×
Log in to StudySoup
Get Full Access to Physics - Textbook Survival Guide
Join StudySoup for FREE
Get Full Access to Physics - Textbook Survival Guide

What is the thermal efficiency of an engine that operates

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 46P Chapter 20

University Physics | 13th Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

4 5 1 397 Reviews
19
0
Problem 46P

What is the thermal efficiency of an engine that operates by taking n moles of diatomic ideal gas through the cycle shown in F? ig. P20.38??

Step-by-Step Solution:

Solution 46P Step 1: The process, 1 2 is an isochoric process. Therefore, the work done by the system will be zero. According to ideal gas equation, at point 1, P1 1nRT 1 Where, “n” is the number of moles of the gas P - Pressure at point 1 1 V 1Volume at point 1 R - Universal gas constant T 1Temperature at point 1 T = P V /nR 1 1 1 From the graph, we can see that, P = P and V = V 1 0 1 0 Therefore, T = P V1/nR 0 0 Similarly, T = 2P2 /nR 0 0 The heat change in the system, Q = nC T V Where, C - SpeVfic heat capacity at constant volume T - Change in temperature of the system In process 1 2, T = T - T = 2P V /n2- P 1 /nR = P0V0nR 0 0 0 0 For a diatomic gas, C = 5/2 R V Therefore, Q = n×(5/2)R×P V /nR = 5P V 02 0 0 0 Step 2: The process, 2 3 is an isobaric process. Therefore, the work done by the system will not be zero. Work done, W = PdV Where, dV - Change in volume dV = (V - V ) = ( 2V - V ) = V 3 2 0 0 0 Therefore, W = 2P V 0 0 Step 3: The process, 3 4 is an isochoric process. Therefore, the work done by the system will be zero. According to ideal gas equation, at point 3, P3 3nRT 3 T3 P V3nR3 From the graph, we can see that, P = 2P and V = 2V3 0 3 0 Therefore, T = 4P 3 /nR 0 0 Similarly, T = 2P V /nR 4 0 0 The heat change in the system, Q = nC T V Where, C - Specific heat capacity at constant volume V T - Change in temperature of the system In process 3 4, T = T - T = 2P V /nR - 4P V /nR = - 2P V /nR 3 4 0 0 0 0 0 0 For a diatomic gas, C = 5/2 R V Therefore, Q = n×(5/2)R× -2P V /nR = - 5P V0 0 0 0

Step 4 of 5

Chapter 20, Problem 46P is Solved
Step 5 of 5

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The full step-by-step solution to problem: 46P from chapter: 20 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “What is the thermal efficiency of an engine that operates by taking n moles of diatomic ideal gas through the cycle shown in F? ig. P20.38??” is broken down into a number of easy to follow steps, and 26 words. This full solution covers the following key subjects: cycle, diatomic, efficiency, engine, gas. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 46P from 20 chapter was answered, more than 259 students have viewed the full step-by-step answer.

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

What is the thermal efficiency of an engine that operates

×
Log in to StudySoup
Get Full Access to Physics - Textbook Survival Guide
Join StudySoup for FREE
Get Full Access to Physics - Textbook Survival Guide
×
Reset your password