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Essential Physics II
英語で物理学の
エッセンス II
Lecture 4: 19-10-15
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Last week’s homework:
2015 / 10 / 26
(１０月２６日)
This week’s homework:
Also due:
2015 / 10 / 26
(１０月２６日)
on http://masteringphysics.com
Last lectures...
Temperature
difference, T
internal energy
increases, U
Eheat
moving energy: heat
Eheat
Ideal gases:
Ideal gas law:
pV = nRT
# mols
T measures K for random motion of molecules:
R = 8.314 J/K · mol
3
1 ¯2
kT = mv
2
2
Last lectures...
Quiz
An ideal gas has volume,V and pressure, p.
The thermal speed of the gas molecules is v.
If both volume and pressure are doubled to 2V and 2p, what is
the thermal speed of the gas molecules?
(a)
v
(b)
2v
(c)
4v
(d)
v/2
(e)
v/4
Last lectures...
Quiz
An ideal gas has volume,V and pressure, p.
The thermal speed of the gas molecules is v.
If both volume and pressure are doubled to 2V and 2p, what is
the thermal speed of the gas molecules?
p1 V 1
T1 =
Ideal gas law: pV = nRT
(a) v
nR
p2 V 2
4p1 V1
= 4T1
T2 =
=
(b) 2v
nR
nR
r
(c) 4v
3
1
3kT
Since: kT = mv¯2
v̄ =
m
2
2
(d) v/2
r
r
3k(4T1 )
3kT2
=
2
v
¯
=
v
¯
=
1
2
(e) v/4
m
m
Heat and Work
Joule’s apparatus
Quiz
You stir water (vigorously) with a spoon.
The water’s temperature, T ...
(A) increases
(B) decreases
(C) unchanged
No temperature difference between spoon and water:
No heat is transferred. Spoon does work on the water.
Mechanical work
increases water’s internal energy
T =0
Joule’s apparatus
Joule’s apparatus
Quiz
What is the temperature increase, T ?
(Assume all energy is converted to heat)
coil = 1800 J/kg · K
2 kg
(A) 10 C
1m
(B) 1 C
(C) 0.1 C
(D) 0.01 C
1 litre ~ 1 kg
(E) 0 C
Joule’s apparatus
Quiz
What is the temperature increase, T ?
(Assume all energy is converted to heat)
coil = 1800 J/kg · K
2 kg
Potential energy:
U = mg h
= (2 kg)(9.81 m/s2 )(1 m)
1m
= 19.62 J
Heat energy: Q = moil c T
1 litre ~ 1 kg
19.62 J
T =
' 0.01 K
(1 kg)(1800 J/kg · K)
= 0.01 C
(max)
Joule’s apparatus
Temperature can be increased by...
Heating:
Temperature difference between flame and water
T
Heat energy
Q = mc T
internal energy
higher T
Doing work:
Water is stirred
Mechanical energy
W
internal energy
higher T
Same final state: T is higher
First Law of Thermodynamics
Change in
internal energy
U =Q+W
heat transferred
work done on
the system
1st law of thermodynamics:
The change in internal energy of a system depends only on the
net heat transferred to the system and the net work done on the
system, independent of the particular processes involved.
How the energy moves is not important
First Law of Thermodynamics
U =Q+W
Rate of energy flow
dU
dQ dW
=
+
dt
dt
dt
Rate of change of
internal energy
Rate of heat transfer
Rate at which work
is done on the system
First Law of Thermodynamics
A power plant supplies
energy at a rate of 3.0 GW.
steam
U
3 GW
Q
Ex.
1 GW
W
drives turbine for electricity
Used steam is cooled in the water
If the power plant produces electrical energy at a rate of 1.0 GW,
what is the rate of heat transfer to the water?
dU
dQ dW
=
+
dt
dt
dt
dQ
dU
=
dt
dt
=
dW
=
dt
2.0 GW
3.0 GW
( 1.0 GW)
extracting
energy
doing
work
First Law of Thermodynamics
Quiz
A heat source supplies heat to a gas at a rate of 187.0 W
The gas does work at a rate of 130.9 W
What rate does the internal energy (dU/dt) of the gas change?
