Thermodynamics -chap. 15
note for AP B physics:
if it says ON gas, use W= -PDV
if it says BY gas, use W= PDV
always: U= Q + W
The Ideal Gas Law
(review chap.14)

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
NA = Avogrado’s number = 6.02 x 1023
Summary of Kinetic Theory:
The relationship between energy and temperature
(for monatomic ideal gas)
ave KE/molecul e 
Root-mean-square speed :
v rms 

v
2

1
m v
2

2
3k B T
M
3
2

3RT

k BT
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)
Thermodynamics

a system (gas) interacts with surroundings via thermal
processes (heat and work transferred into gas) :
system
P,V,T
Example:

surroundings
gas piston (car engine, bike pump, syringe)
system = gas at pressure P and temperature T and volume V
surroundings = piston and walls of container
state of system: values for P,V, and T describe gas
The 4 Laws of Thermodynamics
0. Law: Two objects at thermal equilibrium will have
no heat flow between them and the same temperature.
T 1 = T2
1st law: Total internal energy = heat put in + work done on it
U=
Q + W
2nd law: Heat will spontaneously flow from hot to cold
Carnot engine: ideal efficiency= (Hot- cold temp) /hot temp
(otherwise: efficiency… use heat’s Q, not temp’s T)
3rd law: You can’t get to absolute zero! (T= OK)
(can’t remove all the energy w/o still adding some)
First Law of Thermodynamics
“The internal energy of a system tend to increase
when HEAT is added and work is done ON the
system.”
Suggests a CHANGE or subtraction
D U  Q  W  D U  Q Add  W on
Q > 0 : heat is added to system
Q < 0 : heat is subtracted from system
W > 0 : work done on system by surroundings
W < 0 : work done on surroundings by system
Internal Energy (DU) and Heat Energy (Q)
All of the energy inside a
system is called INTERNAL
ENERGY, DU.
When you add HEAT(Q), you
are adding energy and the
internal energy
INCREASES.
Both are measured in joules.
But when you add heat,
there is usually an increase
in temperature associated
with the change.
DU  DT
if D T  , D U 
if D T  0 , D U  0
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 the area underneath the P,V graph.
Work done by a gas
Suppose you had a piston filled with a
specific amount of gas. As you add
heat, the temperature rises and
thus the volume of the gas
expands. The gas then applies a
force on the piston wall pushing it a
specific displacement. Thus the gas
does NEGATIVE work
(uses its own internal energy to do
work).
Work is the AREA of a P vs. V graph
W= -PDV = -P* (Vf- Vi)
the work done on the gas
Compress gas:
gas expands itself
smaller volume (DV=-)
larger volume (DV=+)
positive work more (W=+) negative work (W=-)
gas has more energy
gas loses energy
move left on P-V diagram move right on P-V diagram
-Work: clockwise
+Work: counterclockwise
Negative or positive Work Done ON gas?
(P=constant)
System: Gas
Surroundings:
Piston, walls
W
W
Gas
P,V1,T1
Dy
Isobaric process
Gas
P,V2,T2
W = -F s = -P A D y = -P DV
DV > 0
expanding gas system does work on surroundings
(isobaric process: pressure kept constant even though increase volume
by adding heat – which offsets some but not all work done by gas)

answer= negative work on gas
What area is the total work done from 1 … to 4?
P
Isobaric Process: P=constant (horizontal line)
Isochoric Process: V=constant (vertical line)
P
1
W =- PDV (<0)
P
1
2
V
2
1
P
1
V
W =- PDV (>0)
2
4
3
DV < 0
V
DV = 0
P
1
2
3
Wtot < 0
4
DV > 0
2
3
DV = 0
3
V
W = -PDV = 0
4
2
3
W = -PDV = 0
4
3
Wtot = ??
4
P
4
1
V
V
P
1
4
2
If we go
the other
way
then
3
Wtot > 0
V
Now try this:
a.
b.
C.
D.
1 to 2 to 3 to 1 (one cycle)
What’s the total change in internal energy U?
Is positive or negative work done?
Is heat absorbed or released?
Write an equation for work using P’s and V’s
P
P1
P3
1
2
3
V1
V2
V
DU = 0 since no change overall (DT=0)
overall net work is negative (clockwise)
Heat is absorbed (Q>0) since work <0 and U =0
Area = (V2-V1)x(P1-P3)/2
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
Net Work = area under P-V curve
Area the same in both cases!
9 V(m3)
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 (neg work done on gas)
5. Q = DU - W = 1500 J - -1000 J = 2500 J > 0 heat gained by gas
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
(no change in volume)
5. Q = DU - W = 1500 – 0 = 1500 J
=> It requires less heat to raise T at const. volume than at const. pressure.
Example
Sketch a PV diagram and find
the work done on the gas
during the following stages.
(a)
A gas is expanded from a
volume of 1.0 L to 3.0 L at a
constant pressure of 3.0
atm.
W ON   P D V   3 x10 ( 0 . 003  0 . 001 ) 
5
(b)
-600 J
The gas is then cooled at a
constant volume until the
pressure falls to 2.0 atm
W   PDV  0
since D V  0
Example continued
a)
The gas is then
compressed at a constant
pressure of 2.0 atm from a
volume of 3.0 L to 1.0 L.
W ON   P D V   2 x10 (. 001  . 003 ) 
5
b)
+400 J
The gas is then heated
until its pressure
increases from 2.0 atm to
3.0 atm at a constant
volume.
W   PDV  0
since D V  0
Example continued
What is the net work
ON gas?
-600 J + 400 J = -200 J
Rule of thumb: If the system
rotates CW, the NET work is
negative.
If the system rotates CCW, the
NET work is positive.
Side note: work by gas = +200J
NET work is
the area inside
the shape.
Example
A series of thermodynamic processes is shown in the pV-diagram.
In process ab 150 J of heat is added to the system, and in
process bd , 600J of heat is added. Fill in the chart. On gas.
150
0
150 J
600 -240
360 J
750 -240
510 J
-90
600
0
600
-90
510 J
Classification of Thermal Processes
(using work on gas definition)




