PHY 113 C General Physics I
11 AM – 12:15 PM MWF Olin 101
1.
2.
3.
4.
Plan for Lecture 23:
Chapter 22: Heat engines
Thermodynamic cycles; work and heat
efficiency
Carnot cycle
Otto cycle; diesel cycle
Brief comments on entropy
11/19/2013
PHY 113 C Fall 2013 -- Lecture 23
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11/19/2013
PHY 113 C Fall 2013 -- Lecture 23
2
Comment about Exam 3:
• Part I – take home portion (1 problem):
available Thursday 11/21/2013 after class;
must be turned in before Part II – in-class
portion (3 problems): Tuesday 11/25/2013
• Some special arrangements for early exams
have been (or will be) arranged by prior
agreement
• Of course, all sections of the exam are to be
taken under the guidelines of the honor code
11/19/2013
PHY 113 C Fall 2013 -- Lecture 23
3
Important equations for macroscopic and microscopic
descriptions of thermodynamic properties of matter
First Law of Thermodynamics : ΔEint  Q  W
Vf
Thermodynamic Work :
W    PdV
Vi
Ideal Gas Law : PV  nRT
Microscopic analysis of gas molecules :
(assume N molecules of mass m0 or n moles of molar mass M )
2 1
2 1
2 
2 
PV  N  m0 vrms   n Mvrms   nRT
3 2
 3 2

1
2
 Mvrms
 RT
3
nRT
For molecules with   C P / CV : Eint 
 1
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PHY 113 C Fall 2013 -- Lecture 23
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Webassign – Assignment 20
The rms speed of an oxygen molecule (O2) in a container
of oxygen gas is 563 m/s. What is the temperature of the
gas?
From the kinetic analysis of gas molecules :
(assume N molecules of mass m0 or
n moles of molar mass M )
2 1
2 1
2 
2 
PV  N  m0 vrms   n Mvrms   nRT
3 2
 3 2

1
2
 Mvrms
 RT
3
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PHY 113 C Fall 2013 -- Lecture 23
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Webassign – Assignment 20
In a constant-volume process, 213 J of energy is
transferred by heat to 0.99 mol of an ideal monatomic gas
initially at 299 K.
(a) Find the work done on the gas.
For constant volume process, W=0.
(b) Find the increase in internal energy of the gas.
DEint = Q + 0 = 213J + 0 = 213 J
(c) Find its final temperature.
Q  nCV T f  Ti 
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PHY 113 C Fall 2013 -- Lecture 23
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Webassign – Assignment 20
A 2.00-mol sample of a diatomic ideal gas expands slowly
and adiabatically from a pressure of 5.06 atm and a
volume of 12.2 L to a final volume of 29.6 L.
(a) What is the final pressure of the gas?
(b) What are the initial and final temperatures?
(c) Find Q for the gas during this process.
(d) Find ΔEint for the gas during this process.
(e) Find W for the gas during this process.
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PHY 113 C Fall 2013 -- Lecture 23
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Digression:
Adiabatic process (Q=0)
DEint  W
n
RDT   PDV
γ-1
PV  nRT
DPV  PDV  nRDT
nRDT  γ-1PDV  DPV  PDV
DV DP
γ

V
P
 V fγ 
 Pf 


  ln γ  ln   PiVi γ  Pf V fγ
V 
 Pi 
 i 
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PHY 113 C Fall 2013 -- Lecture 23
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Webassign – Assignment 20
A 2.00-mol sample of a diatomic ideal gas expands slowly
and adiabatically from a pressure of 5.06 atm and a
volume of 12.2 L to a final volume of 29.6 L.
For diatomic ideal gas:   1.4
(a) What is the final pressure of the gas?
PiVi γ  Pf V fγ
Pf  Pi Vi / V f   5.06atm12.2/29.6 

1.4
b) What are the initial and final temperatures?
PV=nRT
c) Find Q for the gas during this process. Q=0
d) Find ΔEint for the gas during this process. ΔEint=W
e) Find W for the gas during this process.
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PHY 113 C Fall 2013 -- Lecture 23
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Webassign – Assignment 20
(a) How much work is required to compress 4.95 mol of
air at 19.6°C and 1.00 atm to one-tenth of the original
volume by an isothermal process?
(b) How much work is required to produce the same
compression in an adiabatic process?
(c) What is the final pressure in part (a)?
(d) What is the final pressure in part (b)?
11/19/2013
PHY 113 C Fall 2013 -- Lecture 23
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Webassign – Assignment 20
(a) How much work is required to compress 4.95 mol of
air at 19.6°C and 1.00 atm to one-tenth of the original
volume by an isothermal process?
Vf
W    PdV
Vi
nRT
For an isothermal process : P 
V
Vf
nRT
W  
dV  nRT ln
V
 Vi
Vi
Vf



1
W  4.95  8.314  292.75  ln 
 10 
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PHY 113 C Fall 2013 -- Lecture 23
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Webassign – Assignment 20
(b) How much work is required to compress 4.95 mol of
air at 19.6°C and 1.00 atm to one-tenth of the original
volume by an adiabatic process? Note: assume 1.4
For an adiabatic process : DEint  W  nC V T f  Ti 
For an adiabatic process : T f V f 1  TiVi  1
 1
0.4
 Vi 
 10 


T f  Ti
 292.74   
V 
1
 f 
R
CV 
 2.5 R
 1
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PHY 113 C Fall 2013 -- Lecture 23
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Thermodynamic cycles for designing ideal engines
and heat pumps
http://auto.howstuffworks.com/engine1.htm
Engine process:
P (1.013 x 105) Pa
Pf
Work of engine :
B
C
Heat input to system : Q  Qin  Qout
Efficiency :
A
Pi
Vi
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Weng  W
D