(a) 56.1 W
dU
dQ dW
=
+
dt
dt
dt
(b) 318 W
= 187 W
(c) -56.1 W
(d) 187 W
= 56.1W
130.9W
First Law of Thermodynamics
Quiz
A gas expands at constant T to twice (x2) its original volume.
During the expansion, the gas absorbs 200 kJ of heat.
What is the change in internal energy of the gas during the
expansion?
What measures internal energy (U) ?
(a)
0 kJ
(b)
100 kJ
(c)
200 kJ
(d)
400 kJ
Temperature (T).
Does T change?
No.
Therefore does U change?
No.
First Law of Thermodynamics
Quiz
A gas expands at constant T to twice (x2) its original volume.
During the expansion, the gas absorbs 200 kJ of heat.
How much work does the gas do during the expansion?
U =Q+W
(a)
0 kJ
0 = 200 kJ + W
(b)
100 kJ
(c)
200 kJ
work done on the gas: W =
(d)
400 kJ
work done by the gas:
200 kJ
200 kJ
Reversible & Irreversible
U =Q+W
First law applies to any system
but...
Ideal gas: pV = nRT
Only need 2 variables
[ ( p , V ) or ( V , T ) or ( T , p ) ]
can use pV diagram to show state:
p
p2 , V2 , T2
p1 , V1 , T1
V
Reversible Process
Gas
T
T
System: Gas ball immersed (covered by) water in
equilibrium
Water
If water T increases slowly:
Gas and water change T together
p
Always in equilibrium
p2 , V2 , T2
p1 , V1 , T1
V
Change follows curve
quasi-static process
If water T decreases slowly:
Change follows same curve
reversible process
p
p2 , V2 , T2
p1 , V1 , T1
V
Irreversible Process
Gas
T
T
System: Gas ball immersed (covered by) water in
equilibrium
Water
If water T increases quickly:
e.g. pour boiling water over cold gas ball
Gas and water NOT in equilibrium
Different p and T in different regions
p
p2 , V2 , T2
not well-defined values
irreversible process
p1 , V1 , T1
V
Work
Ideal gas: pV = nRT
A
Insulated cylinder
(no heat loss, Eout = 0 )
Volume can change
Pressure, p
Cross-section area, A
A
Reversible process
How much work is done on the gas?
p
Work
A
Force from gas: Fgas = pA
A
Gas does work:
x
p
Wgas = Fgas x
= pA x
p
=p V
Newton’s 3rd Law
Work done on the gas:
W =
Wgas =
But... p maybe changes as V changes: dW =
Z
Z V2
Total work: W = dW =
pdV
= - (area under pv curve )
p V
pdV
small volume change: p ~ constant
p
V1
(+ if gas is compressed V2 < V1 , - if expands V1 < V2 )
V
Work
Quiz
2 gas cylinders start and end in the same state.
(pi , Vi ) ! (pf , Vf )
p
They move between the 2 states by
different processes.
What is the same for both cylinders?
(a)
V
work done on or by the gas
(b)
heat added or removed
(c)
change in internal energy
quantities that do not depend
on the HOW you reach a state
(path taken) = state variables
depends only on current p,V
Work
W =
Z
V2
pdV
V1
What happens when 1 of
is constant (no change) ?
work done on gas during
volume change
p
(pressure)
V
(volume)
T
(temperature)
Q
(heat)
Isothermal ( T = constant )
T = constant
p
1
p/
V
pV = nRT
T1 T2 T3
System changes along isotherm:
curve of constant T: T3 > T2 > T1
V
p
Work done on gas = -(area)
W =
=
Z
V2
pdV =
V1
nRT
Z
V2
V1
Z
Z
dV
V
V2
V1
nRT
dV
V
1
dx = ln x
x
=
nRT ln V
ln x2
|VV21
ln x1 = ln
=
✓
x2
x1
◆
nRT ln
V1
✓
V2
V1
◆
V2
V
Isothermal ( T = constant )
Internal energy, U
from
from
molecules’ kinetic energy
(motion)
Gas T
Therefore, constant T = constant U
1st law of thermodynamics:
since: W = nRT ln
Q=
✓
V2
V1
U =0=Q+W
◆
W = nRT ln
✓
V2
V1
◆
Q=
isothermal process
W
Isothermal
How much work does the bubble do
as it rises to the surface?
