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

Which of the following processes always results in an increase in the energy of a system? a) The system loses heat and does work on the surroundings. b) The system gains heat and does work on the surroundings. c) The system loses heat and has work done on it by the surroundings. d) The system gains heat and has work done on it by the surroundings. e) None of the above.

Short Answer

Expert verified
a) The system loses heat and does work on the surroundings b) The system gains heat and does work on the surroundings c) The system loses heat and has work done on it by the surroundings d) The system gains heat and has work done on it by the surroundings e) None of the above Answer: d) The system gains heat and has work done on it by the surroundings.

Step by step solution

01

Option a: The system loses heat and does work on the surroundings.

In this case, both heat (q) and work (w) are taken away from the system, which means that q is negative and w is also negative. According to the first law of thermodynamics, ΔU = q + w. Since both q and w are negative, their sum will also be negative, meaning that the internal energy of the system decreases.
02

Option b: The system gains heat and does work on the surroundings.

Here, heat (q) gets added to the system, making it positive. However, the system is doing work (w) on the surroundings, taking energy away from it and making w negative. In this case, the change in internal energy is ΔU = q + w. Depending on the magnitudes of q and w, the internal energy may increase, decrease, or remain the same, so this option does not always result in increased energy.
03

Option c: The system loses heat and has work done on it by the surroundings.

In this situation, the system loses heat (q), resulting in a negative value for q. However, work (w) is being done on the system by the surroundings, which adds energy to the system and makes w positive. The change in internal energy is given by ΔU = q + w. Similar to option b, this option does not always result in increased energy since it depends on the magnitudes of q and w.
04

Option d: The system gains heat and has work done on it by the surroundings.

Here, the system gains heat (q) and has work done on it by the surroundings (w), which means that both q and w are positive. According to the first law of thermodynamics, ΔU = q + w. Because both q and w are positive, their sum will also be positive, leading to an increase in the system's internal energy. Therefore, this option will always result in an increase in the energy of a system.
05

Option e: None of the above.

We have found that option d always results in increased energy, so this option is not correct. In conclusion, the correct answer is option d) The system gains heat and has work done on it by the surroundings. This is the process that always results in an increase in the energy of a system.

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.

Internal Energy
Internal energy is a key concept in the study of thermodynamics and refers to the total energy contained within a system. It is composed of several types of energies, including kinetic energy, due to the motion of the system's molecules, and potential energy, which is related to the intermolecular forces within the system.

Understanding how internal energy changes provides insight into the behavior of a system. When heat is added or work is done on a system, its internal energy increases. Conversely, if a system loses heat or does work on its surroundings, its internal energy decreases. This concept is fundamental in analyzing thermodynamic processes and is mathematically represented by the first law of thermodynamics as \( \Delta U = q + w \), where \( \Delta U \) is the change in internal energy, \( q \) is the heat added to or removed from the system, and \( w \) is the work done on or by the system.
Thermodynamic Processes
Thermodynamic processes involve the transfer of energy to or from a system and can include changes in temperature, volume, pressure, or internal energy. Such processes are often described in terms of heat transfer and work done.

There are several types of thermodynamic processes, including isothermal (constant temperature), isobaric (constant pressure), isochoric (constant volume), and adiabatic (no heat transfer). Each process type dictates how the internal energy, pressure, and volume of a system change, and it is important to understand these processes to predict the behavior of a system under various conditions. The ideal behavior of gases during these thermodynamic processes is often depicted by PV diagrams, which graph volume against pressure and demonstrate the work done on or by the system during a process.
Heat Transfer
Heat transfer is the movement of thermal energy from one object or system to another, driven by a temperature difference. It occurs through three main mechanisms: conduction, which involves direct molecular interaction; convection, which is the transfer of heat by the movement of fluids; and radiation, which is the transfer of energy by electromagnetic waves.

Understanding heat transfer is essential in thermodynamics, as it directly affects the internal energy of a system. The relationship between heat transfer and work in thermodynamic processes is encapsulated by the first law of thermodynamics. It is critical for students to grasp that heat transfer is not the same as temperature; instead, it is a measure of thermal energy flow, and it is considered positive when it enters a system and negative when it exits.
Work-Energy Principle
The work-energy principle is fundamental in both physics and thermodynamics. It states that work done on an object is equal to the change in its kinetic energy. In the context of thermodynamics, work can also relate to other forms of energy transfer, such as the expansion or compression of a gas in a system.

Work is a form of energy transfer that can either increase or decrease the internal energy of a system, depending on whether it is done on or by the system. Positive work is done on the system, leading to an increase in internal energy, while negative work is done by the system, leading to a decrease in internal energy. This principle is integral to solving problems that involve energy conversions and demonstrates the direct relationship between work and the internal energy state of a system.

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

20.9a) The maximum efficiency of a Carnot engine is \(100 \%\) since the Carnot cycle is an ideal process. b) The Carnot cycle consists of two isothermal processes and two adiabatic processes. c) The Carnot cycle consists of two isothermal processes and two isentropic processes (constant entropy). d) The efficiency of the Carnot cycle depends solely on the temperatures of the two thermal reservoirs.

20.14 Imagine dividing a box into two equal parts, part \(A\) on the left and part \(B\) on the right. Four identical gas atoms, numbered 1 through 4 , are placed in the box. What are most probable and second most probable distributions (for example, 3 atoms in \(\mathrm{A}, 1\) atom in \(\mathrm{B}\) ) of gas atoms in the box? Calculate the entropy, \(S\), for these two distributions. Note that the configuration with 3 atoms in \(\mathrm{A}\) and 1 atom in \(\mathrm{B}\) and that with 1 atom in A and three atoms in B count as different configurations.

An ideal gas is enclosed in a cylinder with a movable piston at the top. The walls of the cylinder are insulated, so no heat can enter or exit. The gas initially occupies volume \(V_{1}\) and has pressure \(p_{1}\) and temperature \(T_{1}\). The piston is then moved very rapidly to a volume of \(V_{2}=3 V_{1}\). The process happens so rapidly that the enclosed gas does not do any work. Find \(p_{2}, T_{2},\) and the change in entropy of the gas.

A key feature of thermodynamics is the fact that the internal energy, \(E_{\text {int }}\) of a system and its entropy, \(S\), are state variables; that is, they depend only on the thermodynamic state of the system and not on the processes by which it reached that state (unlike, for example, the heat content, \(Q\) ). This means that the differentials \(d E_{\text {int }}=T d S-p d V\) and \(d S=\) \(T^{-1} d E_{\text {int }}+p T^{-1} d V,\) where \(T\) is temperature (in kelvins), \(p\) is pressure, and \(V\) is volume, are exact differentials as defined in calculus. What relationships follow from this fact?

A volume of \(6.00 \mathrm{~L}\) of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergoes the following processes, all done reversibly: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the entropy change for each process.

See all solutions

Recommended explanations on Physics 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