Transcript lecture37

IV. Kinetic theory
(continued – see previous lecture)
5. Heat capacitance
a) Monoatomic gas
3
E  K tr  nRT
2
3
nRdT
2
dQ  nCV dT
dK tr 
}
3
CV  R
2
1 dQ
CV 
- molar specific heat at constant v olume
n dT
b) Equipartition principle. Degrees of freedom
f
CV  R
2
f
E  nRT
2
Diatomic gas, including rotations:
5
CV  R
2
Diatomic gas, including rotations and vibrations:
7
CV  R
2
Molar Specific Heats of Gases at Constant Volume CV
At Room Temperature
Monatomic
He
Ar
CV (J/mol·K)
12.47
12.47
Diatomic
H2
N2
O2
20.42
20.76
21.10
Polyatomic
CO2
SO2
28.46
31.39
3R/2 = 12.47 J/mol·K
5R/2 = 20.79 J/mol·K
7R/2 = 29.10 J/mol·K
Question
How many rotational degrees
of freedom does the water
molecule have?
1. 1
2. 2
3. 3
4. 4
c) A hint of quantum theory
b) Heat capacitance of solids
E  3nRT
CV  3R
5. Distribution of molecular speeds (Maxwell distribution)
3
dN  Nf (v)dv
 m  2 2  mv2 / 2 k BT
 v e
f (v)  4 
 2k BT 
Molar Specific Heats of Elemental Solids at Constant Volume due
to Lattice Vibrations
The rule of Dulong and Petit only holds at high temperatures.
At low T, quantum mechanical effects reduce CV.
V. The first low of Thermodynamics
(conservation of energy)
ΔU = ΔQ - ΔW
dU = dQ - dW
1.
•Macro- a micro- parameters
•Equation of state
m
for monatomic ideal gas: PV  RT

•Thermodynamic equilibrium
2. Internal energy (kinetic plus potential energy of particles)
dU = dQ + dWext
dWext = -dW
dW = P dV
dU = dQ - dW = T dS - P dV
dQ = T dS
3. Work
dW  Fdx  PAdx  PdV
dW  PdV
•Work is path dependent
•Heat is path dependent
•Internal energy is path independent
V2
W   P(V )dV
V1
P
P
1
2
1
ΔW = P(V2 - V1)>0
V1
V2
ΔW
V
P
2
V1
V2
P
2
1
V1
3
2
ΔW = P(V2 - V1)<0
V2
V
ΔW
4
1
V
V1
V2
V
ΔW12 = 0
ΔW23 > 0
ΔW34 > 0
ΔW42 < 0
Example1: A quantity of air is taken from state a to state b along a path
that is a straight line in the PV-diagram, where Va=0.070 m3, Vb=0.110 m3,
Pa=1.00*105 Pa, Pb=1.40*105 Pa. Assume that the gas is ideal.
a) What is the work, W, done by the gas in this process?
W = ½ (Pa + Pb )(Vb - Va )
P
b
W = ½ (1.00 + 1.40)* 105 Pa*(0.1100 - 0.0700 ) m3
a
W = 4.8* 103 J
V
Va
Vb
b) What happen with temperature and internal energy of this gas?
For ideal gas:
PaVa PbVb