T = 300 K
8 mm
p = 350 kPa
Ex.
25 m
Assume T = constant = 300 K
patm = 101 kPa
Isothermal process:
W =
nRT ln
??
✓
V2
V1
??
◆
Isothermal
Ex.
How much work does the bubble do
as it rises to the surface?
T = 300 K
8 mm
p = 350 kPa
25 m
Assume T = constant = 300 K
patm = 101 kPa
Isothermal process:
W =
nRT ln
4 3
ideal gas: pV = nRT = ⇡r p
3
4 3
⇡r
3
V2
V1
◆
p1 V1 = constant = p2 V2
constant
V2
p1
= 3.5
=
V1
p2
p1 V1 = constant = p2 V2
4 3
W = ⇡r p ln 3.5 =
3
✓
0.94 J
p1 V 1 =
Isothermal
Quiz
An ideal gas expands isothermally at 300 K.
Its volume increased from 0.020 m3 to 0.040 m3 .
The final pressure is 120 kPa.
(A)
3.3 kJ
(B)
1.7 kJ
(C)
3.3 kJ
(D)
1.7 kJ
(E)
0.0 kJ
The heat transfer to the gas....?
R = 8.314 J/mol · K
Isothermal
Quiz
An ideal gas expands isothermally at 300 K.
Its volume increased from 0.020 m3 to 0.040 m3 .
The final pressure is 120 kPa.
The heat transfer to the gas....?
pV = nRT
(A)
constant
3.3 kJ
(B)
1.7 kJ
(C)
3.3 kJ
(120 ⇥ 103 Pa)(0.040 m3 ) = nRT
✓ ◆
V2
W = nRT ln
V1
✓
3
(D)
(E)
1.7 kJ
= (120 ⇥ 10 Pa)(0.040 m ) ln
= 3.3 kJ
0.0 kJ
3
Q=
W = 3.3 kJ
0.040
0.02
◆
Constant volume,V
constant volume: isometric
isochoric
isovolume
volume does not change
1st law of thermodynamics:
W =0
U =Q
Introduce molar specific heat at constant volume, CV :
Q = nCV
# moles
(constant-volume)
T
C per unit mass (last lecture)
CV per mol
U = nCV
T
(any process)
U only depends on T:
U
= constant
T
Isobaric (p = constant)
p
System changes along isobar:
curve of constant P
Work done on gas = -(area)
W =
p(V2
V1 ) =
p V
V1
1st law of thermodynamics: Q =
Since:
U = nCV
T
U
W =
Q = nCV
U +p V
T +p V
Introduce molar specific heat at constant pressure, CP :
Q = nCP T
nCp T = nCV
T +p V
isobaric process
V2
V
Isobaric (p = constant)
How are CV and Cp related?
Ideal gas law: pV = nRT
p V = nR T
Therefore:
nCp T = nCV
T +p V
So: Cp = CV + R
nCp T = nCV
molar specific heat
For solids and liquids, expansion is small
Therefore, work is small: Cp ⇠ CV
T + nR T
Adiabatic (Q = 0)
Q
No heat flow : Q = 0
Q
Q
Q
Quickly occurring processes ~ adiabatic
(finished before heat transfer occurs)
Q
e.g.
combustion engine
1st law of thermodynamics:
U =W
adiabatic process
Adiabatic (Q = 0)
Work / internal energy
W done by gas lowers gas U
If V increases, gas does W
U decreases
T decreases
Since Q = 0
Since pV = nRT
p decreases
Isothermal changes only V and p
p
Adiabat : steeper than isotherms
V1
V2
V
pV
= constant
= Cp /CV
Adiabatic (Q = 0)
pV
= constant
Rewrite this equation in terms of T :
(a) V
=T
(b) T V
= constant
(c) T V
(d) T
1
1
= constant
V = constant
Quiz
Adiabatic (Q = 0)
pV
= constant
Rewrite this equation in terms of T :
Ideal gas: pV = nRT
constant
nRT
V
V
TV
= constant
1
= constant
nRT
p=
V
Quiz
Adiabatic (Q = 0)
Ex.
