/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 95 Concept Questions An ideal gas e... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Concept Questions An ideal gas expands isothermally, doing work, while heat flows into the gas. (a) Does the internal energy of the gas increase, decrease, or remain the same? (b) Is the work done by the gas greater than, less than, or equal to the heat that flows into the gas? Account for your answers. Problem Three moles of neon expand isothermally from 0.100 to \(0.250 \mathrm{~m}^{3}\). Into the gas flows \(4.75 \times 10^{3} \mathrm{~J}\) of heat. Assuming that neon is an ideal gas, find its temperature.

Short Answer

Expert verified
The internal energy remains the same; the work done equals the heat added; temperature is 812.4 K.

Step by step solution

01

Understanding Isothermal Process

An isothermal process is a thermodynamic process in which the temperature of the system remains constant. For an ideal gas, this means that the internal energy change (\(\Delta U\)) is zero since internal energy depends only on temperature for an ideal gas.
02

Apply the First Law of Thermodynamics

The first law of thermodynamics states: \[ \Delta U = Q - W \] Since the process is isothermal for an ideal gas, \(\Delta U = 0\). Therefore, the heat added to the system (\(Q\)) is equal to the work done by the system (\(W\)), which means \(Q = W\).
03

Calculate Temperature Using Ideal Gas Law

The ideal gas law is given by \( PV = nRT \), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature. For an isothermal expansion, \( \Delta U = 0 \). We use: \[ W = nRT \ln \left(\frac{V_f}{V_i}\right) \] Given \(Q = 4.75 \times 10^3 \mathrm{~J}\) and since \(Q = W\), we find the temperature \(T\) by solving: \[ 4.75 \times 10^3 = 3 \times 8.314 \times T \ln \left(\frac{0.250}{0.100}\right) \]
04

Solve for Temperature

Rearrange the equation to solve for \(T\): \[ T = \frac{4.75 \times 10^3}{3 \times 8.314 \times \ln(2.5)} \] Using a calculator, compute \(T\): \[ T \approx 812.4 \mathrm{~K} \]

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 91Ó°ÊÓ!

Key Concepts

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

Ideal Gas
An ideal gas is a simplified model used in physics and chemistry to describe the behavior of gases under various conditions. It assumes that:
  • The gas particles do not interact with each other.
  • The volume of individual gas particles is negligible compared to the volume of the container.
  • All collisions between particles and container walls are perfectly elastic, meaning there is no loss of kinetic energy.
For many real gases, this model works well under high temperatures and low pressures where intermolecular forces become less significant. The ideal gas law, expressed as \(PV = nRT\), relates pressure \(P\), volume \(V\), and temperature \(T\) of a gas, where \(n\) is the number of moles and \(R\) is the gas constant. This formula is very useful in predicting the behavior of gases under changing conditions.
Internal Energy
Internal energy, in the context of an ideal gas, primarily depends on temperature. It constitutes the total kinetic energy of all the particles in the gas.
For an ideal gas, changes in internal energy are directly related to changes in temperature. During an isothermal process, where the temperature remains constant, the internal energy change (\(\Delta U\)) is zero because the temperature doesn't change.
This implies that any energy entering or leaving the system as heat must be balanced by work done by or on the system. This concept is fundamental when analyzing thermodynamic processes like those of gases expanding or compressing under different conditions.
First Law of Thermodynamics
The first law of thermodynamics, also known as the law of energy conservation, is fundamental in understanding energy exchange. It is expressed as \( \Delta U = Q - W \), which states that the change in internal energy \(\Delta U\) of a system is equal to the heat \(Q\) added to the system minus the work \(W\) done by the system.
For an isothermal process involving an ideal gas, there is no change in internal energy (\(\Delta U = 0\)). Therefore, all the heat added to the system is used to perform work, so \(Q = W\). This foundational principle helps describe how energy is transferred as heat and work within a system, ensuring energy's conservation.
Isothermal Expansion
Isothermal expansion is a process where a gas expands while maintaining a constant temperature. This requires careful management of heat exchange so the internal energy remains unchanged.
During this process for an ideal gas, while the volume increases, the pressure decreases. The equation \(W = nRT \ln\left(\frac{V_f}{V_i}\right)\) captures this work done as the gas expands.
Since temperature is constant, the internal energy doesn't change, aligning with the principle \(\Delta U = 0\). The heat added equals the work done by the gas. This kind of process is crucial to understanding heat engines and refrigeration cycles, where the efficiency of heat energy conversion is optimized.

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

Two Camot air conditioners, \(\mathrm{A}\) and \(\mathrm{B}\), are removing heat from different rooms. The outside temperature is the same for both, but the room temperatures are different. The room serviced by unit \(\mathrm{A}\) is kept colder than the room serviced by unit B. The heat removed from both rooms is the same, (a) Which unit requires the greater amount of work? (b) Which unit deposits the greater amount of heat outside? Explain. The outside temperature is \(309.0 \mathrm{~K}\). The room serviced by unit \(\mathrm{A}\) is kept at a temperature of \(294.0 \mathrm{~K}\), while the room serviced by unit \(\mathrm{B}\) is kept at \(301.0 \mathrm{~K}\). The heat removed from either room is \(4330 \mathrm{~J}\). For both units, find the magnitude of the work required and the magnitude of the heat deposited outside. Verify that your answers are consistent with your answers to the Concept Questions.

Heat is added to two identical samples of a monatomic ideal gas. In the first sample the heat is added while the volume of the gas is kept constant, and the heat causes the temperature to rise by \(75 \mathrm{~K}\). In the second sample, an identical amount of heat is added while the pressure (but not the volume) of the gas is kept constant. By how much does the temperature of this sample increase?

A system does \(164 \mathrm{~J}\) of work on its environment and gains \(77 \mathrm{~J}\) of heat in the process. Find the change in the internal energy of (a) the system and (b) the environment.

A Carnot engine operates between temperatures of 650 and \(350 \mathrm{~K}\). To improve the efficiency of the engine, it is decided either to raise the temperature of the hot reservoir by \(40 \mathrm{~K}\) or to lower the temperature of the cold reservoir by \(40 \mathrm{~K}\). Which change gives the greatest improvement? Justify your answer by calculating the efficiency in each case.

An air conditioner keeps the inside of a house at a temperature of \(19.0^{\circ} \mathrm{C}\) when the outdoor temperature is \(33.0^{\circ} \mathrm{C}\). Heat, leaking into the house at the rate of 10500 joules per second, is removed by the air conditioner. Assuming that the air conditioner is a Carnot air conditioner, what is the work per second that must be done by the electrical energy in order to keep the inside temperature constant?

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.