Weng
Qin
Vf
PHY 113 C Fall 2013 -- Lecture 23
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Examples process by an ideal gas:
P (1.013 x 105) Pa
Pf
B
C
AB
Q
W
A
Pi
D
DEint
BC
γPf (V f  Vi )
 V f ( Pf  Pi )
-γPi (V f  Vi )
γ -1
γ -1
γ -1
γ -1
0
-Pf(Vf-Vi)
0
Pi(Vf-Vi)
Vi ( Pf  Pi )
Pf (V f  Vi )
 V f ( Pf  Pi )
-Pi (V f  Vi )
γ -1
γ -1
γ -1
γ -1
Vf

Weng
Qin

P

f
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DA
Vi ( Pf  Pi )
Efficiency :
Vi
CD
PHY 113 C Fall 2013 -- Lecture 23
 Pi V f  Vi 
QAB  QBC
14
Example from homework

Efficiency :
Weng
Qin

P

 Pi V f  Vi 
f
QAB  QBC
Also :  W  Weng  Q  Qin  Qout

Qin  Qout
  1
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PHY 113 C Fall 2013 -- Lecture 23
Qin
 1
Qout
Qin
QCD  QDA
QAB  QBC
15
Most efficient thermodynamic cycle -- Carnot
Sadi Carnot 1796-1832
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PHY 113 C Fall 2013 -- Lecture 23
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Carnot cycle:
AB
BC
CD
DA
Isothermal at Th
Adiabatic
Isothermal at Tc
Adiabatic
Efficiency of Carnot cycle

Qin  Qout
Qin
 1
Qout
Qin
Tc
ε  1
Th
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PHY 113 C Fall 2013 -- Lecture 23
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iclicker exercise:
We discussed the efficiency of an engine as

Qin  Qout
Qin
 1
Qout
Qin
Is this result
A. Special to the Carnot cycle
B. General to all ideal thermodynamic cycles
iclicker exercise:
We discussed the efficiency of an engine running
with hot and cold reservoirs as
T
  1 c
Th
Is this result
A. Special to the Carnot cycle
B. General to all ideal thermodynamic cycles
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PHY 113 C Fall 2013 -- Lecture 23
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Note that for a Carnot cycle :
Qout
Qin

WAB
WCD
nRTc ln(VC / VD )

nRTh ln(VB / VA )
For adiabatic process
ThVB 1  TcVC 1
ThVA 1  TcVD 1
 VC / VD  VB / VA

Qout
Qin

Tc
Th
For Carnot cycle:
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Tc
  1
Th
PHY 113 C Fall 2013 -- Lecture 23
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iclicker exercise:
Why should we care about the Carnot
cycle?
A. We shouldn’t
B. It approximately models some
heating and cooling technologies
C. It provides insight into another
thermodynamic variable -- entropy
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PHY 113 C Fall 2013 -- Lecture 23
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PHY 113 C Fall 2013 -- Lecture 23
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Webassign Assignment 21
A heat engine operates between a reservoir at 28°C
and one at 362°C. What is the maximum efficiency
possible for this engine?
The Carnot cycle is the most efficient process operating
between hot and cold temperatures :
Tc
273.15  28
  1  1
Th
273.15  362
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PHY 113 C Fall 2013 -- Lecture 23
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Webassign Assignment 21
An ideal gas is taken through a Carnot cycle. The isothermal
expansion occurs at 260°C, and the isothermal compression
takes place at 50.0°C. The gas takes in 1.28 x103 J of energy
from the hot reservoir during the isothermal expansion.
(a) Find the energy expelled to the cold reservoir in each
cycle.
Qc
Tc
273.15  50
  1  1
 1
Th
273.15  260
Qh
(b) (b) Find the net work done by the gas in each cycle.
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PHY 113 C Fall 2013 -- Lecture 23
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The Otto cycle
Theoretical efficiency :
 V2 
  1   
 V1 
 1
V1/V2 is the
“compression ratio”
-- typically V1/V2 = 8
 =0.56
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PHY 113 C Fall 2013 -- Lecture 23
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The Diesel cycle
Theoretical efficiency :
1  TD  TA 

  1  
  TC  TB 
In principle, higher
efficiency than
comparable Otto cycle.
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PHY 113 C Fall 2013 -- Lecture 23
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Engine vs heating/cooling designs
Heat pump; heating mode :

Qh
W
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
Th
(Carnot)
Th  Tc
Heat pump; cooling mode :

Qc
W

Tc
(Carnot)
Th  Tc
PHY 113 C Fall 2013 -- Lecture 23
27
Brief comments about entropy – macroscopic picture
Carnot cycle
Note that for a Carnot cycle :
Qc
Qh

WAB
WCD

nRTc ln(VC / VD )
nRTh ln(VB / VA )
For adiabatic process
ThVB 1  TcVC 1
ThVA 1  TcVD 1
 VC / VD  VB / VA

Qc
Tc
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Qc
Qh

PHY 113 C Fall 2013 -- Lecture 23

Tc
Th
Qh
Th
28
Brief comments about entropy – continued
For a Carnot cycle :
Qc
Tc

Qh
Th
Define :
dS 
dQ
T
For a Carnot cycle S cycle 
Qh
Th

Qc
Tc
0
 S " state variable" (like DEint )
Other examples of entropy :
Qf
dQ mL fusion
Change of entropy while melting : DS  

T
Tmelting
Qi
For melting 1 kg water at 0o C (273.15o K) DS 
11/19/2013
PHY 113 C Fall 2013 -- Lecture 23
333000 J
 1219 J / K
o
273.15 K
29