/*! 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 35 The temperature of \(2.5 \mathrm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 35

The temperature of \(2.5 \mathrm{mol}\) of a monatomic ideal gas is \(350 \mathrm{K}\). The internal energy of this gas is doubled by the addition of heat. How much heat is needed when it is added at (a) constant volume and (b) constant pressure?

Short Answer

Expert verified
(a) Approximately 10,926 Joules of heat are needed at constant volume. (b) Approximately 18,210 Joules of heat are needed at constant pressure.

Step by step solution

01

Recall the formula for internal energy of a monatomic ideal gas

The internal energy \( U \) of a monatomic ideal gas is given by \( U = \frac{3}{2}nRT \), where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.
02

Calculate the initial internal energy

Using \( n = 2.5 \) mol, \( R = 8.314 \, \text{J/mol路K} \), and \( T = 350 \text{K} \), the initial internal energy is \( U_i = \frac{3}{2} \times 2.5 \times 8.314 \times 350 \). Calculate \( U_i \).
03

Calculate the final internal energy

Since the internal energy is doubled, the final internal energy \( U_f = 2 \times U_i \).
04

Step 4a: Determine heat added at constant volume

At constant volume, the change in internal energy \( \Delta U \) is equal to the heat added, \( Q_v \). Therefore, \( Q_v = U_f - U_i \). Calculate \( Q_v \).
05

Step 4b: Determine heat added at constant pressure

At constant pressure, the heat added is \( Q_p = \Delta U + \Delta nRT \). However, since the gas stays the same \( \Delta n = 0 \), hence, \( Q_p = U_f - U_i + nR \Delta T \). Find \( \Delta T = \frac{\Delta U}{\frac{3}{2}nR} \) and calculate \( Q_p = Q_v + nR \Delta T \).
06

Calculate \( Q_p \) with approximate values

With the calculated \( \Delta T \) from the increase in temperature, substitute it into \( Q_p = Q_v + nR \Delta T \) to find the heat added at constant pressure.

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.

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in thermodynamics. It describes the behavior of an ideal gas in terms of its pressure, volume, temperature, and the number of moles. The law is given by the equation \( PV = nRT \), where:
  • \( P \) is the pressure of the gas,
  • \( V \) is the volume,
  • \( n \) is the number of moles,
  • \( R \) is the universal gas constant (8.314 J/mol路K), and
  • \( T \) is the temperature in Kelvin.
The Ideal Gas Law is widely used because it approximates the behavior of real gases under many conditions. It holds true when gases are not at extremely high pressures or low temperatures, where interactions between particles become significant. This law is helpful in calculating any one thermodynamic variable if the other three are known.
Internal Energy
Internal energy is a key concept in thermodynamics, especially for ideal gases. It refers to the total energy contained within a system due to the kinetic energy of its particles.
  • For a monatomic ideal gas, the internal energy \( U \) is given by the formula: \( U = \frac{3}{2}nRT \).This is derived from the kinetic theory of gases.
  • The equation shows that internal energy is directly proportional to temperature; hence, it increases if the temperature increases.
Because the internal energy depends solely on temperature and the number of moles, it can be doubled by adding heat while keeping this relationship intact. Modifying internal energy is crucial in processes like compression, expansion, and thermal heating of gases.
Heat Transfer
Heat transfer is the process of energy transfer from one body or system to another due to temperature difference. In thermodynamics, heat transfer can change the internal energy of a gas without requiring any work.
  • Heat absorbed or released is often measured in joules or calories.
  • The change in internal energy \( \Delta U \) of the gas is tied to the heat addition when no work is done, especially in ideal gases at constant volume.
  • At constant volume, the heat added equals the change in internal energy, \( Q_v = \Delta U \).At constant pressure, the situation involves both the change in internal energy and work done on or by the system.
This principle is central to understanding how systems exchange thermal energy and adjust to temperature changes.
Monatomic Gas
A monatomic gas consists of single atoms, like helium or argon. These gases have unique properties when it comes to thermodynamics:
  • This type of gas exhibits 3 degrees of freedom corresponding to translational motion in three-dimensional space.
  • Its internal energy formula, \( U = \frac{3}{2}nRT \), results directly from kinetic energy associated with these degrees of freedom.This is simpler than that for polyatomic gases, which have additional rotational and possibly vibrational modes.
Monatomic gases follow the Ideal Gas Law quite closely because interactions between non-bonded atoms are minimal.The simplicity of these gases makes them an excellent model for understanding basic thermodynamic principles.
Constant Volume
In thermodynamic processes, keeping the volume constant has significant implications. Here's why:
  • When the volume is constant, there is no work done by or on the gas as there is no volume change.
  • The entire heat added is used to change the internal energy hence \( Q_v = \Delta U \).
  • This means any heat transfer directly relates to the temperature change.Therefore, calculations are somewhat more straightforward than for processes at constant pressure.
Constant volume scenarios are vital in technical applications like combustion engines, where cylinders are closed off and the explosive expansion increases internal energy.
Constant Pressure
Experiments and processes at constant pressure hold key insights into gas behavior.
  • At constant pressure, when heat is added, the gas does work by expanding.As a result, you must account for the work done when calculating heat transfer.
  • The formula becomes \( Q_p = \Delta U + P\Delta V \).Since \( P\Delta V = nR\Delta T \), where \( \Delta T \) is the change in temperature, it simplifies to \( Q_p = Q_v + nR\Delta T \).
  • This means calculations take into account temperature change as well as the displacement work done by the gas.Such processes are common in industrial applications such as turbines and reactors.
Understanding constant pressure conditions is essential as they closely mimic everyday environmental experiences, where pressure tends to be more stable than volume.

