/*! 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 83 One-half mole of a monatomic ide... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 83

One-half mole of a monatomic ideal gas absorbs 1200 \(\mathrm{J}\) of heat while 2500 \(\mathrm{J}\) of work is done by the gas. (a) What is the temperature change of the gas? (b) Is the change an increase or a decrease?

Short Answer

Expert verified
The temperature decreases by approximately 104.4 K.

Step by step solution

01

Understand the First Law of Thermodynamics

The first law of thermodynamics is given by the formula \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system. Here \( Q = 1200 \, \mathrm{J} \) and \( W = 2500 \, \mathrm{J} \).
02

Calculate the Change in Internal Energy

Using the first law of thermodynamics, calculate the change in internal energy: \( \Delta U = 1200 - 2500 = -1300 \, \mathrm{J} \). This indicates that the internal energy of the gas decreases by 1300 J.
03

Relate Internal Energy Change to Temperature Change

For a monatomic ideal gas, the change in internal energy can also be expressed as \( \Delta U = \frac{3}{2} n R \Delta T \), where \( n \) is the number of moles, \( R \) is the ideal gas constant (8.31 J/(mol路K)), and \( \Delta T \) is the temperature change. Here, \( n = 0.5 \) moles.
04

Solve for Temperature Change

Since we have \( \Delta U = -1300 \, \mathrm{J} \), substitute into \( \frac{3}{2} \times 0.5 \times 8.31 \times \Delta T = -1300 \). Solving for \( \Delta T \), you get \( \Delta T = \frac{-1300}{\frac{3}{2} \times 0.5 \times 8.31} \approx -104.4 \, \mathrm{K} \).
05

Determine Change Direction

The calculated \( \Delta T \) is \(-104.4 \, \mathrm{K} \), indicating the temperature decreases since it is negative.

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 91影视!

Key Concepts

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

Monatomic Ideal Gas
A monatomic ideal gas is a simple type of gas composed of individual atoms rather than molecules. This simplicity makes it easier to study and model in thermodynamics. In a monatomic gas:
  • The particles are single atoms without any bonds or internal structures to store energy.
  • Common examples include noble gases like helium or neon.
  • The only form of kinetic energy is the translation of atoms as there are no rotational or vibrational movements.
Understanding the behavior of a monatomic ideal gas helps predict how it reacts when subjected to different thermodynamic processes.
Internal Energy
Internal energy (\( U \)) refers to the total energy contained within a system. For an ideal gas, this energy is purely due to the motion of its particles. The internal energy is:
  • A function of temperature; higher temperatures mean greater particle motion and thus more energy.
  • A crucial factor in thermodynamic processes as it influences how a system exchanges heat and performs work.
  • For a monatomic ideal gas, the internal energy only depends on its temperature and the number of moles: \( U = \frac{3}{2} nRT \).
Changes in internal energy can be calculated using the first law of thermodynamics: \( \Delta U = Q - W \), where \( Q \) is heat added and \( W \) is work done.
Temperature Change
The temperature change (\( \Delta T \)) in a system indicates how the thermal energy level of the system has shifted as a result of heat transfer and work done. For a monatomic ideal gas experiencing a temperature change:
  • Temperature is directly linked to the average kinetic energy of the gas particles.
  • If the gas absorbs heat, the temperature generally increases; if it expends energy as work, the temperature may decrease.
  • In our exercise, we calculated a negative temperature change, \(-104.4 \, \mathrm{K}\), indicating a cooling effect.
The temperature change can be determined through the relationship \( \Delta U = \frac{3}{2} n R \Delta T \), leveraging the internal energy change.
Ideal Gas Constant
The ideal gas constant (\( R \)) is a key quantity in the ideal gas law and plays a critical role in thermodynamics. Its approximate value is 8.31 \( \mathrm{J/(mol \cdot K)} \). Here's why it's important:
  • It relates the energy scale to the temperature and quantity of gas.
  • Facilitates the calculation of internal energy and temperature changes in ideal gases.
  • Appears in key equations such as \( PV = nRT \), linking pressure, volume, temperature, and mole count.
In calculations involving the internal energy change, \( R \) ensures the units of energy align correctly with moles and temperature, helping us find \( \Delta T \)in the exercise.

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

Suppose that 31.4 \(\mathrm{J}\) of heat is added to an ideal gas. The gas expands at a constant pressure of \(1.40 \times 10^{4}\) Pa while changing its volume from \(3.00 \times 10^{-4}\) to \(8.00 \times 10^{-4} \mathrm{m}^{3}\) . The gas is not monatomic, so the relation \(C_{P}=\frac{5}{2} R\) does not apply. (a) Determine the change in the internal energy of the gas. (b) Calculate its molar specific heat capacity \(C_{p} .\)

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partition. Both sides contain one mole of a monatomic ideal gas \(\left(\gamma=\frac{5}{3}\right),\) with the initial temperature being 525 \(\mathrm{K}\) on the left and 275 \(\mathrm{K}\) on the right. The partition is then allowed to move slowly (i.e., quasi- statically) to the right, until the pressures on each side of the partition are same. Find the final temperatures on the (a) left and (b) right.

One mole of neon, a monatomic gas, starts out at conditions of standard temperature and pressure. The gas is heated at constant volume until its pressure is tripled, then further heated at constant pressure until its volume is doubled. Assume that neon behaves as an ideal gas. For the entire process, find the heat added to the gas.

Multiple-Concept Example 6 deals with the same concepts as this problem does. What is the efficiency of a heat engine that uses an input heat of \(5.6 \times 10^{4} \mathrm{J}\) and rejects \(1.8 \times 10^{4} \mathrm{J}\) of heat?

Heat engines take input energy in the form of heat, use some of that energy to do work, and exhaust the remainder. Similarly, a person can be viewed as a heat engine that takes an input of internal energy, uses some of it to do work, and gives off the rest as heat. Suppose that a trained athlete can function as a heat engine with an efficiency of 0.11 . (a) What is the magnitude of the internal energy that the athlete uses in order to do \(5.1 \times 10^{4} \mathrm{J}\) of work? (b) Determine the magnitude of the heat the athlete gives off.
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Particle Model of Matter

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electrostatics

Read Explanation

Nuclear Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
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.