Announcements
1. Please bring your laptops to lab this week.
2. Physics seminar this week – Thursday, Nov. 20 at 4 PM –
Professor Brian Matthews will discuss relationships
between protein structure and function
3. Schedule –
Today, Nov. 18th: Examine Eint especially for ideal gases
Discuss ideal gas law
Continue discussion of first law of thermodynamics
Thursday, Nov. 20th: Review Chapters 15-20
Tuesday, Nov. 25th: Third exam
6/23/2016
PHY 113 -- Lecture 20
1
From The New Yorker Magazine, November 2003
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PHY 113 -- Lecture 20
2
First law of thermodynamics
DEint = Q - W
Vf
W   PdV
Vi
For an “ideal gas” we can write an explicit relation for Eint.
What we will show:
( ideal gas )
Eint
n
N

RT 
k BT
γ-1
γ-1
g is a parameter which depends on the type of gas
(monoatomic, diatomic, etc.) which can be measured as the
ratio of two heat capacities: g=CP/CV.
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PHY 113 -- Lecture 20
3
Thermodynamic statement of conservation of energy –
First Law of Thermodynamics
DEint = Q - W
Work done by system
Wif depends on path
Heat added to system
Qif depends on path
“Internal” energy of system
DEint = Eint(f)-Eint(i)  independent on path
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PHY 113 -- Lecture 20
4
How is temperature related to Eint?
Consider an ideal gas
 Analytic expressions for physical variables
 Approximates several real situations
Ideal Gas Law:
PV = n R T
volume (m3)
pressure (Pa)
temperature (K)
gas constant (8.31 J/(moleK))
number of moles
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PHY 113 -- Lecture 20
5
Ideal gas – P-V diagram at constant T
PV = n R T
Pi
Pf
Pi Vi = Pf Vf
Vi
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Vf
PHY 113 -- Lecture 20
6
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PHY 113 -- Lecture 20
7
Microscopic model of ideal gas:
Each atom is represented as a tiny hard sphere of mass m
with velocity v. Collisions and forces between atoms are
neglected. Collisions with the walls of the container are
assumed to be elastic.
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PHY 113 -- Lecture 20
8
What we can show is the pressure exerted by the atoms by
their collisions with the walls of the container is given by:
2N 1
2N
2
P
K avg
2 m v avg 
3V
3V
Proof:
Force exerted on wall perpendicular to x-axis by an atom
which collides with it:
Dp
2m v
Fix  - ix  i ix Dt  2d / vix
-vix
Dt
Dt
2mi vix mi vix2
 Fix 

2d / vix
d
vix
number of atoms
2
Fix
mi vix N
d
P 

mi vx2
A
dA V
i
i
average over atoms
volume
6/23/2016
PHY 113 -- Lecture 20
x
9
Ideal gas law continued:
Recall that
N
1N
2 N  1R 2 
2
2
P

m
v

m
v

Macroscopi c relation
: PV  nRT  N m Tv  Nk BT
x
V
3V
3 V  2N A

2 1

Microscopi c model : PV  N  m v 2   Nk BT
3 2

Therefore:  1
3
2 
 m v   k BT
2
 2
 vrms 
Also:
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v
2
3k BT

m
1 m v 2   3 k T


B
2
2


1
3

2 
 Eint  N  m v   N k BT
2
2

PHY 113 -- Lecture 20
10
Big leap!
Internal energy of an ideal gas:
1
3
N
n
Eint  N  m v 2   N k BT 
k BT 
RT
2
γ-1
γ-1
2

derived for monoatomic
ideal gas
Gas
He
N2
H2O
6/23/2016
more general relation for
polyatomic ideal gas
g (theory)
5/3
7/5
4/3
PHY 113 -- Lecture 20
g(exp)
1.67
1.41
1.30
11
Determination of Q for various processes in an ideal gas:
n
Eint 
RT
γ-1
n
DEint 
RDT  Q - W
γ-1
Example: Isovolumetric process – (V=constant  W=0)
n
DEint i  f 
RDTi f  Qi f
γ-1
n
In terms of “heat capacity”: Qi  f 
RDTi  f  nCV DTi  f
γ-1
R
CV 
γ-1
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PHY 113 -- Lecture 20
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Example: Isobaric process (P=constant):
DEint i f
n

RDTi  f  Qi f - Wi f
γ-1
In terms of “heat capacity”:
Qi  f 
n
n
RDTi  f  Pi (V f - Vi  
RDTi  f  nRDTi  f  nC P DTi  f
γ-1
γ-1
R
γR
 CP 
R
γ-1
γ-1
Note: g = CP/CV
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PHY 113 -- Lecture 20
13
More examples:
Isothermal process (T=0)
n
RT
γ-1
n

RDT  Q - W
γ-1
Eint 
DEint
DT=0  DEint = 0  Q=W
V f
dV
W   PdV  nRT 
nRT ln
Vi
Vi V
 Vi
Vf
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Vf
PHY 113 -- Lecture 20



14
Even more examples:
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
V 
 Pi
 i 
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


 PiVi γ  Pf V fγ
PHY 113 -- Lecture 20
15
PiVi g
Adiabat : P  g
V
PiVi
Isotherm : P 
V
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PHY 113 -- Lecture 20
16
Peer instruction question
Suppose that an ideal gas expands adiabatically. Does
the temperature
(A) Increase
(B) Decrease
(C) Remain the same
PiVi γ  Pf V fγ
Ti
PiVi  nRTi  Pi  nR
Vi
TiVi γ-1  T f V fγ-1
 Vi
T f  Ti 
V f
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



γ -1
PHY 113 -- Lecture 20
17
Review of results from ideal gas analysis in terms of the
specific heat ratio g  CP/CV:
n
R
DEint 
RDT  nCV DT ; CV 
γ-1
γ-1
γR
CP 
γ-1
For an isothermal process, DEint = 0  Q=W
Vf
V f 
V f 
W   PdV nRT ln   PiVi ln 
Vi
 Vi 
 Vi 
For an adiabatic process, Q = 0
PiVi γ  Pf V fγ
TiVi γ-1  T f V fγ-1
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PHY 113 -- Lecture 20
18
Extra credit:
Show that the work done by an ideal gas which
has an initial pressure Pi and initial volume Vi
when it expands adiabatically to a volume Vf is
given by:
γ -1
Vf

PiVi   Vi  
W   PdV 
1-  
γ - 1  V f  
Vi


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PHY 113 -- Lecture 20
19
Peer instruction questions
Match the following types of processes of an ideal gas with
their corresponding P-V relationships, assuming the initial
pressures and volumes are Pi and Vi, respectively.
1. Isothermal
2. Isovolumetric
3. Isobaric
4. Adiabatic
(A) P=Pi
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(B) V=Vi
(C) PV=PiVi
PHY 113 -- Lecture 20
(D)PVg=PiVig
20
P (1.013 x 105) Pa
Examples process by an ideal gas:
Pf
B
C
AB
Q
W
A
Pi
Vi
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D
Vf
CD
DA
Vi ( Pf - Pi )
γPf (V f - Vi )
- V f ( Pf - Pi )
-γPi (V f - Vi )
γ -1
γ -1
γ -1
γ -1
0
DEint
BC
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
Efficiency as an engine:
e = Wnet/ Qinput
PHY 113 -- Lecture 20
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PHY 113 -- Lecture 20
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