Homework Help is Here – Start Your Trial Now!
Science
Physics
111. What is the change in internal energy of the gas? a. 2.56 x 104 J b. 1.28 x 105 J AV = ncpAT c. 6.82 x 10 JAU=(7)(20.785) (1172.6) d. 1.02 x 105 J = 170607,437 ? Cp² (+1) R = (2+1) 8.314 = 20.785 A AT = 1465.75 -293.15 AT-1172.

111. What is the change in internal energy of the gas? a. 2.56 x 104 J b. 1.28 x 105 J AV = ncpAT c. 6.82 x 10 JAU=(7)(20.785) (1172.6) d. 1.02 x 105 J = 170607,437 ? Cp² (+1) R = (2+1) 8.314 = 20.785 A AT = 1465.75 -293.15 AT-1172.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
See similar textbooks
icon
Related questions
Question

Having trouble with #111, only asking about that one. It refers to previous questions but those I have already answered. 

€ a. 2.74 x 10³ J b. 7.41 x 10³ J c. 4.39 x 10³ J d. 1.1 x 10³ J 109. How much work is done on the external environment by 7 moles of a monatomic ideal gas initially at 20°C that is ISOBARICALLY expanded at 1 atm to 5 times its original volume? a. 2.56 x 10^ J V₁=nRT₁ = 7x8.314 x 293,15= 0.16838 m³ b. 3.41 x 104 J 101325 c.) 6.82 x 104 J d. 0 J 110. What is the final temperature of the gas (in Celsius)? a. = 8.2.5 x 10 °C TB PB VB = 5TA 5 (293.15K) = 1465.75 -273.15 b. 2.98 x 10³ °C c. 1.67 x 10³ °C nR 1192.6°C d. 1.19 x 10³ °C W = P(VF-V₁) W = 101325 x (4X0-16838) W=P(5V₂-V₁) - P(4V₁) = 16.82×104 J 111.) What is the change in internal energy of the gas? a. 2.56 x 10¹ J AV = nCp AT b. 1.28 x 105 J c. 6.82 x 10¹J AU=(7)(20.785) (1172.6) d. 1.02 x 105 J = 170607 437 ? 112. How much heat is absorbed by the gas during the expansion? Calculate this in two ways: First, use your answers above for the change in internal energy and the work done. Secondly, use the specific heat at constant pressure and the change in temperature. Compare your results (to 3 sig figs) a. 2.56 x 104 J b. 6.82 x 104 J c. -1.02 x 105 J d. 1.71 x 105 J b. 1.83 x 104 J C.1.69 x 104 J d. 8.44 x 10³ J Cp = (2+1) R = (2+1) 8.314 = 20.785 tyd be 113. How much work is done on the external environment by 5 moles of a monatomic ideal gas at 20°C if it constant is ISOTHERMALLY expanded to 4 times its original volume? constant a. 0 J W = nRT x In ( 11 ) A AT = 1465.75/-293.15 AT-1172.6 = 5 x 8.314 x 293.15 x In (4 15 x In (4) = 16,885.07 J 114. What is the change in internal energy of the gas? AU -0 May 10, 2023, 6:49 PM
Transcribed Image Text:€ a. 2.74 x 10³ J b. 7.41 x 10³ J c. 4.39 x 10³ J d. 1.1 x 10³ J 109. How much work is done on the external environment by 7 moles of a monatomic ideal gas initially at 20°C that is ISOBARICALLY expanded at 1 atm to 5 times its original volume? a. 2.56 x 10^ J V₁=nRT₁ = 7x8.314 x 293,15= 0.16838 m³ b. 3.41 x 104 J 101325 c.) 6.82 x 104 J d. 0 J 110. What is the final temperature of the gas (in Celsius)? a. = 8.2.5 x 10 °C TB PB VB = 5TA 5 (293.15K) = 1465.75 -273.15 b. 2.98 x 10³ °C c. 1.67 x 10³ °C nR 1192.6°C d. 1.19 x 10³ °C W = P(VF-V₁) W = 101325 x (4X0-16838) W=P(5V₂-V₁) - P(4V₁) = 16.82×104 J 111.) What is the change in internal energy of the gas? a. 2.56 x 10¹ J AV = nCp AT b. 1.28 x 105 J c. 6.82 x 10¹J AU=(7)(20.785) (1172.6) d. 1.02 x 105 J = 170607 437 ? 112. How much heat is absorbed by the gas during the expansion? Calculate this in two ways: First, use your answers above for the change in internal energy and the work done. Secondly, use the specific heat at constant pressure and the change in temperature. Compare your results (to 3 sig figs) a. 2.56 x 104 J b. 6.82 x 104 J c. -1.02 x 105 J d. 1.71 x 105 J b. 1.83 x 104 J C.1.69 x 104 J d. 8.44 x 10³ J Cp = (2+1) R = (2+1) 8.314 = 20.785 tyd be 113. How much work is done on the external environment by 5 moles of a monatomic ideal gas at 20°C if it constant is ISOTHERMALLY expanded to 4 times its original volume? constant a. 0 J W = nRT x In ( 11 ) A AT = 1465.75/-293.15 AT-1172.6 = 5 x 8.314 x 293.15 x In (4 15 x In (4) = 16,885.07 J 114. What is the change in internal energy of the gas? AU -0 May 10, 2023, 6:49 PM
Expert Solution
steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Ideal Gas law
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON
SEE MORE TEXTBOOKS