Physics 207 – Lecture 23
Lecture 23
Goals:
• Wrap up Chapter 16 and start Chapter 17
• Assignment
v HW-9 due Tuesday, Nov 22
v Wednesday: Read Chapter 17
v Third test on Thursday, December 1
Physics 207: Lecture 23, Pg 1
Ideal Gas Law
l Assumptions
that we will make:
v “hard sphere” model for the atoms
v density is low
v temperature not too high
PV=nRT
n: # of moles
R=8.31 J/mol K: universal gas constant
Physics 207: Lecture 23, Pg 2
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Physics 207 – Lecture 23
l Suppose
that we have a sealed container at a
fixed temperature. If the volume of the container
is doubled, what would happen to pressure?
A)Double
B)Remain the same
C)Halved
D)None of the above
PV=nRT
P V = constant if n, T fixed
Physics 207: Lecture 23, Pg 3
Boltzmann’s constant
PV=nRT
n: # of moles
n=N/NA
P V = n R T=(N/NA) R T
=N (R/NA) T
P V= N kB T
kB: Boltzmann’s constant
Physics 207: Lecture 23, Pg 4
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Physics 207 – Lecture 23
Pressure
PV diagrams
3 atm
1 atm
1 Liter
3 Liters
Volume
Physics 207: Lecture 23, Pg 5
PV diagrams: Important processes
lIsochoric process: V = const (aka isovolumetric)
lIsobaric process: P = const
lIsothermal process: T = const
Physics 207: Lecture 23, Pg 6
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Physics 207 – Lecture 23
l Which
one of the following PV diagrams describe
an isobaric process (P=constant)
B)
1
Pressure
Pressure
2
p1 p 2
=
T1 T2
C)
p1V1 = p2V2
1
Pressure
A)
V1 V2
=
T1 T2
1
2
2
Volume
Volume
Volume
PV constant,
isothermal
V constant,
isochoric
P constant,
isobaric
Physics 207: Lecture 23, Pg 7
Work done on a gas
∆x
∆W = Fext ∆x
= P A ∆x
= P ∆V
Fext=P A
∆V=Vfinal-Vinitial
∆W=-P ∆V
Physics 207: Lecture 23, Pg 8
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Physics 207 – Lecture 23
Work done on a gas
Pressure
∆W=-P ∆V
W= - the area under the P-V curve
W= the area under the F-x curve
1
2
Volume
Physics 207: Lecture 23, Pg 9
an isochoric process (V=const), W=0
2
Pressure
l For
1
Volume
Physics 207: Lecture 23, Pg 10
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Physics 207 – Lecture 23
l How
does the work done on the gas compare for
the below two processes?
Pressure
A) |W1|>|W2|
B) |W1|=|W2|
C) |W1|<|W2|
1
2
Volume
Physics 207: Lecture 23, Pg 11
Work and Energy Transfer
l When
you do work on a system, the energy of the
system increases:
Potential Energy (∆U)
Kinetic Energy (∆K)
Thermal Energy (∆Ethermal)
∆U+∆K+∆Ethermal=∆Esystem=Wexternal
Physics 207: Lecture 23, Pg 12
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Physics 207 – Lecture 23
l This
description is incomplete. Energy can be
transferred without doing any work.
Q: Thermal energy transfer
∆U+∆K+∆Ethermal=∆Esystem=Wexternal+Q
Q>0, environment is at a higher temperature
Q<0, environment is at a lower temperature
Physics 207: Lecture 23, Pg 13
First Law of Thermodynamics
∆U+∆K+∆Ethermal=∆Esystem=Wexternal+Q
l For
systems where there is no change in
mechanical energy:
∆Ethermal =Wexternal+Q
l1
calorie=4.186 Joules
l1
food calorie=1000 calories
Physics 207: Lecture 23, Pg 14
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Physics 207 – Lecture 23
Work on system, W>0
Work by system, W<0
Environment
System
Energy in
Energy out
Heat to system, Q>0
Heat from system, Q<0
Physics 207: Lecture 23, Pg 15
Pressure
Isochoric process (const V)
• Work is: -(the area under the curve)
Pi
W=0
• Using ideal gas law:
Pf
PV=nRT
Tf < Ti
Volume
• From first law of thermodynamics we conclude:
∆Ethermal =W+Q<0
Q<0
Physics 207: Lecture 23, Pg 16
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Physics 207 – Lecture 23
Pressure
Isothermal expansion (const T)
• Temperature does not change
Pi
∆Ethermal=0
• Work done on the gas is:
Pf
Vi
Vf
W<0
Volume
• From first law of thermodynamics we conclude:
∆Ethermal =W+Q=0
Q=-W
Physics 207: Lecture 23, Pg 17
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