How to Understand Your Scores
Review for Final Exam
Resending Nov-9 Message
ƒ BONUS ASSIGNMENT: Here (Deadlne: 12:30 pm, Dec. 3 (Tue))
ƒ FINAL EXAM: Dec 6 (Fri) (3 pm - 5 pm) at MPHY 203
[Fall 2013 Final Exam Schedule] [Formular Sheet for Exam4 and
Final]
ƒ Old PHYS 218 Exam Solutions: 1. 2. 3. Final.
ƒ TENTATIVE PLAN for Final Exam as of Dec 1 (6 PM):
*Naively*, the corresponding letter
grades are:
If you are 85 or better, you are in
*target* area of an A (>=90).
75 or better, B (>=80)
55 or better, C (>=60)
Problem 1 Understanding motion of objects in 2D (Chap. 3 + 2)
40 or better, D (>=50)
Problem 2 Understanding Newton's laws (Chap.4 + 5 + 6)
Exams – returned during the classes
and picked up in my office.
Problem 3 Understanding a collision of 2 bodies and a motion of 2
bodies after the collision (Chap. 8 + others such as 7)
Problem 4 Understanding a rotational motion of solid objects (Chap.
9 + 10)
Problem 5 Understanding wave motion (Chap. 12 along with 11)
Problem 6 Understanding pV = nRT (Chap.15)
3
Review for Final Exam
1
Exams 1, 2, 3, and 4
4
7
5
8
6
1
Projectile Motion
Understanding Projectile Motions
– The red ball is dropped at
the same time that the
yellow ball is fired
horizontally. The strobe
marks equal time intervals.
I
– Projectile motion as
horizontal motion with
constant velocity (ax = 0)
and vertical motion with
constant acceleration (ay =
g).
II
III
I
Kinematics (2D)
11
Further Look at Projectile Motion
II
(3) vy = 0
Projectile motion
as horizontal
motion with
constant velocity
(ax = 0) and
vertical motion
with constant
acceleration (ay
= g).
(1) Choose an origin &
an x-y coordinate system
(4) y = 0
(2) ax = 0
ay = g = 9.80 m/s2
vx = constant
9
Projectile Motion
– The red ball is dropped at
the same time that the
yellow ball is fired
horizontally. The strobe
marks equal time intervals.
– Projectile motion as
horizontal motion with
constant velocity (ax = 0)
and vertical motion with
constant acceleration (ay =
g).
I
Kinematics (2D)
12
10
Same Problems?
The Same Problem?
II
II
[Quick Quiz 2]
Is this a motion with a
constant acceleration or
with a varying acceleration?
y=0 m
R
y=? m
R
15
13
(3) Use kinematic eqs.
in x and y separately.
II
II
Vy = 0
(2) ax = 0
ay = g = 9.80 m/s2
Vx = constant
y = 0 y = 1.00m
(1) Choose an origin &
an x-y coordinate system
Kinematics (2D)
16
14
Firing at a More Complex Target
The Same Problem?
™ A moving target presents a real-life scenario.
™ It is possible to solve a falling body as the
target. This problem is a “classic” on
standardized exams.
Example 2: A projectil
projectile
le is launched
la
from ground level too the tto
top of a cliff
which is R = 195 m away and H = 155 m high. The projectile lands on top of
the cliff T = 7.60 s after it is fired. Use 2sinT cosT = sin2T if necessary. The
acceleration due to gravity is g = 9.80 m/s2 pointing down. Ignore air friction.
a. Find the initial velocity of the projectile (magnitude v0 and direction T).
b. Find a formula of tanT in terms of g, R, H and T.
http://www.youtube.com/watch?v=cxvsHNRXLjw
II
III
t = 7.6 s
Kinematics (2D)
19
17
The Same Problem?
A boy on a small hill aims his waterballoon slingshot horizontally,
straight at second boy hanging from a tree
branch a distance d away. At the instant
the water balloon is released, the second
boy lets go and falls from the tree, hoping
to avoid being hit. Show that he made the
wrong move.
III
A boy on a small hill aims his waterballoon slingshot upward, directly at
second boy hanging from a tree branch. At
the instant the water balloon is released,
the second boy lets go and falls from the
tree, hoping to avoid being hit. Show that
he made the wrong move.
I
Example 3.3.
Question:
How fast
must
the
motorcycle
leave the
cliff-top?
Kinematics (2D)
20
18
I
H
200 m, given Æ x?
?
2
Same
Concept
H?
d, given Æ H?
