Under constant pressure P = 18 bar an ideal gas of 1.5 mol undergoes expansion from 1.2L to 23.5L, and Cv,m 20.79J/mol/K. Please calculate q, w, AU, and AH Instruction: = (1) all units are KJ (not J); (2) Please round to 1 decimal place (not E notation). For example: 23.010 should be entered as 23.0; If the answer is negative, please add the negative sign. If answer is zero, please just enter 0 without decimal. (1) w= (2) q= (3) AU= (4) ΔΗ=
Under constant pressure P = 18 bar an ideal gas of 1.5 mol undergoes expansion from 1.2L to 23.5L, and Cv,m 20.79J/mol/K. Please calculate q, w, AU, and AH Instruction: = (1) all units are KJ (not J); (2) Please round to 1 decimal place (not E notation). For example: 23.010 should be entered as 23.0; If the answer is negative, please add the negative sign. If answer is zero, please just enter 0 without decimal. (1) w= (2) q= (3) AU= (4) ΔΗ=
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
Related questions
Question
100%
can u help, the 2nd pic is an example of how to solve
Transcribed Image Text:QUESTION 1
Under constant pressure P = 18 bar an ideal gas of 1.5 mol undergoes expansion from 1.2L to 23.5L, and CV,m
20.79J/mol/K. Please calculate q, w, AU, and AH
Instruction:
=
(1) all units are KJ (not J);
(2) Please round to 1 decimal place (not E notation). For example: 23.010 should be entered as 23.0; If the answer is
negative, please add the negative sign. If answer is zero, please just enter 0 without decimal.
(1) W=
(2) q=
(3) AU=
(4) ΔΗ=
Transcribed Image Text:Change in internal energy during the process is calculated as follows:
AU = nC, (T2 - T₁)
= 1.5 mol xx8.314
= - - 1683.585 J
AU (J) = 1.68E3
Step 3
For ideal gas,
Cp-C₁ = R
-
mol.K
(303 - 393) K
Cp=5R/2 (R+3R/2)
Change in enthalpy is calculated as follows:
AH = nCp (T₂ - T₁)
= 1.5 mol xx 8.314 (303 - 393) K
= -2805.975 J
mol.K
AH (J) = -2.81E3
For constant pressure process, the heat transfer is equal to the change in internal energy.
Thus,
q (J) = -2.81E3
Step 4
Work done during the process is calculated applying firs law of thermodynamics as follows:
9 -w = AU
-2.81 E3w= - 1.68 E3
-w = 1.13E3
w (J) = -1.13E3
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps
Recommended textbooks for you
Introduction to Chemical Engineering Thermodynami…
Chemical Engineering
ISBN:
9781259696527
Author:
J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:
McGraw-Hill Education
Elementary Principles of Chemical Processes, Bind…
Chemical Engineering
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
Elements of Chemical Reaction Engineering (5th Ed…
Chemical Engineering
ISBN:
9780133887518
Author:
H. Scott Fogler
Publisher:
Prentice Hall
Introduction to Chemical Engineering Thermodynami…
Chemical Engineering
ISBN:
9781259696527
Author:
J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:
McGraw-Hill Education
Elementary Principles of Chemical Processes, Bind…
Chemical Engineering
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
Elements of Chemical Reaction Engineering (5th Ed…
Chemical Engineering
ISBN:
9780133887518
Author:
H. Scott Fogler
Publisher:
Prentice Hall
Industrial Plastics: Theory and Applications
Chemical Engineering
ISBN:
9781285061238
Author:
Lokensgard, Erik
Publisher:
Delmar Cengage Learning
Unit Operations of Chemical Engineering
Chemical Engineering
ISBN:
9780072848236
Author:
Warren McCabe, Julian C. Smith, Peter Harriott
Publisher:
McGraw-Hill Companies, The