University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
IDEAL GAS
The molecules making up a gas become more and more widely separated as
pressure is decreased, and the volume of the molecules themselves becomes a
smaller and smaller fraction of the total volume occupied by the gas. Furthermore,
the forces of attraction between molecules become ever smaller because of the
increasing distances between them. In the limit, as the pressure approaches zero,
the molecules are separated by infinite distances. Their volumes become negligible
compared with the total volume of the gas, and the intermolecular forces approach
zero. At these conditions all gases are said to be ideal.
PV   RT
R  lim p0
P // pressure ( Nm-2)
V- // specific volume ( m3 mol-1 )
R// universal gas constant 8.314 Jmol-1K-1
T// temperature K
PV  ( PV )* 22711.6cm 3 .bar.mol 1


T
T
273.15
R may be expressed in various units commonly used values are given in App. A
R=8.314 J / mol. K
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
83.14 cm 3. bar/ mol . K
0.082 lit .atm /mol . K
1.987 cal/mol. K
10.73 ft 3 .psi / mol . R
CLOSED SYSTEM PROCESS
1 – Constant volume process (isometric) or isochoric
T1 P1

T2 P2
U  Q  W
W 0
∆U = Q = ∫ CV dT
2- Constant pressure process (isobaric)
T1 V1

T2 V2
W  PV
U  Q  W  Q  PV
Q  H   CPdT
Because U = f (T) for ideal gas both H and CP also depend on temperature alone.
This is evident from the definition H = U + PV or H = U + RT
dH = d U + R d T
CP d T = CV d T + R d T
CP = CV + R
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
This equation does not imply that CP and CV are themselves constant for an ideal
gas , but only that they vary with temperature in such a way that their difference is
equal to the constant R.
3- Constant temperature process (isothermal)
P1 V2

P2 V1
U  Q  W
2
W   PdV
1
2
U  Q   PdV
1
for perfect gas U  0
W  Q  nRT ln
V2
P
 nRT ln 1
V1
P2
4- Adiabatic process (isotropic): is one for which there is no heat transfer between
system and surroundings
U  Q  W
Q  0  U  W
U  CV (T2  T1 ) for perfect gas
dU =-dW
CV d T = - p dV = 
dT
R dV

T
CV V
RT
dV
V
*
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
H  U  PV  U  RT
dH  dU  RdT
C P dT  Cv dT  RT
CP = CV + R
CP
R
 1
CV
CV
  1
R
R
  1
Or
CV
CV
We can write equation * as below formula:
If ɤ is constant, so integration yields: ln
Or
T2  V1 
 
T1  V2 
dT
dV
 (  1)
T
V
T2
V
 (  1) ln 2
T1
V1
 1
**
This equation relates temperature and volume for mechanically reversible adiabatic
process involving an ideal gas with constant heat capacities.
P1V1 P2V2
V1 P2T1



For ideal gas
T1
T2
V2 P1T2
 V1 
 
 V2 
 1
 PT 
  2 1 
 P1T2 
T2  P2T1 


T1  P1T2 
 1
 1
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
T2  P2 

T1  P1 
11
 T2 
 
 T1 
 1
 T1 
 
 T2 
P 
  2 
 P1 
 1
 1
T   P 
  2    2 
 T1   P1 
 1

***
By equating equations (**) and (***)
T2  V1 
 
T1  V2 
V 
  1 
 V2 
 1
 1
P 
  2 
 P1 
P 
  2 
 P1 
 1

 1




Or P1V1  P2V2  PV  Const.
Work for adiabatic process
-dW = dU = CV dT
W = - ∆U = - CV ∆T
W  CV T 
W
 RT RT1  RT2

 1
 1
P1V1  P2V2
 1
(Because RT1 = P1 V1 and RT 2 = P2 V2)
In case, V2 is not known it can be eliminated from last equation
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
 1
 1





P1V1   P2 
RT1   P2   

W
1  

1  
   1   P1 

  1   P1 




The above equations are for ideal gas with constant heat capacities for the
mechanically reversible as well as adiabatic process, The process which are
adiabatic but not reversible are not describes by these equations.
5- Polytropic process
This process which is not isothermal or adiabatic.
Q  U  W
W 
P2V2  P1V1 
1 n
RT2  T1 
W 
for perfect gas
1 n
It is general case and there is not specific conditions other than reversibility. For
one mole , there are :
dU = dQ –dW
∆ U = Q – W (first law )
dW = PdV
W=∫ p dV
dU = CV dT
∆U= ∫ CV dT
dH= CP dT
∆H=∫ CP dT
Q cannot be determined directly, but must be obtained through first law
dQ = CV dT +PdV
Q = ∫ CV dT + ∫ p dV
Work must calculated directly from integral pdV.
These equations has been derived for mechanically reversible , non-flow process.
The work for irreversible process is calculated by two steps procedure by :
1- W is determined for a mechanically reversible process accomplishes the
same change of state .
2- The is result is multiplied or divided by an efficiency
a. W irr = W rev * efficiency ( turbine - expansion )
b. W irr = W rev / efficiency ( compressor – compression )