III
Kinematics (2D)
23
21
Structure of Newtonian Mechanics
Can you explain?
Inertial Reference Frame
(Newton’s 1st Law)
Æ
Magic? or Physics?
Giancoli’s Textbook 3rd Ed.
III
Æ
F=ma
(Newton’s 2nd Law)
Action-Reaction
(Newton’s 3rd Law)
Mass
(m)
The Nature of Force
The Nature of Object
We see “An apple in motion (x direction)
tends to stay in motion (x direction).”
Æ
Motion with constant velocity in x
direction.
Kinematics
Æ Æ
Æ Æ
Æ
Æ
(r, v, a)
The Nature of Motion
http://link.brightcove.com/services/player/bcpid36804639001?bckey=AQ~~,AAAACI
JPQzk~,qiwYyUrE_-dz5lglGrCClkfJDM1jW3zH&bclid=0&bctid=109459228001
Newton’s Laws of Motion
24
22
Force: Acceleration/Equilibrium
Quick Quiz Solution: The friction force appears as it keeps from
sliding back on the truck’s bed. Thus, the direction of the (static)
friction is pointing to the right. If you isolate the box,
and draw the free-body diagram for the box, you
&
find that it is consistent with Newtons’ 2nd law: * Ff
a
& &
¦F 0
& &
a 0
m
Ignore the truck and two people!
FN
Acceleration Æ
Kinetic Equations
(see Chap. 2 &3)
*
a
FN
Ff
Ff
Ff is the force on the box by the truck’s bed.
Fg
Fg
The force on a hokey puck
causes the acceleration
Newton’s Laws of Motion
If the net force on a hokey
puck is zero (equilibrium),
the acceleration is zero.
Newton’s Laws of Motion
27
Quick Quiz
Example 6: Suppose that you are standing on a
train accelerating at 0.20g. What minimum
coefficient of static friction must exist between
your feet and the floor if you are not to slide?
FN
25
A hockey puck is sliding at a
constant velocity across a flat
horizontal ice surface. Which is
the correct free-body diagram?
a = 0.20 g
[A] (b)
xF
Ff
f
FG
Which one is correct?
Newton’s 2nd law:
F = m a Æ Ff = m (0.20g)
where Ff = Ps FN = Ps (mg)
Thus, Ps (mg) = m (0.20g)
So, Ps = 0.20
Newton’s Laws of Motion
Newton’s Laws of Motion
28
26
Draw a Diagram Æ FBD Æ Newton’s Laws
5(e) : In a “Rotor-ride” at a carnival,
People pay money to be rotated in a
Vertical cylindrically walled “room.”
Which diagram correctly shows the
Forces acting on each rider? Explain
Explicitly (in words) the reasoning why
you choose the diagram by labeling
each arrow (in wards: e.g., downward
arrow, horizontal-left arrow etc.).
a
b
Taken from Fig. 5-38 (Giancoli)
c
d
e
Newton’s Laws of Motion
31
29
Weight on Planet
Draw a Di
D
Diagram Æ FBD Æ Newton’s Laws
™ Mass is a measure of “how much material do I
have?”
™ Weight is “how hard do I push down on the floor?”
™ If you were offered to get 9.8N of gold on
earth and 9.8N of gold on moon, which offer do
you take?
32
Newton’s Laws of Motion
30
Example 2: A block (mass m1) is placed on a smooth horizontal
Swinging a ball on the end of string …
surface, connected by a thin cord that passes over a pulley to a
second block (m2), which hangs vertically. Draw the free-body
diagram for each of m1 and m2 . Express the acceleration of the
block in terms of m1, m2, and g.
[Quick Quiz] Which one is correct?
(a)
FT = m1 a
Fg2 – FT = m2 a
(b)
FN
a
FT
FT
m2 g – m1 a = m2 a
Fg1
m2 g = m1 a + m2 a
a = m2 g / (m1 + m2 )
Circular Motion
Fg2
Newton’s Laws of Motion
a
x
35
Example Problem 1
33
F.B.D. of Uniform Circular Motion
™ Model airplane on a string
Which one is correct?
v2
m
R
One revolution every
y 4.00 seconds
v
dis tan ce
time
&
marad
&
2S r
T
m
&
F
&
F
&
F
&
F
v2
r̂
r
FT
CORRECT
INCORRECT
Thinking …
Rotation Æ Center-seeking Force Æ FT
36
The quantity v2/R is not a force it doesn’t belong in the free-body
Circular Motion
(force) diagram
34
Understanding Satellite Motion
&
marad
Example Problem 1
A 1000-kg car rounds a curve on a flat road of radius 50.0 m at
a maximum speed of 14.0 m/s without skidding.