Ta
Tb
K ~T
Tb PbVb 1.40 105 Pa  0.1100m3


 2.2
5
3
Ta PaVa 1.00 10 Pa  0.0700m
K b Tb

 2 .2
K a Ta
Example2: This P-V diagram represents a system consisting of a fixed
amount of ideal gas that undergoes three different processes in going
from state A to state B. Rank work, heat transfer, change in internal,
kinetic, and potential energy for each process.
P
State A
ΔW1 < ΔW2 < ΔW3
2
ΔU1 = ΔU2 = ΔU3 = UB - UA
ΔT1 = ΔT2 = ΔT3 = TB - TA
ΔU = ΔQ - ΔW
ΔQ1 <ΔQ2 <Δ Q3
For ideal gas:
PE=0 and Δ PE=0
T~ K=U
ΔK1 = ΔK2 = ΔK3 = KB - KA
3
I
State B
V
Example3: A system consisting of a quantity of ideal gas is in equilibrium state A.
It is slowly heated and as it expands, its pressure varies. It ends up in equilibrium
state B. Now suppose that the same quantity of ideal gas again starts in state A,
but undergoes a different thermodynamic process (i.e., follows a different path
on a P-V diagram), only to end up again in the same state B as before.
Consider the net work done by the system and the net heat absorbed by the
system during these two different processes. Which of these statements is true?
1. The work done may be different in the two processes,
but the heat absorbed must be the same.
2. The work done must be the same in the two processes,
but the heat absorbed may be different.
3. The work done may be different in the two processes,
and the heat absorbed may be different in the two processes.
4. Both the work done and the heat absorbed must be the same in the two processes,
but are not equal to zero.
5. Both the work done and the heat absorbed by the system must be equal to zero
in both processes.
Note: See example 2
Final Exam: Tues, Dec. 16, 2008, 7:00-9:00 p.m.
~1/3 of questions test understanding of concepts/principles
~2/3 of questions test numerical application of concepts/principles
Bring soft (#2) pencils, erasers, and your scientific calculator
Studying:
• Review during lecture: Friday, Dec. 12
• Practice problems and solutions for Ch. 20 are posted
• Compare your homework solutions with the posted solutions
• Review the lecture questions and the text supplements
• Review and solve the example problems in the text
• Solve problems on the practice exams
-- Seven Phys 221 practice Final Exams are posted plus practice
problems on waves and thermodynamics (Ch. 15-20)
• The formula sheet that will be provided with the exam will be posted
• Meet with me before the exam to clear up any serious problems you
are having with the course material
Physics 221 Final Exam
Tuesday, Dec. 16, 7:00-9:00 p.m.
20 2-point problems + 20 4-point problems
(multiple choice)
Total: 40 problems worth 120 course points
~ 1/3 Ch 17-20
~ 1/2 Comprehensive (prior to Ch. 17)
~ 1/5 Laboratory Final Exam
Physics 221 Final Exam Conflicts
If you are enrolled in any of the three
following courses you are entitled to take
a Physics 221 Final Exam Make-up at a
time and place to be arranged.
Math 195, Math 196, Acct 215
Prior to final exam week, students must
request to take the make-up exam.
Question 1
One cm3 of liquid water at 100 ˚C is boiled off at 1 atm pressure and is
converted to 1670 cm3 of steam at 100 ˚C. The “system” is the water.
The work done by the system when it is converted to steam is __ L·atm.
(1 L = 1000 cm3)
1. 0.17
2. 1.7
3. 17
4. 1700
Question 2: The work done by the system in going from point 1 to point 2 is
___ L·atm.
1.
2.
3.
4.
1
2
3
3
ΔU = ΔQ - ΔW
4. Thermodynamic processes
ΔW = P ΔV
•Adiabatic process: ΔQ = 0
ΔU = - ΔW
•Isochoric process: ΔV = 0
ΔW = 0
•Isobaric process:
ΔW = PΔV
ΔP = 0
ΔU = ΔQ
•Isothermal process: ΔT = 0
•For ideal gas (!) only: ΔT = 0
PV  P0V0
ΔU = 0
V2
V2
V1
V1
W   PdV  P0V0 
•Closed cycle process: ΔT = ΔP = ΔV = ΔU = 0
ΔQ = ΔW
V 
dV
 P0V0 ln  2 
V
 V1 
ΔQ = ΔW
The processes on a PV diagram
(ideal gas)
Question
3.0
The work done by the gas during
this process is ___ L·atm.
1
2.5
p (atm)
2.0
1.
2.
3.
4.
1.5
2
1.0
0.5
0.0
0.0
0.5
1.0
1.5 2.0
V (L)
2.5
3.0
An ideal gas undergoes an isothermal
expansion from V1 = 1.00 L to V2 = 2.72 L
and p2 = 1.00 atm as shown in the figure.
V2
V2
V1
V1
W   PdV  P0V0 
V 
dV
 P0V0 ln  2 
V
 V1 
1.5
2.7
3.9
5.4
Question
The net heat absorbed by
the system during one cycle
is ___ L·atm
1. 2
2. -2
3. 4
4. -4
Example: A monatomic ideal gas undergoes an increase in pressure
from p1 = 1.00 atm to p2 = 3.00 atm at V = 24.0 L as shown in the figure.
The heat absorbed by the gas during this process is ??? L·atm.
For Ideal monatomic gas:
U  K  32 nRT  32 PV
V  0  Q  U 
U 
3
2
3
2
P V
2atm24L  72L  atm
Heat capacitance of an ideal gas
1 dQ
CV 
n dT / V
1 dQ
CP 
n dT
at constant volume
/
For an ideal gas:
P
at constant pressure
dU
dT
dU
/ V  dT
dU
/ P  dT
PV  nRT
dU  dQ  dW  dQ  dU  dW  dQ  dU  PdV
CV 
1 dQ
1 dU

