THERMODYNAMICS
222 PHYS
Part-3
Department of Physics, Faculty of Science
Jazan University, KSA
The first law of thermodynamics
The first law of thermodynamics is an extension of the
law of conservation of
energy
“The change in internal energy of a system is equal to the difference between heat
added to the system and the work done by the system”
ΔU = Q - W
the internal energy of a system can
be changed by doing work on it or by
heating/cooling it.
From the microscopic point of view, this
statement is equivalent to a statement of
conservation of energy.
Note: The first law of thermodynamics tells us that in order to change the internal energy of a
system we must add (or remove) heat and/or do work on (or have work done by) the system.
Or we can say….
Change in internal
energy
=
Energy supplied to
system as heat
U= Q (heat) + w (work)
U like reserves of a bank: bank accepts deposits
or withdrawals in two currencies (Q & w) but
stores them as common fund, U.
U– Change in internal
energy [J]
Q – Heat added or lost
by the system [J]
W – Work done on or
by the system [J]
System
+
Energy supplied to
system as work
q
q
w
U
U
Q
Environment
W
w
The First Law of Thermodynamics, In general :
statement of energy conservation for a thermodynamic system
Heat and work are forms of energy transfer and energy is conserved.
U = Q + Won
change in
total internal
energy
heat added
to system
State Function
or
work done
on the system
The signs on Q and W depend
on the way the internal
energy is changed. If Q and W
are positive the internal
energy increases, but if Q and
W are negative the internal
energy decreases.
Path Functions
U = Q - Wby
The internal energy of a system tends to increase if energy is added
via heat (Q) and decrease via work (W) done by the system.
Recap
From the first Law of Thermodynamics: Energy is Conserved
ΔU = Ufinal - Uinital = Q – W
Q = heat absorbed by the system from the surroundings
W = work done by the system on the surroundings
heat is random molecular motion while work is force times distanced moved
under its influence
Exothermic Processes release heat and have Q<0
Endothermic Processes absorb heat and have Q>0
Energy: The SI unit is joule (J) although we will frequently use calorie ;
1 cal = 4.2 J
The work is not necessarily associated with the volume changes – e.g.,
in the Joule’s experiments on determining the “mechanical equivalent of
heat”, the system (water) was heated by stirring.
Special Cases of the 1st Law of Thermodynamics
Isolated System
The system doesn’t interact with the environment.
What does this mean in terms of the 1st law of thermodynamics?
Q = 0 -------- No heat is transferred into or out of the system.
W = 0 ------- No work is done on the system.
0
0
U  Q W
U i U f
U  0
Cyclic Process The system starts and ends in the same state (same internal energy).
The system is not necessarily isolated.
The function that describes the changes in the state on a PV - diagram would be a
closed curve.
U = 0 ------- No net change in the internal energy.
0
U  Q W
0  Q W
Q  W
Applications of 1st law of thermodynamic
Adiabatic Process
This process considers a system where there is no
loss or gain of energy through heat.
This can be accomplished by:
1. Thermally insulating the chamber
2. Performing the process very rapidly
W
– no time for heat to be transferred.
If we do work to compress the gas the
temperature of the gas should increase.
The increase in the temperature of the gas
corresponds to an increase in the internal
energy of the system.
0
U  Q W
U W
Examples:
• Expansion of hot gases in an internal combustion engine
• Liquefaction of gases in a cooling system
Adiabatic Free Expansion
This is a special case of an adiabatic process,
where the gas expands into free space.
0
0
U  Q W
U  0
Isobaric Process
U  Q W
Constant pressure process.
Q0
W 0
Isovolumetric (Isochoric) Process
U  Q W
Q0
W   PV f  Vi 
Constant volume process.
W 0
U  Q
If volume doesn’t change work cannot be done to compress the gas.
Isothermal Process
U  Q W
Constant temperature process.
U  0
Q  W
If you do work to compress a gas the energy you put in is released through heat.
On a PV – diagram it is common to use isotherms to show how the
temperature changes for a process.
Isotherm – Hyperbolic line of constant temperature on a PV – diagram.
Work in Thermodynamic Processes
As stated previously, pressure, temperature and volume are
considered state variables and are used to define the particular
F
state of the system.
Work and Heat are called transfer variables. These describe
changes in the state. They do not describe the state. We know
how to describe the work done on a system.
For example let us look at applying a force to a piston in order to
compress the gas inside a container.
Area
V
y
F
P
A
W = F y
W= PA y
During a compression:
Work done on a gas is W = P V
positive.
F  PA
V = A y
Quasi-static process
Work is the transfer of energy that takes
place when an object is moved against an
opposing force
If the gas is compressed slowly enough for all
of the system to remain in thermal equilibrium.
External force is equal and
opposite to force gas exerts on