a) Draw the free-body diagram for the car.
b) Find the magnitude and direction of the friction force.
c) Find the coefficient of the friction force.
d) Is the result in c) independent of the mass of the car?
v2
m r̂
r
&
mM
F G 2 r̂
r
•
•
•
•
•
•
&
marad
&
G = 6.67 x 1011 Nm2/kg2
ME = 5.98 x 1024 kg
RE = 6.38 x 103 km
MS = 1.99 x 1030 kg
Mmoon = 7.35 x 1022 kg
Rmoon = 1.74 x 103 km
v2
m r̂
r
Ff
FN
mg
Psmin = Ff/FN
Circular Motion
39
A tetherball problem – Example 6.2
37
Example Problem 2
A 1000-kg car rounds a curve on a banked road of radius 50.0
m at a maximum speed of 14.0 m/s without skidding. The
banking angle is 22o.
a) Draw the free-body diagram for the car.
b) Find the magnitude and direction of the friction force.
c) Find the coefficient of the friction force.
d) Is the result in part c) independent of the mass of the car?
&
marad
m
v2
r̂
r
FN z mg
No Skidding on Banked Curve
Circular Motion
40
Ps = Ff/FN
38
3
Energy Method
Equations of Motion
43
Find h Using Energy Conservation
41
Work Energy Theorem
A 50.0-kg crate is pulled 40.0 m by a constant force
exerted (FP = 100 N and T = 37.0o) by a person. Assume
a coefficient of friction force Pk = 0.110. Determine the
work done by each force acting on the crate and its net
work. Find the final velocity of the crate if d = 40 m and
vi = 0 m/s.
Wnet = 6Wi
= 1302 [J] (> 0)
ISEE
ISEE
Wnet = Kf – Ki
= (1/2) m vf2 0
=
Energy
E
nergy Conservation
Conservation
Work and Energy
44
42
47
45
Same Concept with Ballistic Pendulum
(A) Momentum Conservation
8-85
8-44, 8-83
2
1
(B) Energy Conservation
Momentum Conservationn
48
46
Similar Problem
Same Concept with Ballistic Pendulum
(A) Momentum Conservation
B
Skeet+Pellet
A
C
2
1
(B) Energy Conservation
Momentum Conservation
51
49
4
Newton’s Laws of Motion
52
50
Rotational Inertia
Newton’s Laws for Rotation
(factor) (Mass x Distance^2)
3rd part
[N m]
2nd part
[kg m2]
1st part
[s–2]
Chap. 10
Krot = (1/2) I Z2 (ÆK = (1/2) m v2)
Angular Momentum Conservation
Rotational Motion
Rotational Motion
55
Mechanical Energy Conservation
53
Rotational Energy and Inertia
K = Km + KM
m2
m1
Krot = (1/2) I Z2 (ÆK = (1/2) m v2)
Mass x Distance^2
m3
54
Can you explain why?
Mechanical Energy Conservation
Solid disk (M, R0)
MgH
1
1
M v 2 I cmZ 2
2
2
m
Icm = Large Æ v = Small
Rotational Motion
59
Can you find v?
57
Race of the objects on a ramp
™ This is a classic multiple-choice question from
MCAT-style standardized tests.
™ Refer to Figure 9.23.
Analyze the rolling sphere in terms of forces and
torques: find the magnitudes of the velocity v and ….
Rotational Motion
60
Angular Momentum Conservation
&
Fnet
P10-30 (MP)
&
W n et
Rotational Motion
&
ma
Ktrans + Krot = Krolling
&
ID
Rotational Motion
63
61
64
62
Example 2
P10-33
P10-34 (MP)
,iZi ,fZf
A10.11
A spinning figure
skater pulls his arms
in as he rotates on the
ice. As he pulls his
arms in, what
happens to his
angular momentum L
and kinetic energy K?
A. L and K both increase.
B. L stays the same; K increases.
C. L increases; K stays the same.
D. L and K both stay the same.
67
Q10.1
F1
Which force produces the greatest
torque about the point O (marked by
the blue dot)?
Q10.11
F3
The four forces shown all have the
same magnitude: F1 = F2 = F3 = F4.
A spinning figure
skater pulls his arms
in as he rotates on the
ice. As he pulls his
arms in, what
happens to his
angular momentum L
and kinetic energy K?
