complete cycle of going from state X to state Y and then back to state X is zero because the system ends up in the initial state without regard to the path taken.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Thermodynamics
Thermodynamics is a foundational concept in physics that deals with the study of energy, work, and heat, and how they intersect and transform within different systems. At its core, thermodynamics seeks to understand how energy is transferred from one form to another and how it affects matter on a macroscopic scale.

It operates on several key principles, known as the laws of thermodynamics. The first law, for instance, asserts the conservation of energy within a closed system – energy can neither be created nor destroyed, only transformed or transferred. The second law introduces the concept of entropy, indicating that systems tend to progress towards a state of disorder or randomness.

Understanding the principles of thermodynamics is crucial for various applications across engineering, chemistry, and even biological systems, as it helps to predict the direction of energy flow and the efficiency of energy conversion processes.
Reversible and Irreversible Processes
Reversible and irreversible processes are two contrasting concepts in thermodynamics that describe how systems change state. A reversible process is an idealized or theoretical scenario where a system changes state in such a way that the system and its surroundings can be restored to their original states without any net change in the universe. These processes proceed infinitesimally slowly and are characterized by an equilibrium at every stage.

An irreversible process, on the other hand, is more common in reality. It occurs spontaneously and cannot be undone without leaving a net change in the universe. Irreversible processes involve dissipative effects like friction, turbulence, and inelastic deformation, leading to an increase in entropy.

In practical terms, most processes in nature are irreversible, contributing to the natural progression towards disorder. However, reversible processes are important theoretical models that help scientists understand the maximum efficiency that can be achieved under idealized conditions.
The Conservation of Energy
The conservation of energy principle is a cornerstone of physics, encapsulated in the first law of thermodynamics. It states that energy cannot be created or destroyed in an isolated system; it can only change from one form to another. The total amount of energy remains constant over time. In terms of internal energy, this principle implies that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system on its surroundings.

Mathematically, this can be expressed as \(\text{Change in internal energy (\(\text{Δ}U\))} = \text{Heat added (\(\text{Q}\))} - \text{Work done by the system (\(\text{W}\))}\). The application of this principle allows us to solve various problems in thermodynamics, such as calculating the net energy transfer in mechanical and thermal processes. In the context of the exercise, despite the path taken (reversible or irreversible), the conservation of energy dictates that the system's internal energy returns to its original value when the system returns to its initial state.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the reaction: \(\mathrm{H}_{2(g)}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{HCl} ;\) \(\Delta H=-44 \mathrm{kcal}\) What is the enthalpy of decomposition of HCl? (a) \(+44 \mathrm{kcal} / \mathrm{mol}\) (b) - \(44 \mathrm{kcal} / \mathrm{mol}\) (c) \(-22 \mathrm{kcal} / \mathrm{mol}\) (d) \(+22 \mathrm{kcal} / \mathrm{mol}\)

What will be the change in internal energy when 12 kJ of work is done on the system and 2 kJ of heat is given by the system? (a) \(+10 \mathrm{~kJ}\) (b) \(-10 \mathrm{~kJ}\) (c) +5kJ (d) \(-5 \mathrm{~kJ}\)

Match the following columns and mark the appropriate choice. \(\begin{array}{|l|l|l|l|} \hline {\text { Column I }} & & {\text { Column II }} \\ \hline \text { (A) } & \text { Exothermic } & \text { (i) } & \Delta H=0, \Delta E=0 \\ \hline \text { (B) } & \text { Spontaneous } & \text { (ii) } & \Delta G=0 \\ \hline \text { (C) } & \text { Cyclic process } & \text { (iii) } & \Delta H \text { is negative. } \\ \hline \text { (D) } & \text { Equilibrium } & \text { (iv) } & \Delta G \text { is negative. } \\ \hline \end{array}\) (a) \((\mathrm{A}) \rightarrow(\mathrm{ii}),(\mathrm{B}) \rightarrow(\mathrm{iii}),(\mathrm{C}) \rightarrow(\mathrm{i}),(\mathrm{D}) \rightarrow(\mathrm{iv})\) (b) \((\mathrm{A}) \rightarrow(\mathrm{iv}),(\mathrm{B}) \rightarrow(\mathrm{i}),(\mathrm{C}) \rightarrow(\mathrm{iii}),(\mathrm{D}) \rightarrow\) (ii) (c) (A) \(\rightarrow\) (i), (B) \(\rightarrow\) (ii), (C) \(\rightarrow\) (iv), (D) \(\rightarrow\) (iii) (d) \((A) \rightarrow(i i i),(B) \rightarrow(i v),(C) \rightarrow(i),(D) \rightarrow(i i)\)

What is the entropy change when 1 mole oxygen gas expands isothermally and reversibly from an initial volume of \(10 \mathrm{~L}\) to \(100 \mathrm{~L}\) at \(300 \mathrm{~K} ?\) (a) \(19.14 \mathrm{JK}^{-1}\) (b) \(109.12] \mathrm{K}^{-1}\) (c) \(29.12 \mathrm{JK}^{-1}\) (d) \(10 \mathrm{~K}^{-1}\)

\(\Delta U=q+w\), is mathematical expression for (a) first law of thermodynamics (b) second law of thermodynamics (c) third law of thermodynamics (d) zeroth law of thermodynamics.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free