piston. Work done on gas!
r r
dW  F  d s
dW   Fdy  PAdy  PdV
Vf
W    PdV
Vi
Total Work done to change the
volume of a gas
-F is parallel to y
Work
The work done by an external force to compress a
gas enclosed within a cylinder fitted with a piston:
W = (PA) dx = P (Adx) = - PdV
A – the
piston area
The sign: if the volume is decreased, W is positive
(by compressing gas, we increase its internal
energy)
if the volume is increased, W is negative (the gas
decreases its internal energy by doing some work
on the environment).
x
W = PdV
applies to any
shape of
system
boundary
force
P
W 1 2   VV12 P dV
dU = Q – PdV
Work Done by an Expanding Gas
Gas expands slowly enough to maintain thermodynamic equilibrium.
dW  Fdy  PAdy
dW  PdV
Increase in
volume, dV
Vf
W   PdV
Vi
Total Work done to change
the volume of a gas
+dV Positive Work (Work is
done by the gas)
Energy leaves the system
and goes to the environment.
-dV Negative Work (Work is
done on the gas)
Energy enters the system
from the environment.
This is a PV – diagram showing
several isotherms.
What type of process is
described by each of the
arrows?
PV-diagram.
A PV-Diagram is a plot of pressure vs. volume.
The work done during the process shown by the PV-diagram
can be determined by looking at the area under the curve.
Remember this is the same as the integral expression for the
work.
Example: In the three figures shown (a), (b) and (c), rank the
amount of work done by each of the processes shown from
largest to smallest. How much work is done in each case?
(b) > (c) > (a)
(a) W=-Pi(Vf-Vi)
Vf
(b) W=-Pf(Vf-Vi) (c)
W    P (V )dV
Vi
The amount of work done during a process depends
on the path you take from your initial point to your
final point. In other words it depends on how you
change your pressure and volume!
Shaded area is the work done by the system
P depends on V in general
There are
many ways
to take the
system from
i to f.
The work W
done and Q
depends on
the path.
𝑉𝑓
𝑊=
𝑝𝑑𝑉
𝑉𝑖
W and Q are not State Functions
We can bring the system from state 1 to state 2 along infinite
number of paths, and for each path P (T,V ) will be different.
V2
W12   P(T ,V )dV
P
T
2
1
V1
Since the work done on a system depends not only on the initial and final
states, but also on the intermediate states, it is not a state function.
U = Q + W
P
A
P2
P1
D
V1
PV diagram
B
C
V2
U is a state function, W - is not 
Q is not a state function either.
The work is negative for the “clockwise” cycle; if
the cyclic process were carried out in the reverse
order (counterclockwise), the net work done on the
gas would be positive.
Wnet  WAB  WCD   P2 V2  V1   P1 V1  V2 
V
 P2  P1 V2  V1   0
V
Work from closed cycles
Consider cycle A -> B -> A
WA->B
-WB->A
Work from closed cycles
Consider cycle A -> B -> A
WA->B->A= Area
Reverse the cycle,
make it counter
clockwise
-WB->A
WA->B
Processes and cycles
Change of State: implies one or more properties of the system has
changed.
How these properties would change outside of time is curiously
outside the framework of equilibrium thermodynamics!
The best way to think of them is that the changes are slow relative
to the underlying molecular time scales.
• Process: is a succession of changes of state.
Assuming processes are all sufficiently slow such that each stage of
the process is near equilibrium.
Certain common processes are isos, meaning “equal”:
A. Adiabatic: no heat transferred
B. Isothermal: constant temperature,
C. Isobaric: constant pressure, and
D. Isochoric: constant volume.
An important notion in thermodynamics is that of a
• Cycle: series of processes which returns to the original state.
The cycle is a thermodynamic “round trip.”
Paths on the PV diagram
(1) Adiabatic
????
(2) Isochoric
W=0
(3) Isothermal
????
(4) Isobaric
W = P V
4
2
3
1
p
V
Ideal gas
T1
T2
T4 T3
A. Adiabatic Process
• An adiabatic process transfers no heat
ΔU = Q – W
– But Q = 0 , →
ΔU = – W
• When a system expands adiabatically, W is positive (the system does
work) so ΔU is negative.
• When a system compresses adiabatically, W is negative (work is done
on the system) so ΔU is positive.
Adiabatic processes can occur when the system is well insulated or a very
rapid process occurs so that there is not enough time for a significant heat
to be transferred (e.g., rapid expansion of a gas; a series of compressions
and expansions as a sound wave propagates through air).
•For an ideal gas, and most real gasses,
•dQ = dU + PdV
•dQ = CVdT + PdV,.
•Then, when dQ = 0,
PdV
dT  
CV
Adiabatic Processes
A Polytropic process
Remember: In adiabatic process there is no thermal energy transfer to or
from a system (Q = 0)
A reversible adiabatic process involves a “worked” expansion in which we can
return all of the energy transferred.
A Polytropic process is a thermodynamic process that
obeys the relation:
4
2
Atmospheric processes which lead to changes
in atmospheric pressure often adiabatic: HIGH
pressure cell, falling air is compressed and
warmed. LOW pressure cell, rising air expands
and cooled  condensation and rain.
3
1
p
V
T1
T2
T4T3
Q=0
Adiabatic Processes, Derivation
Adiabatic expansion of a perfect gas
U  W

