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
11/18/2003
PHY 113 -- Lecture 20
1
From The New Yorker Magazine, November 2003
11/18/2003
PHY 113 -- Lecture 20
2
First law of thermodynamics
∆Eint = 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
γ is a parameter which depends on the type of gas
(monoatomic, diatomic, etc.) which can be measured as the
ratio of two heat capacities: γ=CP/CV.
11/18/2003
PHY 113 -- Lecture 20
3
Thermodynamic statement of conservation of energy –
First Law of Thermodynamics
∆Eint = Q - W
Work done by system
Wif depends on path
Heat added to system
Qif depends on path
“Internal” energy of system
∆Eint = Eint(f)−Εint(i) Î independent on path
11/18/2003
PHY 113 -- Lecture 20
4
How is temperature related to Eint?
Consider an ideal gas
Î Analytic expressions for physical variables
Î Approximates several real situations
Ideal Gas Law:
PV=nRT
volume (m3)
pressure (Pa)
temperature (K)
gas constant (8.31 J/(mole ⋅K))
number of moles
11/18/2003
PHY 113 -- Lecture 20
5
Ideal gas – P-V diagram at constant T
PV=nRT
Pi
Pf
Pi Vi = Pf Vf
Vi
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Vf
PHY 113 -- Lecture 20
6
11/18/2003
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.
11/18/2003
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:
∆pix 2mi vix
Fix = −
=
∆t ≈ 2d / vix
-vix
∆t
∆t
2mi vix mi vix2
⇒ Fix ≈
=
2d / vix
d
vix
number of atoms
2
mi vix N
Fix
d
=
mi v x2
P=∑ =∑
A
dA V
i
i
average over atoms
volume
11/18/2003
PHY 113 -- Lecture 20
x
9
Ideal gas law continued:
Recall that
1N
2 N ⎧ 1R 2 ⎫
N
2
2
P
m
v
m
v
=
=
=
x
Macroscopic relation
: PV = nRT = N⎨ m Tv = ⎬Nk BT
3V
3 V ⎩ 2N
V
⎭
A
2
1
Microscopic model : PV = N ⎧⎨ m v 2 ⎫⎬ = Nk BT
3 ⎩2
⎭
Therefore: ⎧ 1
3
2 ⎫
⎨ m v ⎬ = k BT
⎩2
⎭ 2
⇒ vrms =
Also:
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
⎭
11/18/2003
PHY 113 -- Lecture 20
10
Big leap!
Internal energy of an ideal gas:
1
3
N
n
k BT =
RT
Eint = N ⎧⎨ m v 2 ⎫⎬ = N k BT ⇒
2
γ-1
γ-1
⎩2
⎭
derived for monoatomic
ideal gas
Gas
He
N2
H2O
11/18/2003
more general relation for
polyatomic ideal gas
γ (theory)
5/3
7/5
4/3
PHY 113 -- Lecture 20
γ (exp)
1.67
1.41
1.30
11
Determination of Q for various processes in an ideal gas:
n
Eint =
RT
γ-1
n
∆Eint =
R∆T = Q − W
γ-1
Example: Isovolumetric process – (V=constant Î W=0)
n
∆Eint i → f =
R∆Ti → f = Qi → f
γ-1
n
In terms of “heat capacity”: Qi → f =
R∆Ti → f ≡ nCV ∆Ti → f
γ-1
R
CV =
γ-1
11/18/2003
PHY 113 -- Lecture 20
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Example: Isobaric process (P=constant):
∆Eint i → f
n
=
R∆Ti → f = Qi → f − Wi → f
γ-1
In terms of “heat capacity”:
Qi → f =
n
n
R∆Ti → f + Pi (V f − Vi ) =
R∆Ti → f + nR∆Ti → f ≡ nC P ∆Ti → f
γ-1
γ-1
R
γR
⇒ CP =
+R=
γ-1
γ-1
Note: γ = CP/CV
11/18/2003
PHY 113 -- Lecture 20
13
More examples:
Isothermal process (T=0)
n
RT
γ-1
n
=
R∆T = Q − W
γ-1
Eint =
∆Eint
Î∆T=0 Î ∆Eint = 0 Î Q=W
⎛V f
dV
W = ∫ PdV = nRT ∫
=nRT ln⎜⎜
Vi
Vi V
⎝ Vi
Vf
11/18/2003
Vf
PHY 113 -- Lecture 20
⎞
⎟⎟
⎠
14
Even more examples:
Adiabatic process (Q=0)
∆Eint = −W
n
R∆T = − P∆V
γ-1
PV = nRT
∆PV + P∆V = nR∆T
nR∆T = −(γ-1)P∆V = ∆PV + P∆V
∆V ∆P
−γ
=
P
V
⎛ V fγ ⎞
⎛ Pf ⎞
⎟
⎜
⇒ − ln γ = ln⎜⎜ ⎟⎟ ⇒ PiVi γ = Pf V fγ
⎜V ⎟
⎝ Pi ⎠
⎝ i ⎠
11/18/2003
PHY 113 -- Lecture 20
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PiVi γ
Adiabat : P = γ
V
PiVi
Isotherm : P =
V
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PHY 113 -- Lecture 20
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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
11/18/2003
⎞
⎟
⎟
⎠
γ -1
PHY 113 -- Lecture 20
17
Review of results from ideal gas analysis in terms of the
specific heat ratio γ ≡ CP/CV:
n
R
∆Eint =
R∆T = nCV ∆T ; CV =
γ-1
γ-1
γR
CP =
γ-1
For an isothermal process, ∆Eint = 0 Î 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
11/18/2003
PHY 113 -- Lecture 20
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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
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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
11/18/2003
(B) V=Vi
(C) PV=PiVi
PHY 113 -- Lecture 20
(D)PVγ=PiViγ
20
P (1.013 x 105) Pa
Examples process by an ideal gas:
Pf
B
C
A→B
Q
A
Pi
Vi
11/18/2003
D
Vf
W
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
∆Eint
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:
PHY 113 -- Lecture 20
e = Wnet/ Qinput
21
11/18/2003
PHY 113 -- Lecture 20
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