University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
Heat effect
Heat capacity :it is amount of heat required to change a unit mass by a unit
temperature .Heat capacity of ideal gas depends on the type of the gas
1. For mono atom gas γ =1.66 ( Helium , Aaron )
2. For diatomic gases γ = 1.4( Co,N2 ,air)
3. For more than tri-atomic gases γ 1.3 (NH3, CH4 )
The higher molecular weight of gases , the lower value of γ
 U 

CV ( heat capacity at constant volume ) = 
 T V
 H 

CP (heat capacity at constant pressure ) = 
 T  P
The temperature dependence may be shown graphically but the value which get
from graph with less accurate
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
On the other hand temperature dependence usually given by an empirical equation;
the two simplest expressions of practical value are:
CP
C
   T  T 2 and , P  a  bT  cT 2
R
R
Where α , β and , γ and a , b , and c are constants characterized of particular gas ,
by combine the above two equations :
CP
D
 A  BT  CT 2  2
( 7-1 )
R
T
Where C or D is zero , depending on the gas considered .Value of A,B,C and D are
CP
given in table 4.1,since
dimensionless , the unit of C P is governed by choice
R
of R unit .equation (7-1) used for all gases as well as ideal gas.
More accurate but more complex equations are found in literatures.
CP = CV + R
CV
C
 P 1
R
R
CV
CP
, is readily found from equation
R
R
CP
CV
Effect of temperature on
, and
are determined by experiment, most often
R
R
from spectroscopic data and knowledge of molecular structure .
CP
How to use the
equation
R
T2
T2
D

H   C P dT   R  A  BT  CT 2  2 dT
T 
T1
T1 
B
C
1
1
 R[ A(T2  T1 )  (T22  T12 )  (T23  T13 )  D(  )
2
3
T2 T1
Thus the temperature dependence of
EX:
The molar heat capacity of methane in the ideal gas state is given in table 4.1as
CP
 1.702  9.081  10 3 T  2.164  10 6 T 2
R
CP
Where T in Kelvin , develop an equation for
for temperature in ○C
R
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
Solution
TK = t ○C +273.15
CP
 1.702  9.081  10 3 (t  273.15)  2.164  10 6 (t  273.15) 2
R
CP
 4.021  7.899  10 3 t  2.164  10 6 t 2
R
EX:
Calculate the heat required to raise the temperature of 1 mol of methane from 260
to 600 ○C in flow process at constant pressure approximately at 1 bar .
Solution
T 2873.15
T2
C
Q  H 
P
dT  R
T1
(
1.702  9.081  10 3 T  2.164  10 6 T 2 )dT
T 1533.15
Q  2378.8R  2378.8 * 8.314  19780 J
As a matter convenience , we define a mean heat capacity
T2
Cpmean 
 CpdT
T1
T2  T1
When equation (7-1) written by use mean heat capacity equation
Cpmean
c
D
2
 A  BTam  (4Tam
 T1T2 ) 
R
3
T1T2
( 7-2 )
Where Tam= (T1+T2) / 2 is the arithmetic mean temperature.
The general equation for all gases and ideal gas
EX:
Rework the last example by applying equation (7-2)
Tam = (533.15 +873.15) / 2 = 703.15
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
Cpmean
2.164  10 6
 1.702  9.081  10 3 * 703.15 
[4 * (703.15) 2
R
3
 (533.15) * (873.15)]  6.997
Q = ∆ H = 6.997*8.314*(873.15-533.15) = 19780 J
For calculation of T 2 in case given T 1 and Q , used try and error to find T 2 by
T2 
H
 T1
Cpmean
(7-3)
Assume value of T 2 for calculation Cp mean by use equation (7-2) substitution of
resulting value into equation (7-3) provides a new value of T 2 which reevaluate
Cpmean. Iteration continues to convergence on T 2 value.
Heat capacities of solid by used table 4.2 , while heat capacities of liquids from table
4.3.
Heat capacity for mixture
Cpmean   Cpi yi
 Cpmean( a ) y( a )  Cpmean(b ) y(b )  ....
Where : y mole fraction