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

An ideal gas undergoes a constant volume (isochoric) process in a closed system. The heat transfer and work are, respectively \((a) 0,-c_{v} \Delta T\) (b) \(c_{v} \Delta T, 0\) \((c) c_{p} \Delta T, R \Delta T\) \((d) R \ln \left(T_{2} / T_{1}\right), R \ln \left(T_{2} / T_{1}\right)\)

Short Answer

Expert verified
Answer: The correct expressions for heat transfer and work in an isochoric process of an ideal gas are heat transfer Q = n c_v ΔT and work W = 0.

Step by step solution

01

Recall the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy (∆U) of a closed system is equal to the heat transfer (Q) into the system minus the work (W) done by the system: \[\Delta U = Q - W\]
02

Remember the Internal Energy Expression for an Ideal Gas

The internal energy of an ideal gas depends only on its temperature and the number of moles. For a constant-volume process, the change in internal energy can be expressed as: \[\Delta U = n c_v \Delta T\] where \(n\) is the number of moles, \(c_v\) is the molar heat capacity at constant volume, and \(\Delta T\) is the change in temperature.
03

Determine the Work Done in an Isochoric Process

In an isochoric process, the volume remains constant, i.e., \(V_{2} = V_{1}\). Therefore, the work done by the system is zero since there is no volume change: \[W = 0\]
04

Apply the First Law of Thermodynamics to Find the Heat Transfer

Since we know the expressions for \(\Delta U\) and \(W\), we can use the first law of thermodynamics to find the heat transfer: \[\Delta U = Q - W\] Substituting \(\Delta U = n c_v \Delta T\) and \(W = 0\), we obtain: \[Q = n c_v \Delta T\]
05

Compare the Found Expressions with the Given Options

From our analysis, we found that the heat transfer, Q, is given by \(Q = n c_v \Delta T\) and the work, W, is equal to 0. Comparing these expressions with the given options, we can conclude that the answer is option (b): \[(b) \hspace{5mm} c_v \Delta T, 0\]

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.

First Law of Thermodynamics
Understanding the First Law of Thermodynamics is crucial for anyone studying physics or engineering, as it lays the foundation for energy conservation in systems. Simply put, this law states that energy cannot be created or destroyed in an isolated system. The total amount of energy in the system remains constant, although it can change from one form to another.An intuitive way to think about this is by considering a bank account: the money in your account (the system's energy) can be increased by deposits (heat transfer into the system), but it can also be decreased by withdrawals (work done by the system). The overall balance reflects the total energy. Mathematically, we express this concept as:\[\Delta U = Q - W\]Where \(\Delta U\) represents the change in internal energy, \(Q\) is the heat added to the system, and \(W\) is the work done by the system. For an isochoric process, where volume does not change, there is no work performed, hence the equation simplifies to \(\Delta U = Q\). This demonstrates that all the heat added to the system increases its internal energy by the same amount.
Internal Energy of an Ideal Gas
In the realm of thermodynamics, the internal energy is a measure of the total energy contained within a substance, which comprises both the kinetic and potential energy of its particles. For an ideal gas—a theoretical gas that perfectly follows the gas laws—this internal energy is solely a function of temperature and is independent of pressure or volume.Ideal gases are composed of randomly moving particles that do not interact with each other except for perfectly elastic collisions, meaning no energy lost in the process. The expression for the change in internal energy, \(\Delta U\), in a constant-volume process is:\[\Delta U = n c_v \Delta T\]where \(n\) symbolizes the number of moles in the gas, \(c_v\) signifies the molar heat capacity at constant volume (how much energy is needed to raise the temperature of a mole of the substance by one Kelvin without changing its volume), and \(\Delta T\) indicates the change in temperature. When considering an isochoric process, only the transfer of heat affects the internal energy as there is no work done due to volume change.
Heat Capacity at Constant Volume
The heat capacity of a substance at constant volume, often represented as \(c_v\), is a measure of how much heat energy is required to raise the temperature of a substance by a certain amount (typically one degree Celsius or one Kelvin) at a fixed volume. This property plays a vital role in calculating the change in internal energy during processes like the isochoric one we're discussing.For an ideal gas, the heat capacity at constant volume is especially important because it reveals how energy input (as heat) translates to an increase in internal energy—and subsequently temperature—since the gas does not expand or contract. With isochoric processes, we use the heat capacity at constant volume rather than at constant pressure (represented by \(c_p\)).Since work is defined as a force applied over a distance and no distance is covered when volume does not change, the heat capacity at constant volume becomes the sole determinant of how much the internal energy of a system will change with a given amount of heat energy. Therefore, knowing the value of \(c_v\) allows us to predict and calculate the system's response to heat input, making it a fundamental concept in thermodynamics applied to closed systems undergoing isochoric processes.

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

A piston-cylinder device contains \(2.2 \mathrm{kg}\) of nitrogen initially at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). The nitrogen is now compressed slowly in a polytropic process during which \(P V^{1.3}=\) constant until the volume is reduced by one-half. Determine the work done and the heat transfer for this process.

The temperature of 2 kg of neon is increased from 20 to \(180^{\circ} \mathrm{C}\). Calculate the change in the total internal energy of the neon, in \(\mathrm{kJ}\). Would the internal energy change be any different if the neon were replaced with argon?

Design an experiment complete with instrumentation to determine the specific heats of a gas using a resistance heater. Discuss how the experiment will be conducted, what measurements need to be taken, and how the specific heats will be determined. What are the sources of error in your system? How can you minimize the experimental error?

One kilogram of carbon dioxide is compressed from \(0.5 \mathrm{MPa}\) and \(200^{\circ} \mathrm{C}\) to \(3 \mathrm{MPa}\) in a piston-cylinder device arranged to execute a polytropic process with \(n=1.3 .\) Use the compressibility factor to determine the final temperature.

A \(3-m \times 4-m \times 5-m\) room is to be heated by one ton \((1000 \mathrm{kg})\) of liquid water contained in a tank that is placed in the room. The room is losing heat to the outside at an average rate of \(6000 \mathrm{kJ} / \mathrm{h}\). The room is initially at \(20^{\circ} \mathrm{C}\) and 100 \(\mathrm{kPa}\) and is maintained at an average temperature of \(20^{\circ} \mathrm{C}\) at all times. If the hot water is to meet the heating requirements of this room for a 24 -h period, determine the minimum temperature of the water when it is first brought into the room. Assume constant specific heats for both air and water at room temperature.

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