Lecture 3: First Law of Thermodynamics and it’s
consequences
Summary of first two lectures
Today:
o Examples of (ir)reversible paths
o Calculation of heat changes and work involved along
various types of paths.
o Mechanical work
o Phase Transitions
o Chemical reactions
Type
State Variables
Open
Closed
Pressure
System
Intensive
Temperature
Density
Isolated
Extensive
Volume
Composition
ΔE,ΔH
System
Irreversible Paths
State A
State B
Reversible Paths
Equation of state
o
o
o
o
Isothermal
Isobaric
Isochoric
Adiabatic
Example 1 Evaporation of water
Path 1:
Irreversible path
P=1atm
n=1mole(l)
T=25
P=1atm
n= 1 mole (vapor)
T=25C
Reversibly raise
Temperature
Reversibly
cool.
P=1atm
n=1 mole(v)
T=100C
P=1atm
n= 1 mole(l)
T= 100C
Isothermal evaporation
Path 2: Vacuum Evaporation
P=1atm
n= 1 mole(l)
T=25
Irreversible path
Isothermally
Reduce Pressure
P=0.001 atm
n=1 mole(l)
T= 25 C
Isothermal evaporation
P=1atm
n= 1mole (vapor)
T=25C
Isothermally
increase P
P=0.001atm
n=1mole(vapor)
T=25C
Note the final result of irreversible evaporation is the
same. Whether we boil the water as in example 1, or
vacuum evaporate the water as in example 2, the final
state of the system is indistinguishable.
More details, general comments
We have seen from the above examples how we can
calculate the new state of a system when the state
variables change. The specific paths we choose such as
isothermal, isobaric etc are particularly useful for the
calculation of heat and work.
From extensive body of thermodynamic work we know
(1) The p-V work is negligible if the transformation
involves solids or liquid unless we are studying
systems under ultra-high pressure.
(2) In a given phase, e.g., liquid, gas, or solid, the
variation in specific heat (Cp or Cv) is essentially
negligible over wide ranges of temperature. The
only exceptions to this rule are phase transitions
such as melting or boiling, where specific heats
change dramatically.
(3) The p-V work is the most significant work for
gaseous systems. For a system of gas trapped in a
cylinder let us consider few specific paths
Calculation of p-V work
1. Isobaric
W= -P(V2-V1) …(Expansion)
W= +P(V2-V1) …(Compression)
P
T2
T1
V
2. Isochoric W=0 since there is no change in volume.
3. Isothermal reversible. In this case, to perform reversible
work we have to follow the equation of state
W    P.dV
P
Ideal Gas Law
PV  nRT V  P 
T1
V
W   nRT 
nRT
V
V 
dV
 nRT ln  2 
V
 V1 
For an isothermal
reversible path of an ideal gas the change in ΔE=0;q = -W
Phase Change
During a phase transition (liquid-vapor) changes in both
heat and work may occur. Normally, we study such
transition at constant pressure and temperature. (Think of
melting point measurement.) For liquid-liquid or solidliquid or solid-solid phase transition the pV work is
negligible, so ΔE≈ΔH. The energy goes to rearrange
molecular packing different phases. But if phase transition
involves gas, (i.e., boiling) there is considerable pV type of
work. Consider again path 1:
Irreversible path
P=1atm
n=1mole(l)
T=25
P=1atm
n= 1 mole (vapor)
T=25C
Reversibly raise
Temperature
P=1atm
n= 1 mole(l)
T= 100C
Reversibly
cool.
P=1atm
n=1 mole(v)
T=100C
Isothermal evaporation
For water during isothermal evaporation we must provide
heat to cause “boiling” this heat is provided by the
surroundings to the system. It is called as latent heat, qL. It
is about 540calories/gm. This heat is necessary for water
molecules to overcome hydrogen bonding forces that hold
them in the liquid phase.
Details of calculation
We can calculate the work and heat, and hence the
energy changes that occur in any irreversible process by
breaking it down in terms of smaller reversible steps.
Consider again path 1.
Step 1: it is isobaric, and isochoric (volume change is
negligible). So the pV work done is 0.
ΔE1=q1=CLp(T2-T1), ΔH1=ΔE1.
Step 2: This requires us to provide latent heat of
evaporation and also it involves a massive change in
volume; so the pV work is also involved.
ΔE2=qL +W= qL –P(VV-VL); ΔH=ΔE2+Δ(PV)= qL
Step 3: Involves isobaric cooling. Only the volume of
gas decreases as the vapor is cooled.
ΔE3=q3 +W=CVp(T1-T2)-P(VT2-VT1) ;
ΔH3=ΔE3+Δ(PV)=q3
This shows how an irreversible process can be broken
down in terms of series of reversible processes and
hence we can calculate associated changes in the state
variables such as energy and enthalpy.
Adding up the total energies
As we have shown before total energy and enthalpy
changes are:
ET  E1  E2  E3
& H T  H1  H 2  H 3
ET  (C PL  C PV )(T2  T1 )  qL  P (VTV2  VTL2 )
ET  (C PL  C PV )(T2  T1 )  qL  RT
H T  (C PL  C PV )(T2  T1 )  qL
Note at constant pressure qL  H (T2 ) so
H T  (CPL  CPV )(T2  T1 )  H (T2 )
General Rules.
o Calculation of ΔH at arbitrary temperature T2 if the
it’s value is known at one temperature.
H (T2 )  H (T1 )  c p (T2  T1 )
o Calculation of ΔH under isobaric conditions
H  E  PV
H  E   ( PV )  E  PV  PV
but P  0
H  E  VP  qP  W  PV  qP  PV  PV
H  qP   cP dT  cP (T2  T1 )
o Calculation of ΔH under isochoric conditions
H  E  PV
H  E  ( PV )  E  R(T2  T1 )
H  qV  R(T2  T1 )   cV dT  R(T2  T1 )
H  (cV  R)(T2  T1 )
E  cV (T2  T1 )
o Calculation under isothermal condition
H  E   ( PV )
 0   ( RT )  0
Changes involved in chemical reactions
Consider a general reaction
n A A  n B B  nC C  n D D
H  H Pr oducts  H Re ac tan ts
H  nC H  n D H  n A H  n B H
o Heat effects depend on whether the reaction takes
place under conditions of PV work or not. If only
work involved is PV type, then ΔE=qV, ΔH=qP=
ΔE+Δ(PV). In general for gaseous reactant we
assume ideal gas law so: ΔH=ΔE+Δ(PV) becomes
ΔH= ΔE+Δ(nRT)
o If the reaction takes place with accompanying rise
in temperature or evolution of heat (q negative)
then the reaction is called as exothermic. Similarly,
if the reaction results in flow of heat from
surrounding into system (q positive) then the
reaction is called as endothermic.
Temperature dependence of ΔH
H  nC H C  nD H D  n A H A  nB H B
so
d H C 
d H D 
d H A 
d H B 
d H 
 nC
 nD
 nA
 nB
dT
dT
dT
dT
dT
 C P ,C  C P , D  C P , A  C P , B
int egrating
H (T2 )  H (T1 )  C P (T2  T1 )
Note this form is exactly same as we found for ΔH for the
phase change.