Comparing Thermodynamic Processes. In a cylinder, 1.20 mol

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

Problem 68P Chapter 19

University Physics | 13th Edition

  • 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 0 379 Reviews
12
2
Problem 68P

Comparing Thermodynamic Processes. In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 X 105 Pa and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a pV-diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

Step-by-Step Solution:

Solution 68P a.) An isothermal process is a constant-temperature process. Step 1 of 2: Work done by an ideal monatomic gas in expanding isothermally from a volume of (V) to three times its volume(3V) initially at a pressure of 3.6 × 10 5 Pa and temperature of 300k can be computed as vf W = P × dV vi = n × R × T ×ln V [/V i ] = 1.20× 8.31 × 300 ×ln 3V[/V ] = 1.20 × 8.31 × 300 × ln 3 [ ] = 1.20× 8.31 × 300 × 1.098 W = 3286.6 J Step 2 of 2: For an ideal gas the internal energy is directly proportional to temperature, it follows that there is no change in internal energy of the gas during an isothermal process . The First law of thermodynamics then becomes U = 0 = Q W Q = W All the heat added to the system is used to do work. The value of heat transfer is equal to the amount of work done by the gas. Q = W = 3286.6 J Step 3 of 3: Since the gas is expanding isothermally,there is no change in the internal energy of the gas. U= 0 b.) An adiabatic process is defined as one with no heat transfer into or out of a system; Q=0. The First law of thermodynamics then becomes Q = U + W 0 = U + W U = W Work done by an ideal monatomic gas in expanding adiabatically from a volume of (V) to three times its volume(3V) initially at a pressure of 3.6 × 105 Pa and temperature of 300k can be computed as W = n × C × vT Where, n = number of moles of monatomic gas C v Molar heat capacity at constant volume dT = Change in temperature of the gas when it expands adiabatically (T T ) 1 2 Step 1 of 1: For an adiabatic process; 1 T × V = constant T 1 V 11 = T 2 V 21 T = T × (V /V ) 1 2 1 1 2 1.671 = 300 × (V /3 × V ) 0.67 = 300 × (1/3) = 300 × 0.478 T = 1 43.4 K 2 Step 2 of 2: Now work done by the gas can be computed as follows; W = n × C × vT = n × C ×v( T 1 ) 2 = 1.2 × 12.47 × (300 143.4) W = 2 343.36 J Step 3 of 3: Since the gas is expanding adiabatically ,there is no transfer of heat into or out of the system;Q= 0. First law of thermodynamics then becomes U = W The value of the heat transfer(Q) is equal to zero. Q = 0 The value of change in the internal energy of the gas is given by U = W = 2343.36 J c.) An isobaric process is a constant-pressure process. In general, none of the three quantities U,Q, and W is zero in an isobaric process. Work done by an ideal monatomic gas in expanding isobarically from a volume of (V) to three times its volume(3V) initially at a pressure of 3.6 × 105 Pa and temperature of 300k can be computed as Step 1 of 1: W= P × dV = n × R × dT Where, n = number of moles of monatomic gas R = universal gas constant dT = T 2 T 1 Step 2 of 2: For an isobaric process; V 1 / V 2 = T 1 T 2 T 2 = (V /2V ) 1 T 1 = (3V / V ) × 300 = 3 × 300 T = 9 00 K 2 Step 3 of 3: Now, work done by the gas can be computed as W = n × R × (T 2 T 1 = 1.2 × 8.31 × (900 300) = 1.2 × 8.31 × 600 W = 5 983.2 J

Step 4 of 5

Chapter 19, Problem 68P is Solved
Step 5 of 5

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

This full solution covers the following key subjects: gas, least, Case, greatest, absolute. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 68P from chapter: 19 was answered by Patricia, our top Physics solution expert on 05/06/17, 06:07PM. The answer to “Comparing Thermodynamic Processes. In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 X 105 Pa and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a pV-diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?” is broken down into a number of easy to follow steps, and 100 words. Since the solution to 68P from 19 chapter was answered, more than 673 students have viewed the full step-by-step answer. University Physics was written by Patricia and is associated to the ISBN: 9780321675460. This textbook survival guide was created for the textbook: University Physics, edition: 13.

×
Log in to StudySoup
Get Full Access to University Physics - 13 Edition - Chapter 19 - Problem 68p

Forgot password? Reset password here

Join StudySoup for FREE
Get Full Access to University Physics - 13 Edition - Chapter 19 - Problem 68p
Join with Email
Already have an account? Login here
Reset your password

I don't want to reset my password

Need help? Contact support

Need an Account? Is not associated with an account
Sign up
We're here to help

Having trouble accessing your account? Let us help you, contact support at +1(510) 944-1054 or support@studysoup.com

Got it, thanks!
Password Reset Request Sent An email has been sent to the email address associated to your account. Follow the link in the email to reset your password. If you're having trouble finding our email please check your spam folder
Got it, thanks!
Already have an Account? Is already in use
Log in
Incorrect Password The password used to log in with this account is incorrect
Try Again

Forgot password? Reset it here