Physical Chemistry
Lecture 11
Reversibility and Work, Enthalpy
Mechanical work
Work on the system
w = −
∫P
ext
dV
path
Depends on how the external pressure is
varied in the process, i.e. it depends on the
path
Three experimentally known possibilities



P > Pext along the path: volume expands; w < 0
P = Pext, no change in the volume; w = 0
P < Pext, volume decreases; w > 0
Only ones seen
Volume never increases when Pext > P
Work and reversibility
Consider a process that
takes an ideal gas from
one state to another in
several different ways
As the number of steps
goes up, the work
approaches a constant
value
In the limit of infinite
number of steps, P=Pext
at every point in the
process
Reversibility
System remains at equilibrium with the
surroundings at every point in a process
A special, unique path
The system does maximum work in going
from the initial to the final state along a
reversible path
Can never be achieved with real processes
Convenient for calculation of state-function
changes, since these are independent of path
Heat transfer at constant
pressure
Chemists frequently carry out processes
at constant pressure, rather than at
constant volume
∆UP ≠ qP
Heat transfer is easy to measure
What is heat transfer at constant P?
Enthalpy – a new state
function
H, enthalpy, = U + PV
A state function because U, P and V are
Differential form
dH
= dU + PdV + VdP
= dqrev − PdV + PdV
= dqrev
+ VdP
+ VdP
At constant pressure
dH
= dqrev
At constant pressure, the reversible heat
transferred is the change in enthalpy
Differential of the enthalpy
In any process, one may evaluate the change
in a state function by integration
state 2
∆H
= H (T2 , P2 ) − H (T1 , P1 ) =
∫ dH
state1
Define the differential in terms of other
changes
dH
 ∂H 
= 
 dT
 ∂T  P
 ∂H 
+ 
 dP
 ∂P T
Must know the functional form of the partial
derivatives to do the integral
Derivatives of the enthalpy
Enthalpy change at constant pressure =
qP, therefore
 ∂H 


 ∂T  P
= CP
The other derivative is determined from
the equation of state
 ∂H 


 ∂P T
= V
 ∂V 
− T

 ∂T  P
Molar heat capacities, CPm, of
organic liquids at 298.15 K
Compound
CPm (J K-1 mol-1)
Compound
CPm (J K-1 mol-1)
n-Hexane
195.0
Benzene
135.6
n-Heptane
224.7
Toluene
156.1
n-Octane
254.0
Ethylbenzene
186.2
n-Nonane
284.5
p-Xylene
182.4
n-Decane
314.7
1,2,3-Trimethylbenzene
216.7
n-Undecane
345.2
1,2,4-Trimethylbenzene
215.9
n-Dodecane
375.3
1,3,5-Trimethylbenzene
209.6
n-Tridecane
406.7
n-Tetradecane
438.5
n-Pentadecane
469.4
n-Hexadecane
501.7
Data from the Handbook of Chemistry and Physics, 63rd Edition, CRC Press, Boca Raton, Florida
Relationship between the heat
capacities
Example of the use of calculus
Use definition of the enthalpy
H
= U
+ PV
Derivative with respect to temperature
 ∂V 
 ∂U 
P
+
= 



 ∂T  P
 ∂T  P
 ∂V 
 ∂P   ∂V 
= CV + T 

 − P
 
 ∂T  P
 ∂T V  ∂T  P
 ∂P   ∂V 
= CV + T 

 
 ∂T V  ∂T  P
 ∂H 


 ∂T  P
CP
 ∂V 
+ P

 ∂T  P
Calculation of enthalpy
changes in processes
Constant-pressure process
∆H
 ∂H 
= ∫
 dT
∂T  P
T1 
T2
=
T2
∫C
P
(T ) dT
T1
Isothermal process
∆H
=
 ∂H 
∫P  ∂P T dP =
1
P2

 ∂V  
∫P V − T  ∂T  P dP
1
P2
Summary
Calculation of state-function changes in
reversible processes, even if the actual
process is irreversible
State functions depend on two variables
Integrate the two components
separately and sum to obtain total
change
All changes calculated with a similar
formalism