### Question Description

chemical engineering thermodynamics

### Unformatted Attachment Preview

## 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 0.0011 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...

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