Chemical Thermodynamics
2013/2014
4th Lecture: Manipulations of the 1st Law and Adiabatic Changes
Valentim M B Nunes, UD de Engenharia
Relations between partial derivatives
Partial derivatives have many useful properties, and we can use it to
manipulate the functions related with the first Law to obtain very
useful thermodynamic relations. Let us recall some of those properties.
If f is a function of x and y, f = f(x,y), then
 f 
 f 
df    dx    dy
 x  y
 y  x
If z is a variable on which x and y depend, then
 f   f   f   y 
         
 x  z  x  y  y  x  x  z
2
Relations between partial derivatives
The Inverter:
 x 
1
  
 y  z  y 
 x  z
The Permuter:
Euler’s chain relation:
 x 
 x   z 
      
 z  y  y  x
 y  z
 x   y   z 
       1
 y  z  z  x  x  y
Finally, the differential df = g dx + h dy is exact, if:
 g   h 
    
 y  x  x  y
3
Changes in Internal Energy
Recall that U = U(T,V). So when T and V change infinitesimally
 U 
 U 
dU  
 dV  
 dT
 V T
 T V
The partial derivatives have already a physical meaning (remember
last lecture), so:
dU   T dV  CV dT
4
Change of U with T at constant pressure
What does
this mean?
Using the relation of slide 2 we can writhe:
 U   U   U   V 
 U 
 V 

 
 
 
 or 
  CV   T 

 T  p  T V  V T  T  p
 T  p
 T  p
We define the isobaric thermal expansion coefficient as
1  V 
p   
V  T  p
Finally we obtain:
 U 

  CV   p TV
 T  p
= 0 for an
ideal gas
Proofs relation between Cp and Cv for an
ideal gas!
Closed system
at constant
pressure and
fixed
composition!
5
Change of H with T at constant volume
Let us choose H = H(T,p). This implies that
 H 
 H 
 H 
 dp or dH  C p dT  
 dp
dH  
 dT  
 T  p
 p T
 p T
Now we will divide everything by dT, and impose constant volume
 H   p 
 H 

  

C



p
 T V
 p T  T V
What is the meaning of this
two partial derivatives?
6
Change of H with T at constant volume
Using the Euler’s relation
Rearranging
 p   T   V 
  1
  
 
 T V  V  p  p T
 V 


 pV
 T  p
 p 

  
 V 
 V 
 T V





P

p

T

T
What does
this mean?
We define now the isothermal compressibility coefficient
1  V 

kT   
V  p T
To assure that
kT is positive!
So, we find that
p
 p 
  
 T V kT
7
Change of H with T at constant volume
Using again the Euler’s relation and rearranging
 H 
 T   H 
1

  
   

p  H  T  p
 p   T 
 p T

  

 T  H  H  p
or
 H 
 T 

    C p
 p T
 p  H
We finally obtain
  p  JT
 H 

  1 
kT
 T V 
What is this? See
next slide! For
now we will call it
µJT

C p

8
The Joule-Thomson Expansion
Consider the fast expansion of a gas trough a throttle:
W  piVi  p f V f
If Q = 0 (adiabatic) then
U f Ui  piVi  p f V f or U f  p f V f  Ui  piVi
So, by the definition of enthalpy
Isenthalpic process!
H f  Hi
 JT
 T 
  
 p  H
9
The Joule-Thomson effect
For an ideal gas, µJT = 0. For most real gases Tinv >> 300 K. If
µJT >0 the gas cools upon expansion (refrigerators). If µJT <0 then
the gas heats up upon expansion.
10
Adiabatic expansion of a perfect gas
From the 1st Law, dU = dq + dw. For an adiabatic process dU = dw
and dU = CvdT, so for any expansion (or compression):
W  CV T
For an irreversible process, against constant pressure:
W   pextV  CV T
pextV
T  
CV
The gas cools!
11
Adiabatic expansion of a perfect gas
For a reversible process, CVdT = -pdV along the path. Now, per
mole, for an ideal gas, PV = RT, so
CV dT
RdV

T
V
Tf
Vf
Tf
Vf
1
1
CV  dT   R  dV  CV ln
  R ln
T
V
Ti
Vi
Ti
Vi
 Tf
ln
 Ti



CV / R
T f  Vi 
Vi
 ln

 
Vf
Ti  V f 
R / CV
For an ideal gas, Cp-Cv = R, and introducing   C p / CV then
 Vi
Tf  
V
 f
 1

 Ti


The gas cools!
12
Adiabatic expansion of a perfect gas
We can now obtain an equivalent equation in terms of the pressure:
piVi
nRTi
piVi  V f 


  
p f V f nRTf
p f V f  Vi 

R / CV

piVi  p f V f
As a conclusion,
pV
 is constant along a reversible adiabatic.
For instance, for a monoatomic ideal gas,
  5/3
13
Adiabatic vs isothermal expansion
14