/*! 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 38 When a dilute gas expands quasi-... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When a dilute gas expands quasi-statically from 0.50 to \(4.0 \mathrm{L},\) it does \(250 \mathrm{J}\) of work. Assuming that the gas temperature remains constant at \(300 \mathrm{K},\) (a) what is the change in the internal energy of the gas? (b) How much heat is absorbed by the gas in this process?

Short Answer

Expert verified
(a) For an isothermal process involving an ideal gas, the change in internal energy is ∆U = 0 J. (b) The heat absorbed by the gas during the process is Q = 250 J.

Step by step solution

01

Identify the given values

We are given the following values: - Initial volume (V1) = 0.50 L - Final volume (V2) = 4.0 L - Work done (W) = 250 J - Temperature (T) = 300 K
02

Apply the First Law of Thermodynamics

According to the First Law of Thermodynamics, the change in internal energy (∆U) is related to the heat absorbed (Q) and the work done (W) by the following equation: ∆U = Q - W We know the work done, W = 250 J, and we need to find ∆U and Q.
03

Evaluate the change in internal energy for an isothermal process

Since the temperature remains constant during the process, the expansion is an isothermal process. For an isothermal process involving an ideal gas, the change in internal energy is zero: ∆U = 0 Now that we have the change in internal energy, we can find the heat absorbed by the gas.
04

Calculate the heat absorbed

Using the First Law of Thermodynamics relation and the value of ∆U, we can now find the heat absorbed (Q): 0 = Q - 250 J Solving for Q, we get: Q = 250 J
05

State the results

(a) The change in the internal energy of the gas during the quasi-static expansion is ∆U = 0 J. (b) The heat absorbed by the gas during the process is Q = 250 J.

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.

Isothermal Process
An isothermal process is a thermodynamic transformation in which a system's temperature remains constant. This implies that any heat energy entering or leaving the system does not change the system's internal temperature. In our exercise, a dilute gas expands while maintaining a steady temperature of 300 K.

What is compelling about an isothermal expansion, especially for an ideal gas, is that even though the gas does work and potentially absorbs heat, its internal energy remains unchanged. The reasoning behind this lies in the kinetic theory of gases, which states that the internal energy of an ideal gas is dependent solely on its temperature. Thus, even as a gas expands or compresses, if it does so isothermally, its internal energy does not vary.

To better understand this concept, imagine a gas trapped in a cylinder with a movable piston. The gas expands slowly against the piston, and heat is simultaneously supplied to the gas at such a rate that the temperature remains constant. The key takeaway for students is that an isothermal process, by definition, maintains a system's temperature, and for an ideal gas, this also means the internal energy stays the same.
Internal Energy
Internal energy, designated as U, is the total energy contained within a thermodynamic system. It encompasses all forms of energy in the system, which for an ideal gas, relates mainly to the kinetic energy of its molecules. Since temperature is a measure of the average kinetic energy of the molecules in a substance, the internal energy in an ideal gas is directly proportional to its temperature.

In the context of our textbook exercise, the isothermal expansion of a gas has a fascinating implication: since the process occurs at a constant temperature, the internal energy does not change throughout the process. Mathematically, this is expressed as Δ±« = 0. This is a critical point for students to understand - during isothermal processes for an ideal gas, any work done by the gas or heat added to the gas does not result in a change in the gas' total internal energy.
Heat Absorption
Heat absorption in a thermodynamic process refers to the transfer of energy into a system in the form of heat. This process is denoted by the symbol Q. In our exercise, heat absorption is what allows the gas to do work on its surroundings during its expansion, while sticking to an isothermal condition.

Using the First Law of Thermodynamics, which states that the change in internal energy of a system (Δ±«) is equal to the heat added to the system (Q) minus the work done by the system (W), and considering that Δ±« is zero for an isothermal process, we can deduce that the heat absorbed by the gas is equal to the work done by it, Q = W. In the case of the exercise, the gas absorbs 250 Joules of heat.

It is also important for learners to understand that heat transfer can occur in various ways, such as conduction, convection, and radiation. In an isothermal process of an ideal gas, all of the heat absorbed goes into doing work and none into changing the internal energy of the system, due to the constancy of temperature.

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 car tire contains \(0.0380 \mathrm{m}^{3}\) of air at a pressure of \(2.20 \times 10^{5} \mathrm{Pa}\) (about 32 psi). How much more internal energy does this gas have than the same volume has at zero gauge pressure (which is equivalent to normal atmospheric pressure)?

Five moles of a monatomic ideal gas in a cylinder at \(27^{\circ} \mathrm{C}\) is expanded isothermally from a volume of \(5 \mathrm{L}\) to 10 L. (a) What is the change in internal energy? (b) How much work was done on the gas in the process? (c) How much heat was transferred to the gas?

An amount of n moles of a monatomic ideal gas in a conducting container with a movable piston is placed in a large thermal heat bath at temperature \(T_{1}\) and the gas is allowed to come to equilibrium. After the equilibrium is reached, the pressure on the piston is lowered so that the gas expands at constant temperature. The process is continued quasi-statically until the final pressure is \(4 / 3\) of the initial pressure \(p_{1}\). (a) Find the change in the internal energy of the gas. (b) Find the work done by the gas. (c) Find the heat exchanged by the gas, and indicate, whether the gas takes in or gives up heat.

A helium-filled toy balloon has a gauge pressure of 0.200 atm and a volume of 10.0 L. How much greater is the internal energy of the helium in the balloon than it would be at zero gauge pressure?

A dilute gas expands quasi-statically to three times its initial volume. Is the final gas pressure greater for an isothermal or an adiabatic expansion? Does your answer depend on whether the gas is monatomic, diatomic, or polyatomic?

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.