Physics 231
Topic 14: Laws of Thermodynamics
Alex Brown
Dec
7-11
2015
MSU
Physics
231 Fall
2015
1
8th 10 pm correction for 3rd exam
9th 10 pm attitude survey
(1% for participation)
10th 10 pm concept test
timed (50 min)
(1% for performance)
11th 10 pm last homework set
17th 8-10 pm final (Thursday)
VMC E100
MSU Physics 231 Fall 2015
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Clicker Question!
Ice is heated steadily and becomes liquid and then vapor.
During this process:
a) the temperature rises continuously.
b) when the ice turns into water, the temperature
drops for a brief moment.
c) the temperature is constant during the phase
transformations
d) the temperature cannot exceed 100oC
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Key Concepts: Laws of
Thermodynamics
Laws of Thermodynamics
1st Law: U = Q + W
2nd Law: Heat flows from hotter  cooler
Thermodynamic Processes
Adiabatic (no heat flow)
Work done in different processes
Heat Engines & Refrigerators
Carnot engine & efficiency
Entropy
Relationship to heat, energy.
Statistical interpretation
Covers chapter 14 in Rex & Wolfson
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Engine based on a container of an idea gas
where the P, V and T change (n is fixed)
piston
area A
y
P,V,T
n fixed
1) Put in contact with a source of
heat at high T during which
heat energy flows in and piston
is pushed up.
2) Put in contact with a source of
heat at low T during which
piston is pushed down and heat
flows out.
3) Comes back to it original state
(e.g. same value of P, V, and T)
4) End result is that we have
turned heat energy into work
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Process visualized with a P-V diagram for the gas inside
P
isobaric line: pressure is constant
volume changes
iso-volumetric line: volume is constant
pressure changes
i
V
n fixed
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PV = n R T
(ideal gas equation from chapter 12)
P = n R T/V = c T/V
(c = constant)
P
lines with constant T
T1
T2
T3
iso-thermal lines
T4
T1 < T2 < T 3 < T 4
V
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A Piston Engine
piston
area A
y
Pi Vi Ti
Piston is moved downward
slowly so that the gas remains
in thermal equilibrium:
Volume decreases (obviously)
Temperature increases
Work is done on the gas
Pf Vf Tf
Ti < Tf
vin
vout
vout > vin (speeds)
work is done on the gas
and temperature increases
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piston
Isobaric Compression
area A
The pressure does not change while
pushing down the piston (isobaric
compression).
y
Pi Vi Ti
P
P
Pf Vf Tf
f
i
Vf
Vi V
W = work done on the gas by
pushing down on the piston
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piston
Isobaric Compression
area A
The pressure does not change while
lowering the piston (isobaric
compression).
y
Pi Vi Ti
P
P
f
Vf
Pf Vf Tf
i
Vi V
W = work done on the gas
W = F d = - P A y
(P=F/A)
W = - P V = - P (Vf-Vi) (in Joule)
Sign of the work done on the gas:
+ if V < 0
- if V > 0
work is the area under the curve in a
P-V diagram with V decreasing
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piston
Non-isobaric Compression
area A
In general, the pressure can change
when lowering the piston.
y
Pi Vi Ti
P
Pf
The work (W) done by the piston on the
gas when going from an initial state (i)
to a final state (f) is the area under
the line on the P-V diagram with
V decreasing.
f
i
Pi
Vf
Vi V
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Work Done on Gases:
Getting the Signs Right!
P
i
V
If the arrow goes from right to left (volume becomes smaller)
positive work is done by pushing the piston down on the gas (W > 0)
the internal (kinetic) energy of the gas goes up
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Work Done on Gases:
Getting the Signs Right!
P
i
V
If the arrow goes from left to right (volume becomes larger)
W < 0 and Wg = -W > 0
positive work (Wg) is done by the gas on the piston.
the internal energy of the gas goes down
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iso-volumetric process
P
v
Work done on/by gas: W = Wg = - PV = 0
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Clicker Quiz!
A gas is enclosed in a cylinder with a moveable piston. The
figures show 4 different PV diagrams. In which case is the
work done by the gas largest?
Work: area under PV diagram
Work done by the gas: volume must become larger, which
leaves (a) or (c). Area is larger for (a).
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M=50 kg
A=100 cm2 = 0.010 m2
mass and area of the lid
Patm
a) What is the pressure PA?