Diesel Engine
Compressed gas raises T ! 500 C
and ignites (starts burning) fuel
Vb
What is compression ratio,
?
Va
Initial T = 20 C and
1
TV
Vb
1
Va
1
= constant
Ta
=
Tb
1
= 1.4
1
Tb Vb
= Ta Va
before
after
✓ ◆1/ 1 ✓
◆1/0.4
Vb
Ta
500 + 273 K
=
= 11
=
Va
Tb
20 + 273 K
Adiabatic (Q = 0)
Since
W =
Z
V2
pdV
and
pV
= constant = C
V1
What is the work done by an adiabatic gas?
Z V2
Z V2
C
C
W =
CV dV
dV =
p=
V
V1 V
V1

+1 V2
CV
pV = C
=
+ 1 V1
W =
p2 V 2
p1 V 1
1
Ideal gas processes
Cyclic Processes
Systems that return periodically to the same state
(p, V, T )
e.g. refrigerator
liquid looses
heat outside
liquid removes
heat from inside
p
A
System moves between ‘A’ and ‘B’
B
V
Cyclic Processes
Cyclic processes can involve:
Isothermal
constant-volume
Isobaric
Adiabatic
processes
p
A
B
Net (total) work done on the gas
V
Ex.
Cyclic Processes
An ideal gas has
= 1.4 , V = 4.0 L, T = 300 K and p = 100 kPa.
1. Compressed adiabatically to 0.25 V
A!B
2. Cooled at constant-volume back to 300K
3. Expands isothermally to V
B!C
C!A
How much work is done on the gas?
Adiabat A ! B
p
B
pV
=C
pB V B
pB = pA (VA /VB )
C
A
(696.4⇥103 )(1⇥10
W =
= 741 J
V
W =
pB V B
???
= pA V A
pA V A
1
= 696.4 kPa
3
) (100⇥103 )(4⇥10
1.4 1
3
)
Ex.
Cyclic Processes
An ideal gas has
= 1.4 , V = 4.0 L, T = 300 K and p = 100 kPa.
1. Compressed adiabatically to 0.25 V
A!B
2. Cooled at constant-volume back to 300K
3. Expands isothermally to V
B!C
C!A
How much work is done on the gas?
p
Constant-volume B ! C
B
volume does not change
C
A
V
W =0
Ex.
Cyclic Processes
An ideal gas has
= 1.4 , V = 4.0 L, T = 300 K and p = 100 kPa.
1. Compressed adiabatically to 0.25 V
A!B
2. Cooled at constant-volume back to 300K
3. Expands isothermally to V
B!C
C!A
How much work is done on the gas?
p
B
C
Isotherm: C ! A
W = nRT ln
constant
⇣
VA
VC
Ideal gas: pV = nRT
3
= pA VA = (100 ⇥ 10 )(4 ⇥ 10
= 400 J
A
W = (400 J)(ln 4) = 555 J
V
⌘
3
)
Ex.
Cyclic Processes
An ideal gas has
= 1.4 , V = 4.0 L, T = 300 K and p = 100 kPa.
1. Compressed adiabatically to 0.25 V
A!B
2. Cooled at constant-volume back to 300K
3. Expands isothermally to V
B!C
C!A
How much work is done on the gas?
p
WABCA = WAB + WBC + WCA
B
= 741 J + 0 J
C
= 186 J
A
V
555 J
Quiz
Cyclic Processes
p
What is the work done?
5.0 atm
isothermal
1 L = 0.001 m3
1 atm = 101, 325 Pa
1.0 atm
10 L
50 L
V
(a)
8120 J
(c)
(b)
0J
(d)
4050 J
4101 J
Quiz
Cyclic Processes
p
What is the work done on the
ideal gas?
