Science
thermochemistry, engineering homework help

Final Answer

attached is my complete answer

Problem 1:

Solution:
An expression for the fugacity of the gas is derived by:
Pf



e 

V

 RT dP

Pi

1  aP1.5
dP
P
1

P



1

    a P  dP
P

1
P

P

2


  ln P  aP 3/2 
3

1

2
2

 

  ln P  aP 3/2    ln1  a 13/2 
3
3

 

2
 ln P  a  P 3/2  1
3
Now, for a  0.001 , P  72 atm , the fugacity will be approximately 1.5 times the pressure
because

2
e  ln P  a  P 3/2  1
3
2
 ln 72   0.001  723/2  1
3
 4.6833
   ln  4.6833  1.544  1.5
The work required to compress 1 mole of gas isothermally and reversely from Patmos  1 atm to

P  72 atm is calculated by:

V2

W   PdV
V1

 RT 1  aP1.5  

  Pd 
P1


P


P2
a 
 1
  RTP   2  0.5 
P1
2P 
 P
P2

P2

1

 RT  aP1.5  ln P 
3
 P1

 1
1.5
1.5
 1

  8.314  298    0.001 72   ln 72     0.0011  ln1  
 3

 3
 10092 J  10.1 kJ
Compared to ideal gas, the work is given by:

P
W  nRT ln  1 
 P2 
 1 atm 
 1 mole   8.314 J  K 1  mol 1   298 K  ln 

 72 atm 
 10596 J  10.6 kJ
If the critical pressure is 3.05 atm then using the generalized fugacity chart below

Hence, the critical temperature is estimated to be Tr  1.15 K .
Problem 2:

Solution:
Given the equation:

1
NO  Cl2  NOCl
2
With P  1 atm, T  1200 K
The composition of effluent stream is calculated by:
The combined rate = effluent + stream = 8500 + 1.5 = 8501.5 m^3 / sec

1/2

eq
eq
 PNO
  PCl2 
 o   o 
 P  P 
K p T  
eq
 PNOCl

 o 
 P 

1

 


 P 

2

1  o 
1
 4.00    P   4.00  

2  
2



 4.00    P

1  o
 4.00    P

2 

1/2

 
 P 
 o
 P 
 

1

 


 P 

2

1  o 
1
 4.00    P   4.00  

2  
2
K p T  


 4.00    P

1  o
 4.00    P

2 


 P
 o
P


1/2







1/2



P 


2  4.00   / 2  P o 


4.0...

Ace_Tutor (5145)
UCLA

Anonymous
Return customer, been using sp for a good two years now.

Anonymous
Thanks as always for the good work!

Anonymous
Excellent job

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