PA
b) If the inside temperature is raised
the lid moves up by 5 cm. How much work
is done by the gas?
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M=50 kg
A=100 cm2 = 0.010 m2
mass and area of the lid
Patm
a) What is the pressure PA?
PA
b) If the inside temperature is raised
the lid moves up by 5 cm. How much work
is done by the gas?
a) PA = Patm + Mg/A = 1.50 x 105
b) Wg = PA V = 75.0 J
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For ideal gas PV=nRT
One mole of an ideal gas initially at 0° C undergoes an
expansion at constant pressure of one atmosphere to
four times its original volume.
a) What is the new temperature?
b) What is the work done by the gas?
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For ideal gas PV=nRT
One mole of an ideal gas initially at 0° C undergoes an
expansion at constant pressure of one atmosphere to
four times its original volume.
a) What is the new temperature?
b) What is the work done by the gas?
a) Use PV = nRT to get
Tf = (Vf/Vi) Ti = 1092 K
b) W = -PV – P(4Vi-Vi) = -3PVi = -3P(nRTi/P)
Wg = -W = 3nRTi = 6806 J
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First Law of
Thermodynamics
By transferring heat to an object
the internal energy can increased
By performing work on an object
the internal energy can increased
The change in internal energy depends on
the work done on the object and the
amount of heat transferred to the object.
Internal energy (KE+PE) where KE is the
kinetic energy associated with translational,
rotational, vibrational motion of atoms
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First Law of Thermodynamics
U = Uf - Ui = Q + W
U = change in internal energy
Q = energy transfer through heat
(+ if heat is transferred to the system)
W = energy transfer through work
(+ if work is done on the system)
This law is a general rule for conservation of energy
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Applications to ideal gas in a closed container
(number of moles, n, is fixed)
PV = n R T
(chapter 12)
U = (d/2) n R T
(chapter 12) (d=3 monatomic)
(d=5 diatomic)
So U = (d/2) P V
(useful for P-V diagram)
(d/2) n R = constant
So U = (d/2) n R T
Example for P-V diagram (in class)
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First Law: Isobaric Process
A gas in a cylinder is kept at 1.0x105 Pa. The cylinder is brought in
contact with a cold reservoir and 500 J of heat is extracted from the
gas. Meanwhile the piston has sunk and the volume decreased by
100cm3. What is the change in internal energy?
Q = -500 J
V = -100 cm3 = -1.0x10-4 m3
W = - P V = 10 J
U = Q + W = - 500 + 10 = - 490
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First Law: General Case
P (Pa)
6
f
In ideal gas (d=3) is compressed
i
3
1
A) What is the change in internal energy
B) What is the work done on the gas?
C) How much heat has been
transferred to the gas?
4 V(m3)
A) U = (3/2)PV
U = 3/2(PfVf - PiVi) = 3/2[6x1 - 3x4] = -9 J
B) Work: area under the P-V graph: (9 + 4.5) = 13.5
(positive since work is done on the gas)
C) U = Q+W so Q = U-W = -9 - 13.5 = -22.5 J
Heat has been extracted from the gas.