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 monatomic ideal gas has an initial temperature of \(405 \mathrm{K}\). This gas expands and does the same amount of work whether the expansion is adiabatic or isothermal. When the expansion is adiabatic, the final temperature of the gas is \(245 \mathrm{K}\). What is the ratio of the final to the initial volume when the expansion is isothermal?

Suppose that the gasoline in a car engine burns at \(631{ }^{\circ} \mathrm{C},\) while the exhaust temperature (the temperature of the cold reservoir) is \(139^{\circ} \mathrm{C}\) and the outdoor temperature is \(27^{\circ} \mathrm{C}\). Assume that the engine can be treated as a Carnot engine (a gross oversimplification). In an attempt to increase mileage performance, an inventor builds a second engine that functions between the exhaust and outdoor temperatures and uses the exhaust heat to produce additional work. Assume that the inventor's engine can also be treated as a Carnot engine. Determine the ratio of the total work produced by both engines to that produced by the first engine alone.

How long would a \(3.00-\mathrm{kW}\) space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coefficient of performance \(=3.00\) ) does when it freezes \(1.50 \mathrm{kg}\) of water at \(20.0^{\circ} \mathrm{C}\) into ice at \(0.0^{\circ} \mathrm{C} ?\)

The sublimation of zinc (mass per mole \(=0.0654 \mathrm{kg} / \mathrm{mol}\) ) takes place at a temperature of \(6.00 \times 10^{2} \mathrm{K},\) and the latent heat of sublimation is \(1.99 \times 10^{6} \mathrm{J} / \mathrm{kg}\). The pressure remains constant during the sublimation. Assume that the zinc vapor can be treated as a monatomic ideal gas and that the volume of solid zinc is negligible compared to the corresponding vapor. Concepts: (i) What is sublimation, and what is the latent heat of sublimation? (ii) When a solid phase changes to a gas phase, does the volume of the material increase or decrease, and by how much? (iii) As the material changes from a solid to a gas, does it do work on the environment, or does the environment do work on it? How much work is involved? (iv) In this problem we begin with heat \(Q\) and realize that it is used for two purposes: First, it makes the solid change into a gas, which entails a change \(\Delta U\) in the internal energy of the material, \(\Delta U=U_{\mathrm{gas}}-U_{\mathrm{solid}} .\) Second, it allows the expanding material to do work \(W\) on the environment. According to the conservation-of-energy principle, how is \(Q\) related to \(\Delta U\) and \(W ?\) (v) According to the first law of thermodynamics, how is \(Q\) related to \(\Delta U\) and \(W\) ? Calculations: What is the change in the internal energy of zinc when \(1.50 \mathrm{kg}\) of zinc sublimates?

Find the change in entropy of the \(\mathrm{H}_{2} \mathrm{O}\) molecules when (a) three kilograms of ice melts into water at \(273 \mathrm{K}\) and \((\mathrm{b})\) three kilograms of water changes into steam at \(373 \mathrm{K}\). (c) On the basis of the answers to parts (a) and (b), discuss which change creates more disorder in the collection of \(\mathrm{H}_{2} \mathrm{O}\) molecules.
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Oscillations

Read Explanation

Measurements

Read Explanation

Electrostatics

Read Explanation

Physical Quantities And Units

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.