5.0 atm
isothermal
constant V
1.0 atm
1 atm = 101, 325 Pa
isobaric
10 L
isobaric: W =
=
50 L
p(V2
V
V1 )
(101, 325 Pa)( 40 ⇥ 10
= 4053 J
constant V:
1 L = 0.001 m3
W =0
3
m3 )
Quiz
Cyclic Processes
p
What is the work done on the
ideal gas?
5.0 atm
constant V
1.0 atm
isothermal
isobaric
1 L = 0.001 m3
1 atm = 101, 325 Pa
V
10 L
50 L ✓ ◆
V2
isothermal: W = nRT ln
V1
constant
pV = (5.0 ⇥ 101, 325 Pa)(10 ⇥ 10
3
)✓
= nRT
◆
50
3
W = (5.0 ⇥ 101, 325 Pa)(10 ⇥ 10 ) ln
10
= 8153.8 J
Quiz
Cyclic Processes
p
What is the work done on the
ideal gas?
5.0 atm
constant V
1.0 atm
isothermal
isobaric
10 L
Total work:
50 L
= 4053 + 0
=
4100.8 J
1 L = 0.001 m3
1 atm = 101, 325 Pa
V
8153.8 J
Specific heats:
Last lecture:
&
CV
Cp
1 ¯2
3
mv = kT
2
2
average K of molecule
(for 1 molecule)
gas temperature
Total internal energy of n mols:
U = nNA
✓
1 ¯2
mv
2
3
= nNA kT
2
3
= nRT
2
From
U = nCV
T
1
CV =
n
3
U
= R
2
T
◆
R
Specific heats:
and:
Cp
=
CV
Therefore:
=
CV
5
3R
3
2R
5
=
= 1.67
3
H2
Ne
N2
Ar
O2
Great! But....
Cp
CV + R
=
CV
He
5
=
3
&
SO2
N O2
7
=
5
' 1.3
Specific heats:
1 ¯2
3
mv = kT
2
2
CV
&
Cp
to find this...
Assumed: Gas molecules have no internal structure
nothing inside
He
OK for monatomic molecules:
Ne
1 molecule = 1 atom
Ar
y
Can only move in 3 directions:
= 3 degrees of freedom
z
x
Specific heats:
CV
&
Cp
Diatomic molecules can move
in 5 directions
x
y
translational motion
5 degrees of freedom
1 molecule =
2 atoms
z
✓
rotational motion
Equipartition Theorem
In thermodynamic equilibrium,
1
Average energy / molecule = kT for each degree of freedom
2
Equipartition theorem
e.g. monatomic molecule, 3 degrees of freedom:
✓
◆
1
3
U =3
kT
= kT
2
2
since all energy is kinetic:
1 ¯2
3
U = K = mv = kT
2
2
Equipartition Theorem
e.g. diatomic molecule, 5 degrees of freedom:
✓
◆
1
5
kT = kT
Av. energy / molecule: U = 5
2
2
Total internal energy of n mols:
1
Therefore: CV =
n
(as before)
and:
5
U = nNA kT
2
5
= nRT
2
U
5
= R
T
2
Cp
CV + R
7
=
=
= = 1.4
CV
CV
5
Equipartition Theorem
e.g. polyatomic molecule, 6 degrees of freedom:
3 translational (x, y, z)
3 rotational
U = 3nRT
CV = 3R
4
=
' 1.33
3
Equipartition Theorem
At very high T, diatomic molecules can also vibrate:
Adds kinetic energy (K)
... and potential energy
+ 2 degrees of freedom
Total: 7 degrees of freedom
(only at high T)
Equipartition Theorem
Ex.
A gas mixture has 2.0 mol of oxygen ( O2 ) and 1.0 mol of argon (Ar)
Find the volume specific heat, CV
O2 :
Ar :
5 degrees of freedom
5
5
U = nRT = 2.0RT = 5RT
2
2
3 degrees of freedom
3
3
3
U = nRT = 1.0RT = RT
2
2
2
3
13
Total internal energy: U = 5RT + RT =
RT
2
2
U
13
=
R
T
2
1
13
1 U
=
R = 2.2R
CV =
2.0 + 1.0 2
n T