n C v dT   P dV  
nRT
dV
V
R

R
; ln(T 2 / T1 )  
ln(V 2 /V 1 )  ln (V 1 /V 2 )Cv
Cv

T2
dT
R V 2 dV



C v V1 V
T1 T
T 2 V 1 
R Cp
R  C p  Cv ;

1   1 
 
Cv Cv
T 1 V 2 
 TV ( 1)  constant ;
,Q
P2V 2
P 1V 1

T1
T2
V 1 
T 2 PV
2 2


 
T1 PV
1 1
V 2 

Hence ,
P2  V 1 
    PV
P1 V 2 

 constant
(  1)



For a reversible
process,
CVdT = -pdV
along the path.
Now, per mole,
for an ideal
gas, PV = RT
(  1)
cP 5
3R
5R

 Monatomic: cV  2 ; c P  2 ;   c  3
V

cP 7
5R
7R

 Diatomic: cV  2 ; c P  2 ;   c  5
V







Adiabatic Process
For an ideal gas, PV=nRT, so
Another way
PV
PdV V dP
, and dT 
nR
nR
PdV
PdV V dP
Then, 

CV
nR
T 
Then,
PdV PdV  VdP

0
CV
nR

 nR  CV  VdP
PdV 
0

C
nR
nR
 V

Thermo & Stat Mech - Spring 2006 Class 3
 nR  CV 
0  PdV 
 V dP but nR  CV  C P
 CV

Cp
C 
0  PdV  P  V dP   PdV V dP
where,  
CV
CV 
  PdV V dP  0
dV
dP

 0, which can be integrated,
V
P
 lnV  ln P  constant

lnV
PV

 ln P  ln  PV

 constant

  constant
Thermo & Stat Mech - Spring 2006 Class 3
29
PV   constant
With the help of PV  nRT
this can also be expressed as,
TV  1  constant
 for “Ideal Gasses”
T
 1  constant
P
  1
2
f
2
 1.67
3
2
diatomic:
  1   1.40
5
2
polyatomic:   1   1.33
6
monatomic:   1 
Thermo & Stat Mech - Spring 2006 Class 3
30
Adiabatic Process in an Ideal Gas
Q12  0
work in adiabatic process
V2
W12    P(V , T )dV
V1
W12

PV
   P(V , T )dV    1 1 dV
V1
V1 V
V2
W12

V2
P1 V1   V1 
1   

  1   V2 

 1
P




 1+2/31.67 (monatomic), 1+2/5 =1.4 (diatomic),
and
1+2/6 1.33 (polyatomic)
(again, neglecting the vibrational degrees of freedom)
2
PV= NkBT2
1
V2
V1
PV= NkBT1
V
B. Isothermal Process
• An isothermal process is a constant temperature process.
• Any heat flow into or out of the system must be slow
enough to maintain thermal equilibrium
• For ideal gases, if ΔT is zero, ΔU = 0
• Therefore, Q = W
– Any energy entering the system (Q) must leave as work
(W)
Isothermal processes
• Work done when PV = nRT = constant  P = nRT / V
3
p
T1
T2
T4 T3
V
Isothermal change T = 0
U = 0,
Boyle’s Law (1627 -1691)
T1= T2 ------- P1V1 = P2 V2
PV = n R T
V 2 
 nRT 
Q W   PdV   
 dV  nRT ln  
 V 
V 1 
V2
V1
V2
V1
V 2 P1
P1V 1  P2V 2 