Isobaric : P = constant
Isochoric : V = constant:
Isothermal : T = constant( DU = 0):
Adiabatic : no heat flow (Q = 0)
W = -P DV
W=0
W = -Q
W = - DU
* Remember: work is area under P-V curve
(positive work if compress gas to less volume)
Thermodynamic Processes - Isothermal
To keep the temperature
constant both the
pressure and volume
change to compensate.
(Volume goes up,
pressure goes down)
“BOYLES’ LAW”
Same temp so DU=0
DU= Q + W= 0 so
Q= -W
DV = + since expands
W = -PV = neg work
Q= -W = positive heat
added to maintain temp
while gas uses own energy
to expand to larger volume
Thermodynamic Processes - Isobaric
More heat Q is added to
the gas than work
energy used by the gas
internal energy (U)
decreases since PV is
smaller.
W= + (compress)
Q= - release heat
DU = - (colder at end)
U= Q + W
Q<W
Thermodynamic Processes - Isovolumetric
No work since constant volume
W= 0
DU = Q
Q = + since higher temp
DU = + (higher int. energy)
Thermodynamic Processes - Adiabatic
ADIABATIC- (GREEK- adiabatos"impassable")
In other words, NO HEAT can
leave or enter the system.
Q= 0 by definition
U= Q + W
U = W only
W = - since gas expands
hence U < 0 (tem drops)
gas loses temp as expands
since don’t provide heat to
offset work done by gas
In Summary
Q=-W
Q+W
W
Second Law of Thermodynamics
“Heat will not flow spontaneously from a colder body to
a warmer body AND heat energy cannot be
transformed completely into mechanical work.”
The bottom line:
1) Heat always flows from a hot body to a cold body
2) Nothing is 100% efficient
Engines
Heat flows from a HOT
reservoir to a COLD
reservoir
Q H  W  QC
W output  Q H  Q C
QH = remove from, absorbs = hot
QC= exhausts to, expels = cold
Engine Efficiency
In order to determine the
thermal efficiency of
an engine you have to
look at how much
ENERGY you get OUT
based on how much
you energy you take IN.
In other words:
e thermal 
W
Q hot

Q H  QC
QH
 1
QC
QH
Rates of Energy Usage
Sometimes it is useful to express the
energy usage of an engine as a
RATE.
For example:
The RATE at which heat is absorbed!
The RATE at which heat is expelled.
QH
t
QC
t
The RATE at which WORK is DONE
W
t
 POWER
Efficiency in terms of rates
e thermal 
QH
W

QH

e
QH
t

t
QH
t
P
t
P 
W
QC
t

P
QH
t
Is there an IDEAL engine model?
Our goal is to figure out just how efficient
such a heat engine can be: what’s the most
work we can possibly get for a given amount
of fuel?
The efficiency question was first posed—and solved—by Sadi Carnot in 1820,
not long after steam engines had become efficient enough to begin replacing
water wheels, at that time the main power sources for industry. Not surprisingly,
perhaps, Carnot visualized the heat engine as a kind of water wheel in which
heat (the “fluid”) dropped from a high temperature to a low temperature,
losing “potential energy” which the engine turned into work done, just like a
water wheel.
Carnot Efficiency
Carnot a believed that there was an
absolute zero of temperature, from
which he figured out that on being
cooled to absolute zero, the fluid would
give up all its heat energy. Therefore, if
it falls only half way to absolute zero
from its beginning temperature, it will
give up half its heat, and an engine
taking in heat at T and shedding it at ½T
will be utilizing half the possible heat,
and be 50% efficient. Picture a water
wheel that takes in water at the top of a
waterfall, but lets it out halfway
down. So, the efficiency of an ideal
engine operating between two
temperatures will be equal to the
fraction of the temperature drop towards
absolute zero that the heat undergoes.
Carnot Efficiency
Carnot temperatures must be
expressed in KELVIN!!!!!!
The Carnot model has 4 parts
•An Isothermal Expansion
•An Adiabatic Expansion
•An Isothermal Compression
•An Adiabatic Compression
The PV diagram in a way shows us that the ratio of the heats are symbolic to
the ratio of the 2 temperatures
Example
A particular engine has a power output of 5000 W and an
efficiency of 25%. If the engine expels 8000 J of heat in each
cycle, find (a) the heat absorbed in each cycle and (b) the time
for each cycle
P  5000 W
e  1
QC
QH
e  0 . 25
0 . 25  1 
8000
QH
Q c  8000 J
Q H  10,667 J
W  Q H  Q C  W  Q H  8000
W 
P 
2667 J
W
t
 5000 
W
t
t  0.53 s
Example
The efficiency of a Carnot engine is 30%. The engine absorbs 800
J of heat per cycle from a hot temperature reservoir at 500 K.
Determine (a) the heat expelled per cycle and (b) the
temperature of the cold reservoir
e
W
 0 . 30 
QH
W
W 
800 J
W  Q H  Q C  W  800  Q C
QC 
560 J
eC  1 
TC 
TC
TH
350 K
 0 . 30  1 
TC
500
240 J