PHY101: Lecture 30
Thermodynamics


New material: Chapter 15
The dynamics of thermal processes
(equilibrium, work, energy)
Physics 101: Lecture 30, Pg 1
The Ideal Gas Law
(review)

P V = N kB T
N = number of molecules
» N = number of moles (n) x NA molecules/mole
kB = Boltzmann’s constant = 1.38 x 10-23 J/K

PV=nRT
R = ideal gas constant = NAkB = 8.31 J/mol/K
Physics 101: Lecture 30, Pg 2
Summary of Kinetic Theory:
The relationship between energy and temperature
(for monatomic ideal gas)
1
3
2
ave KE/molecul e  m v  k BT
2
2
Root-mean-square speed :
v rms 

v2
3k BT
3RT


M

Careful with units:
R = 8.31 J/mol/K
 = molar mass in kg/mol
vrms = speed in m/s
Internal Energy U
= number of molecules x ave KE/molecule
= N (3/2) kBT
= (3/2) n RT = (3/2) P V (ideal gas)
Physics 101: Lecture 30, Pg 3
Thermodynamics

Dynamics of thermal processes, i.e. when a system interacts
with its surroundings via thermal processes :
system
P,V,T

surroundings
The state of the system (e.g. gas) is determined by P,V
and T.
Example: gas filled container with piston on open end of
container:
system = gas at pressure P and temperature T and volume V
surroundings = piston and walls of container
Physics 101: Lecture 30, Pg 4
The Laws of Thermodynamics

Like Newton’s laws govern the dynamics of an object in motion, the
three laws of thermodynamics govern the dynamics of thermal
processes.
0. Law: Thermal equilibrium
(Mechanics: S F = 0, a=0)
Indicator of thermal equilibrium: DT=0
(like a=0)
Two systems are in thermal equilibrium, if there is no net flow
of heat between them if brought in thermal contact (i.e.
the two systems are already in their most probable state and
have thus no motivation to spontaneously change their state).
If two systems are individually in equilibrium with a third system,
they are in thermal equilibrium with each other.
Physics 101: Lecture 30, Pg 5
The Laws of Thermodynamics

1. Law of Thermodynamics (energy conservation)
change in
internal energy
of the system
DU
heat added to/subtracted
work done
=
from the system
- by/on the system
=
Q
-
W
Q > 0 : heat is added to system
Q < 0 : heat is subtracted from system
W > 0 : work done by system on surroundings
W < 0 : work done on system by surroundings
Physics 101: Lecture 30, Pg 6
Thermodynamic Systems and P-V Diagrams
Ideal gas law: P V = n R T
 For n fixed, P and V determine the “state” of the system
T = P V/ (n R)
U = (3/2) n RT = (3/2) P V
 Examples:
P
 which point has highest T ?
1
2
P1
»2
 which point has lowest U ?
3
P
3
»3
 to change the system from 3 to 2,
V1
V2
energy must be added to system.

V
Work done in a thermal process is given by the area underneath the
P,V graph.
Physics 101: Lecture 30, Pg 7
Work Done by a System (P=constant)
System: Gas
Surroundings:
Piston, walls
W
Dy
W
Gas
P,V1,T1
Isobaric process
Gas
P,V2,T2
W = F s = P A D y = P DV
W > 0 if DV > 0
expanding system does work on surroundings (W positive)
W < 0 if DV < 0
contracting system : work is done on the system by
surroundings (W negative)
W = 0 if DV = 0
system with constant volume does no work
Physics 101: Lecture 30, Pg 8
Classification of Thermal Processes




Isobaric : P = constant, W = P DV
Isochoric : V = constant, W = 0 => DU = Q
Isothermal : T = constant, W = n R T ln(Vf/Vi) (ideal gas)
Adiabatic : Q = 0 => W = - DU= -3/2 n R DT (ideal gas)
Physics 101: Lecture 30, Pg 9
Isobaric Process: P=constant
Isochoric Process: V=constant
P
1
Wtot = ??
4
P
1
W = PDV (>0)
P
1
2
V
2
4
3
1
P
1
V
Wtot > 0
W = PDV (<0)
2
4
3
DV < 0
V
1
V
W = PDV = 0
2
4
3
DV = 0
3
V
DV = 0
P
2
3
4
DV > 0
3
W = PDV = 0
P
4
2
V
P
1
4
2
If we go
the other
way
then
3
Wtot < 0
V
Physics 101: Lecture 30, Pg 10
Now try this: What is the total work done
by system when going from
state 1 to state 2 to state 3 and
back to state 1 ?
P
P1
P3
1
2
Area = (V2-V1)x(P1-P3)/2
3
V1
V2
V
Physics 101: Lecture 30, Pg 11
First Law of Thermodynamics
Example
P
2 moles of monatomic ideal gas is taken
P
from state 1 to state 2 at constant pressure
P=1000 Pa, where V1 =2m3 and V2 =3m3. Find
T1, T2, DU, W, Q.
1
V1
1. P V1 = n R T1  T1 = P V1/(nR) = 120K
2
V2
V
2. P V2 = n RT2  T2 = P V2/(nR) = 180K
3. DU = (3/2) n R DT = 1500 J or DU = (3/2) P DV = 1500 J
4. W = P DV = 1000 J > 0 Work done by the system (gas)
5. Q = DU + W = 1500 J + 1000 J = 2500 J > 0 heat gained by system (gas)
Physics 101: Lecture 30, Pg 12
First Law of Thermodynamics
Example
2 moles of monatomic ideal gas is taken
from state 1 to state 2 at constant volume
V=2m3, where T1=120K and T2 =180K. Find Q.
1. P V1 = n R T1  P1 = n R T1/V = 1000 Pa
2. P V2 = n R T1  P2 = n R T2/V = 1500 Pa
P
P2
2
P1
1
V
V
3. DU = (3/2) n R DT = 1500 J
4. W = P DV = 0 J
5. Q = DU + W = 0 + 1500 = 1500 J
=> It requires less heat to raise T at const. volume than at const. pressure.
Physics 101: Lecture 30, Pg 13
Concept Question
Shown in the picture below are the pressure versus volume graphs for
two thermal processes, in each case moving a system from state A to
state B along the straight line shown. In which case is the work done by
the system the biggest?
1. Case 1
P(atm)
P(atm)
2. Case 2
3. Same
correct
A
B
4
2
4
A
B
2
Case 1
3
Case 2
9 V(m3)
3
9 V(m3)
Net Work = area under P-V curve
Area the same in both cases!
Physics 101: Lecture 30, Pg 14