V 1 P2
 P1 
 Q W  n R T ln  
 P2 
Therefore
V 2
ln 
V 1
 W

 nRT
V 2
 
V 1
 W




 nRT 
 e

 P2
and 
 P1
 W




 nRT 
 e

c. Isochoric Process
•
An isochoric process is a constant volume process. When the volume of a
system doesn’t change, it will do no work on its surroundings. W = 0
ΔU = Q
Heating gas in a closed container is an isochoric process
P
isochoric ( V = const )
W12  0
dU  Q12
3
Q12  Nk B T2  T1   0
2
2
 CV T 
PV= NkBT2
PV= NkBT1
1
V1,2
V
D. Isobaric Process
•
•
An isobaric process is a constant pressure process.
ΔU, W, and Q are generally non-zero, but calculating the work done by an
ideal gas is straightforward
W = P·ΔV
Water boiling in a saucepan is an example of an isobar process
2
W 12    PdV   P V 2 V 1   0
P
1
2
5
Q12  Nk B T2  T1   0  CP T 
2
dU  W12  Q12
1
V1
PV= NkBT2
PV= NkBT1
V2
V
Isobaric Processes & Specific Heat
Isobaric Process : constant P
Isotherms
W   P  V 2 V 1   P V
Q  U W
 U  P V
CP = molar specific heat at constant pressure
1  dQ 
CP  
n  dT  P
Q  n CP T
isobaric processes
n CP T  n CV T  P V
Ideal gas, isobaric :
 n CV T  n R T

CP  CV  R
Ideal gas
Summary, Work and ideal gases
W    PdV  0
Isochoric
W  P  dV  P V f V i 
Isobaric
Vf
W    PdV  nRT 
Vi
𝑊=0
V f 
dV
 nRT ln  
V
V i 
Isothermal
Isochoric
 1

 Adiabatic (and


PV
P1 V1
V1
1     reversible)
W12    P(V , T )dV   
dV  W12 
V1
V1 V
  1   V2  


V2
V2

1 1

Enthalpy
It is a thermodynamic quantity equivalent to the total heat
content of a system.
H  U  PV
V2
V2
V1
V1
U 2  U 1  Q W  Q   PdV  Q p  P  dV  Q p  P (V 2 V 1 )
Q p  (U 2  PV 2 )  (U 1  PV 1 )  H 2  H 1
H  Q p
constant P, closed system, P-V work only
The heat Qp absorbed in a constant-pressure
process equals the system’s enthalpy change.
The universal gas constant
ΔH  nCP ΔT
ΔE  ΔU  nCV ΔT
heat capacity at constant pressure C p
dQP
dH
CP 
(
)P  ΔH  Q p  TT12 CP dT
dT
dT
heat capacity at constant volume Cv
dQV
dU
CV 
(
)V  ΔU  QV  TT12 CV dT
dT
dT
CP - CV  R
note : (C p 
Cp
n
, Cv 
Cv
)
n
dH dU d (U  nRT )  dU
C p  CV 


 nR
dT dT
dT
C p CV


 C p  CV  R
n
n
EXOTHERMIC & ENDOTHERMIC REACTIONS
Exothermic process: a change (e.g. a chemical reaction) that releases
heat.
A release of heat corresponds to a decrease in enthalpy.
Exothermic process: H < 0 (at constant pressure).
e.g: Condensation, crystallization of liquids
Endothermic process: a change (e.g. a chemical reaction) that requires
(or absorbs) heat.
Absorption of heat corresponds to an increase in enthalpy.
Endothermic process: H > 0 (at constant pressure).
e.g: Evaporation, fusion, melting of solids
Vaporisation
Energy has to be supplied to a liquid to enable it to overcome forces that hold
molecules together (Endothermic process (H positive))
Melting
Energy is supplied to a solid to enable it to vibrate more vigorously until molecules
can move past each other and flow as a liquid-- Endothermic process (H
positive)
Freezing
Liquid releases energy and allows molecules to settle into a lower energy state and
form a solid --Exothermic process (H negative)
(we remove heat from water when making ice in freezer)
Changes in Thermal Systems
Example of a Reversible Process:
Cylinder must be pulled or pushed slowly enough that
the system remains in thermal equilibrium
Changes in Thermal Systems
Example of an Irreversible Process:
The gas expands freely when
the valve is opened.