One mole of a monatomic ideal gas the finished graph is the graph from part a StateP (atm)V (L)т (к)1216.8723.6216.872.05173.0943.594173.0911.422275.4357 1. One mole of a monatomic ideal gas is held initially at a pressure of 6 atm and 3 L. The gasundergoes isothermal expansion to 5 L followed by adiabatic expansion to 7 L. The gas is thenisothermally compressed to back to 4 L and adiabatically compressed back to 2 L.b. Complete the table belowProcessQ (kJ)w (kJ)Δυ (kJ)AH (kJ)AS (J/K)10 22 0 33 0 44 0 1C.Calculate the maximum efficiency for this cycle.

a)Step 1-2 →isothermal expansionStep 2-3→adiabatic expansionStep 3-4→isothermal compressionStep…

Answered: ndergoes isothermal expansion to 5 L…

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

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Question

One mole of a monatomic ideal gas
the finished graph is the graph from part a

State
P (atm)
V (L)
т (к)
1
216.87
2
3.6
216.87
2.05
173.09
4
3.59
4
173.09
11.42
2
275.4
357

1. One mole of a monatomic ideal gas is held initially at a pressure of 6 atm and 3 L. The gas
undergoes isothermal expansion to 5 L followed by adiabatic expansion to 7 L. The gas is then
isothermally compressed to back to 4 L and adiabatically compressed back to 2 L.
b. Complete the table below
Process
Q (kJ)
w (kJ)
Δυ (kJ)
AH (kJ)
AS (J/K)
10 2
2 0 3
3 0 4
4 0 1
C.
Calculate the maximum efficiency for this cycle.

Expert Solution

Step 1

$a) S t e p 1 - 2 \to i s o t h e r m a l e x p a n s i o n S t e p 2 - 3 \to a d i a b a t i c e x p a n s i o n S t e p 3 - 4 \to i s o t h e r m a l c o m p r e s s i o n S t e p 4 - 5 \to a d i a b a t i c c o m p r e s s i o n$

$T h e f o l l o w i m g r e l a t i o n s c a n b e u s e d f o r t a b u l a t i n g t h e v a l u e s I s o t h e r m a l e x p a n s i o n / c o m p r e s s i o n (i s e n t h a l p i c p r o c e s s) Q = - W W = - n R T \ln \frac{V_{f i n a l}}{V_{i n i t i a l}} ∆ H = 0 ∆ U = 0 ∆ S = n R \ln \frac{P_{i n i t i a l}}{P_{f i n a l}} w h e r e n i s n u m b e r o f m o l e s o f m o n o a t o m i c i d e a l g a s R i s g a s c o n s \tan t = 8.314 J / K . m o l F o r a d i a b a t i c e x p a n s i o n / c o m p r e s s i o n (i s e n t r o p i c p r o c e s s) Q = 0 W = \frac{K ({V_{f i n a l}}^{1 - γ} - {V_{i n i t i a l}}^{1 - γ})}{1 - γ} w h e r e λ = \frac{C_{p}}{C_{V}} \approx 1.66 f o r m o n o a t o m i c g a s e s K = P V^{λ} ∆ H = 0 ∆ U = - W ∆ S = 0$

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