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Types of Processes
PP
A: Iso-volumetric V=0
B: Adiabatic
Q=0
C: Isothermal T=0
D: Isobaric
P=0
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Iso-volumetric Process (V = 0)
1) PV = n R T
2) U = W + Q
3) U = (d/2) n R T
V = 0
W = 0 (area under the curve is zero)
4) U = Q = (d/2) n R T
5) P/T = constant
When P = + (like in the figure)
T = + (5)
U = + (4)
Q = + (4) (heat added)
When P = T = U = Q = - (heat extracted)
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Isobaric Process (P = 0)
1) PV = n R T
2) U = W + Q
3) U = (d/2) n R T
P = 0
4) W = - PV = - n R T
5) Q = U - W = [(d+2)/2] n R T
6) V/T = constant
When V = - (like in the figure)
T = - (6)
W = + (4) (work done on gas)
U = - (3)
Q = - (5) (heat extracted)
When V = +
T = +
W = - (work done by gas)
U = +
Q = + (heat added)
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molar heat capacities
Constant volume
Q = (d/2) n R T = Cv n T
where Cv = (d/2) R
Constant pressure
molar heat capacity at constant volume
Q = [(d+2)/2] n R T = CP n T
where CP = [(d+2)/2] R
For all
molar heat capacity at constant pressure
U = (d/2) n R T = Cv n T
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Isothermal Process (T = 0)
1) PV = n R T
2) U = W + Q
3) U = (d/2) n R T
T = 0
U = 0
Q = -W
PV = constant
work done on gas is the
area under the curve:
𝑉𝑓
W = −nRT ln
𝑉𝑖
When V = - (like in the figure)
P = + (like in the figure)
W = + (work done on gas, from area)
Q = - (heat extracted, Q = -W)
When V = +
P = W = - (work done by gas)
Q = + (heat added)
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Adiabtic Process (Q = 0)
1) PV = n R T
2) U = W + Q
3) U = (d/2) n R T
Q = 0 (system is isolated)
W = U (work goes into internal energy)
P (V) = constant  = Cp/Cv = (d+2)/d
>1
When V = - (like in the figure)
P = + (like in the figure)
T = + (see figure)
U = + (3)
W = + (work done on gas, area)
When V = +
P = T = U = W = - (work done by gas)
dashed lines are isotherms
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Process
Isobaric
U
nCv T
Q
nCp T
W
-P V
Adiabatic
nCv T
0
U
Isovolumetric
nCv T
U
0
nCv T=0
-W
-nRTln(Vf/Vi)
nCv T
U-W
(PV Area)
negative if V
expands
Isothermal
General
Ideal gas (monatomic d = 3)
Cv = (d/2) R
Cp = [(d+2)/2] R
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(diatomic d = 5)
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piston
area A
y
P,V,T
First Law: Adiabatic process
A piston is pushed down rapidly. Because the
transfer of heat through the walls takes a
long time, no heat can escape.
During the moving of the piston, the
temperature has risen 1000C. If the
container contained 10 mol of an ideal gas,
how much work has been done during the
compression? (d=3)
U = (3/2) nRT
Q = 0 and
U = Q + W
W = U = (3/2) nR T = (3/2)(10)(8.31)(100) = 1.25x104 J
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Clicker Quiz!
A vertical cylinder with a movable cap
is cooled. The process corresponding
to this is:
a)
b)
c)
d)
e)
CB
AB
AC
CA
Not shown
After the cooling of the gas and the lid
has come to rest, the pressure is the
same as before the cooling process.
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Adiabatic process
An molecular hydrogen gas goes from
P1 = 9.26 atm and V1 = 0.0118 m3 to
P2 and V2 via an adiabatic process.
If P2 = 2.66 atm, what is V2 ?
H2 (d=5) and adiabatic:
PV =Constant with  = Cp/Cv = (d+2)/d= 7/5
P1 (V1)1.4 = P2 (V2)1.4
(V2)1.4 = (P1 /P2)(V1)1.4 = 0.0069
V2 = 0.00690.714 = 0.029
(1/1.4) = 0.714)
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Cyclic processes (monatomic with d=3)
P (Pa)
A
5.0
1.0
C
10
B
50
V (m3)
In a cyclic process,
The system returns to its original state.
Therefore, the internal energy must be the same after
completion of the cycle [U = (3/2) PV and U=0]
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Cyclic Process: Step by Step (1)
Process A to B
Negative work is done on the gas:
(the gas is doing positive work).
P (Pa)
A
5.0
1.0
W= - Area under P-V diagram
C
10
B
50
V (m3)
= - [ (50-10)(1.0-0.0)
+½(50-10)(5.0-1.0) ]
= - 40 - 80
= - 120 J (work done on gas)
Wg = 120 J (work done by gas)
U = 3/2 (PBVB - PAVA)
= 1.5[(1)(50) - (5)(10)] = 0
The internal energy has not changed
U=Q+W so Q = U-W = 120 J
Heat that was added to the system was used to do the work!
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Cyclic Process: Step by Step (2)
P (Pa)
A
5.0
1.0
Process B-C
W = Area under P-V diagram
C
10
= - [(50-10)(1.0-0.0)]
W=40 J
Work was done on the gas
B
50
V (m3)
U = 3/2(PcVc-PbVb)
= 1.5[(1)(10) - (1)(50)] = - 60 J
The internal energy has decreased by 60 J
U=Q+W so
Q = U-W = - 60 - 40 J = - 100 J
100 J of energy has been transferred out of the system.
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Cyclic Process: Step by Step (3)
P (Pa)
A
5.0
1.0
Process C-A
W=-Area under P-V diagram
W=0 J
No work was done on/by the gas.
