SOLUTIONS MANUAL
THE THERMODYNAMICS OF
PHASE AND REACTION EQUILIBRIA
Second Edition
Ismail Tosun
, CHAPTER 2
Problem 2.2
The schematic diagram is given below:
P
T P
For a throtting process e = 0. Therefore, integration of Eq. (2.3-12) leads to
Z 2 Z 2" Ã
e
! #
0= e + e (1)
400 1
The given equation of state can be rearranged as
e = + (2)
Therefore à ! à !
e e
= e = (3)
The use of Eq. (3) in Eq. (1) yields
Z 2
e = ( 1 2) (4)
400
Substitution of the numerical values
Z 2 h i
5
(25 895 + 0 033 ) = 3 73 × 10 (5 0 1) × 106
400
From the MATHCAD worksheet given below 2 = 404 7 K.
Mathcad worksheet of Problem 2.2.
3
THE THERMODYNAMICS OF
PHASE AND REACTION EQUILIBRIA
Second Edition
Ismail Tosun
, CHAPTER 2
Problem 2.2
The schematic diagram is given below:
P
T P
For a throtting process e = 0. Therefore, integration of Eq. (2.3-12) leads to
Z 2 Z 2" Ã
e
! #
0= e + e (1)
400 1
The given equation of state can be rearranged as
e = + (2)
Therefore à ! à !
e e
= e = (3)
The use of Eq. (3) in Eq. (1) yields
Z 2
e = ( 1 2) (4)
400
Substitution of the numerical values
Z 2 h i
5
(25 895 + 0 033 ) = 3 73 × 10 (5 0 1) × 106
400
From the MATHCAD worksheet given below 2 = 404 7 K.
Mathcad worksheet of Problem 2.2.
3