n dT / V n dT
CP 
1 dQ
1 dU 1 PdV


 CV  R
/
P
n dT
n dT n dT
CP

CV
CV 
f
R
2
CV 
CP  CV  R
R
 1
CP  R

 1
monoatomic gas : f  3  C v  32 R, C P  52 R,   53  1.67
diatomic gas at room temperatu re : f  5  C v  52 R, C P  72 R,   75  1.40
Relating heat capacities at constant volume and pressure
U ~T
U ~ T
Adiabatic processes for an ideal gas
dU = dQ -PdV
For an adiabatic process: dQ = 0
nRT
nCV dT  dV
V
For an ideal gas: PV=nRT
CV 
1 dU
 dU  nCV dT
n dT
C  CV
R
 P
 (  1)
CV
CV
dT
dV
 - - 1
T
V
ln T    1 ln V  const
ln T  ln V  1  const


ln TV  1  const
PV  nRT
TV  1  const
T1V1 1  T2V2 1
PV   const
P1V1  P2V2
Example:
P
1
P1  2atm
V1  1l
2
T1  60 C
V2  2l
a T1  T2
b Q  0
P2  ?
V1
(a)
V2
P1V1 P2V2

T1
T2
V
V1
P2  P1
V2
P1V1  P2V2
T1  T2
P2  2atm
(b)
P1V1 P2V2

T1
T2
P1V1  P2V2
 V1 
P2  P1  
 V2 
V 
T2  T1  1 
 V2 

 1

1
P2  2  atm
2
1
T2  60 C  
2

1l
 1atm
2l
Example
The volume of the air inside the
cylinder of an engine decreases
from 1.00 L to 0.100 L during
adiabatic compression. The initial
pressure of the air is 1.00 atm and
initial temperature is 27 ˚C.
Given: p1  1.00 atm; V1  1.00 L;
V2  0.100 L; r  V1 / V2  10.0;
T1  27 °C = 300 K.
Adiabatic compression:
p1V1  p2V2

 V1 
V1
p2  p1   p1    p1r 
V2
 V2 

The final pressure of the air
is ___ atm. ( = 1.40 for air)
= (1.00 atm)(10.0)1.40  25.1 atm.
T1V1 1  T2V2 1
The final temperature of the
air is ___ ˚C.
 V1 
T2  T1  
V 
 1
 T1r  1
2
= (300 K)(10.0)1.40 1  754 K = 481 °C.
Adiabatic processes for an ideal gas (2)
TV  1  const
T1V1 1  T2V2 1
PV   const
P1V1  P2V2
Work
V2
W   PdV  P1V1

V1
 P1V1
V2

V1
dV


V


1
V21  V11 
1 
1

( P1V1 V21 - P1V1 )
1 
dU  dW  nCV dT
PV=nRT
1
W
(P1V1  P2V2 )
 1
W  nCV (T1  T2 )
CV
1
W 
(P1V1  P2V2 ) 
(P1V1  P2V2 )
R
 1
Question
The volume of the air inside the
cylinder of an engine decreases
from 1.00 L to 0.100 L during
adiabatic compression. The
initial pressure of the air is 1.00
atm and the final pressure is 25.1
atm.
The work done by the gas
during this process is ___
L·atm. ( = 1.40 for air)
1.
2.
3.
4.
-8
-4
2
6
W
1
(P1V1  P2V2 )
 1
Question
Some air undergoes an isobaric
expansion at 1.00 atm pressure
from V1 = 5.00 L to V2 = 9.00 L.
(CV = 5R/2)
The heat absorbed by the gas
during this process is ___ L·atm.
pV  nRT
Q  nC p T  C p R PV
C p  CV  R
Q  C p R PdV  (7 / 2)1.00atm(4.00 L)
Q  14 L  atm