C
10
B
50
V (m3)
U = 3/2(PcVc-PbVb)=
= 1.5[ (5)(10) - (1)(10) ] = 60 J
The internal energy has increased by 60 J
U=Q+W so Q = U-W = 60-0 J = 60 J
60 J of energy has been transferred into the system.
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Summary of the process
Quantity
Process
P (Pa)
A
5.0
1.0
B
C
10
50
A-B
V (m3)
Work on
gas (W)
Heat(Q)
U
A-B
-120 J
120 J
0J
B-C
40 J
-100 J
-60 J
C-A
0J
60 J
60 J
-80 J
80 J
0
SUM
(net)
B-C
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C-A
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P (Pa)
A
5.0
1.0
What did we do?
C
10
B
50
Quantity
Process
Work on
gas (W)
Heat(Q)
A-B
-120 J
120 J
0J
B-C
40 J
-100 J
-60 J
C-A
0J
60 J
60 J
-80 J
80 J
0
SUM
(net)
V (m3)
U
The gas performed net work (80 J) (Wg = -W)
while net heat was supplied (80 J):
We have built an engine that converts heat
energy into work!
When the path on the P-V diagram is clockwise work is done by the gas
(engine) – heat engine
The work done by the gas is equal to the area of the loop
Wg = (5-1)(50-10)/2 = 80
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P (Pa)
Quantity
Process
A
5.0
1.0
B
C
10
50
Work on
gas (W)
Heat(Q)
U
A-B
-120 J
120 J
0J
B-C
40 J
-100 J
-60 J
C-A
0J
60 J
60 J
-80 J
80 J
0
SUM
V (m3) (net)
Qh = 180 heat input from hot source
Qc = 100 heat output to cold source (wasted heat)
Wg = -W = Qh – Qc = 80 work output by gas (engine)
efficiency e = Wg /Qh = 80/180 = 0.4444
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Generalized Heat Engine
Water turned to steam
Wg = Qh - Qc
Heat reservoir Th
efficiency: Wg/Qh
Qh (heat input)
Wg
e = 1 - Qc/Qh
Work is done
Work
engine
The steam
moves a piston
Qc (heat output)
The steam is
condensed
Cold reservoir Tc
The efficiency is determined
by how much of the heat you
supply to the engine is turned
into work instead of being lost
as waste.
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Reverse Direction: The Fridge
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Heat
Pump
(fridge)
heat is expelled to outside
heat reservoir Th
Qh
a piston compresses the coolant work is done
engine
Qc
the fridge is cooled
cold reservoir Tc
W
work
Coefficient of performance
COP = |Qc|/W
Qc: amount of heat removed
W: work input
W= Qh - Qc
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P (Pa)
A
5.0
1.0
On the P-V diagram the heat pump
(fridge) is given by a path that goes
counter clockwise.
B
C
10
50
The area inside the loop is the amount
of work done on the gas to remove
heat from the cold source.
V (m3)
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Clicker Quiz!
P (Pa)
3x105
Consider this clockwise cyclic
process.
Which of the following is true?
1x105
1
a)
b)
c)
d)
e)
3 V (m3)
This is a heat engine and the work done by the gas is +4x105
This is a heat engine and the work done by the gas is +6x105
This is a heat engine and the work done by the gas is –4x105
This is a fridge and the work done on the gas is +4x105 J
This is a fridge and the work done on the gas is +6x105 J
Clockwise: work done by the gas, so heat engine
Work by gas=area enclosed = (3-1) x (3x105-1x105) = 4x105 J
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What is the most efficient
engine we can make given a
hot and a cold reservoir?
What is the best path to
take on the P-V diagram?
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AB isothermal expansion
BC adiabatic expansion
W-, Q+
W-, T-
Carnot engine
Q=0
T=0
Th
T=0
Q=0
W+, T+
DA adiabatic compression
W+, Q-
Tc
CD isothermal compression
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inverse Carnot cycle
Carnot cycle
Work done by engine: Weng
Weng = Qh - Qc
Efficiency: ecarnot = 1-(Tc/Th)
e = 1-(Qc/Qh) also holds since
this holds for any engine
A heat pump or a fridge!
By doing work we can
transport heat
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Carnot engine
e =1 - (Qc/Qh )
always
ecarnot =1 - (Tc/Th ) carnot only!!
In general: e < ecarnot
The Carnot engine is the most efficient way to operate
an engine based on hot/cold reservoirs because the
process is reversible: it can be reversed without loss
or dissipation of energy
Unfortunately, a perfect Carnot engine cannot be built.
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Example
The efficiency of a Carnot engine is 30%. The engine absorbs
800 J of heat energy per cycle from a hot reservoir at
500 K. Determine
a) the energy expelled per cycle and
b) the temperature of the cold reservoir
c) how much work does the engine do per cycle?
a) Generally for an engine: efficiency: e = 1 – (Qc/Qh)
Qc = Qh(1-e) = 800(1-0.3) = 560 J
b) for a Carnot engine: efficiency: e = 1 - (Tc/Th )
Tc = Th(1-e) = 500(1-0.3) = 350
c) W = Qh – Qc = 800 – 560 = 240 J
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The 2nd law of thermodynamics
1st law: U=Q+W
In a cyclic process (U=0) Q=-W: we cannot do more work
than the amount of energy (heat) that we put inside
2nd law in equivalent forms:
- Heat flows spontaneously ONLY from hot to cold masses
- Heat flow is accompanied by an increase in the entropy
(disorder) of the universe
- Natural processes evolve toward a state of maximum
entropy
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Entropy
Lower Entropy
Higher Entropy
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Reversing Entropy
We can only reverse the
increase in entropy if we
do work on the system
Do work to compress
the gas back to a
smaller volume
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Entropy
The CHANGE in entropy (S):
∆S =
Q
T
(J/K unit)
Adiabatic process Q=0 and S = 0
If heat flows out (Q < 0) then S < 0 entropy decreases
If heat flows in (Q > 0) then S > 0 entropy increases
For a Carnot engine, there is no change in entropy
over one complete cycle
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Entropy and Work
Entropy represents an
inefficiency wherein
energy is “lost” and cannot
be used to do work.
Shot = -Qhot/Thot
Scold = Qcold/Tcold
= -24000J / 400K = -60 J/K
= +24000J / 300K = +80 J/K
Shot + Scold = -60 J/K + 80 J/K = +20 J/K
Entropy increases!
Cold mass: Gained heat, can do more work.
Hot mass: Lost heat, can do less work.
Cold mass gained less potential to do work than host mass lost.
Net loss in the ability to do work.
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Review: calorimetry
If we connect two objects with different temperature
energy will transferred from the hotter to the cooler
one until their temperatures are the same.
If the system is isolated:
Energy flow into cold part
= Energy flow out of hot part
mc cc ( Tf - Tc) = mh ch (Th - Tf)
the final temperature is:
Tf = mc cc Tc + mh ch Th
mc cc + mh ch
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Phase Change
GAS(high T) Q=cgasmT
Gas 
liquid
Q=mLv
Q=csolidmT Solid (low T)
liquid (medium T)
Q=cliquidmT
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liquid 
solid
Q=mLf
58
Heat transfer via conduction
Tc
Th
Conduction occurs if there is a
temperature difference between
two parts of a conducting medium
Rate of energy transfer P
A
P = Q/t (unit Watt = J/s)
P = k A (Th-Tc)/x = k A T/x
k: thermal conductivity
Unit: J/(m s oC)
x
Metals
Gases
Nonmetals
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k~300 J/(m s oC)
k~0.1 J/(m s oC)
k~1
J/(m s oC)
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Multiple Layers
Th k
1
Tc
k2
A
L1
Q A(Th  Tc )
P

t  ( Li / ki )
i
L2 (x)
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Net Power Radiated (photons)
An object emits AND receives radiation,
energy radiated per second = net power radiated (J/s)
PNET =  A e (T4-T04)
= Power radiated – Power absorbed
where
T: temperature of object (K)
T0: temperature of surroundings (K)
 = 5.6696x10-8 W/m2K4
A = surface area
e = object dependent constant emissivity (0-1)
for a black body e=1 (all incident radiation is absorbed)
MSU Physics 231 Fall 2015
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Wavelength where the radiant energy
is maximum
where b=2.90×10−3 m K
Wiens displacement constant
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