O
F2
65
F4
A. F1
B. F2
C. F3
A. L and K both increase.
D. F4
B. L stays the same; K increases.
E. not enough information given to decide
C. L increases; K stays the same.
D. L and K both stay the same.
68
66
A10.2
F1
Which of the four forces shown
here produces a torque about O that
is directed out of the plane of the
drawing?
A10.1
F3
F3
The four forces shown all have the
same magnitude: F1 = F2 = F3 = F4.
O
F2
F1
Which force produces the greatest
torque about the point O (marked by
the blue dot)?
F4
O
F2
F4
A. F1
A. F1
B. F2
B. F2
C. F3
C. F3
D. F4
D. F4
E. more than one of these
E. not enough information given to decide
71
69
Q10.2
Q10.3
A plumber pushes
straight down on the
end of a long wrench
as shown. What is the
magnitude of the
torque he applies about
the pipe at lower right?
F1
Which of the four forces shown
here produces a torque about O that
is directed out of the plane of the
drawing?
F3
O
F2
F4
A. F1
B. F2
A. (0.80 m)(900 N)sin 19°
C. F3
B. (0.80 m)(900 N)cos 19°
D. F4
C. (0.80 m)(900 N)tan 19°
E. more than one of these
D. none of the above
72
70
A10.5
A glider of mass m1 on a frictionless horizontal track is connected
to an object of mass m2 by a massless string. The glider accelerates
to the right, the object accelerates downward, and the string rotates
the pulley. What is the relationship among T1 (the tension in the
horizontal part of the string), T2 (the tension in the vertical part of
the string), and the weight m2g of the object?
A10.3
A plumber pushes
straight down on the
end of a long wrench
as shown. What is the
magnitude of the
torque he applies about
the pipe at lower right?
A. m2g = T2 = T1
B. m2g > T2 = T1
C. m2g > T2 > T1
A. (0.80 m)(900 N)sin 19°
D. m2g = T2 > T1
B. (0.80 m)(900 N)cos 19°
E. none of the above
C. (0.80 m)(900 N)tan 19°
D. none of the above
75
Q10.6
73
Q10.5
A glider of mass m1 on a frictionless horizontal track is connected
to an object of mass m2 by a massless string. The glider accelerates
to the right, the object accelerates downward, and the string rotates
the pulley. What is the relationship among T1 (the tension in the
horizontal part of the string), T2 (the tension in the vertical part of
the string), and the weight m2g of the object?
A lightweight string is wrapped
several times around the rim of a
small hoop. If the free end of the
string is held in place and the hoop
is released from rest, the string
unwinds and the hoop descends.
How does the tension in the string
(T) compare to the weight of the
hoop (w)?
A. m2g = T2 = T1
B. m2g > T2 = T1
A. T = w
C. m2g > T2 > T1
B. T > w
D. m2g = T2 > T1
C. T < w
E. none of the above
D. not enough information given to
decide
76
74
A10.7
A10.6
A solid bowling ball rolls
down a ramp.
A lightweight string is wrapped
several times around the rim of a
small hoop. If the free end of the
string is held in place and the hoop
is released from rest, the string
unwinds and the hoop descends.
How does the tension in the string
(T) compare to the weight of the
hoop (w)?
Which of the following forces
exerts a torque on the bowling
ball about its center?
A. the weight of the ball
B. the normal force exerted by the ramp
C. the friction force exerted by the ramp
A. T = w
D. more than one of the above
B. T > w
E. The answer depends on whether the ball rolls without
slipping.
C. T < w
D. not enough information given to
decide
79
77
Q10.7
A solid bowling ball rolls
down a ramp.
Which of the following forces
exerts a torque on the bowling
ball about its center?
5
A. the weight of the ball
B. the normal force exerted by the ramp
C. the friction force exerted by the ramp
D. more than one of the above
E. The answer depends on whether the ball rolls without
slipping.
80
78
Figure 15.4
Visualizing SHM
A 2.00-kg frictionless block is attached to an ideal spring with force constant k = 315 N/m. initially the
spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0
m/s, and undergoes a simple harmonic motion (SHM). Let’s characterize the SHM of the block. Find:
(a)(5 pts) the period (T) in seconds
(b)(5 pts) the maximum speed (vmax) in m/s
(c)(5 pts) the amplitude (A) of the motion in meters
(d)(5 pts) the maximum magnitude of the force (in N) on the block exerted by the spring during the
motion.
and TIME dependence…
Fx
k x
[Bonus (10 pts)] If you have time, sketch x-t and vt graphs of this motion.
ax
k
x
m
vmax, ax=0
vwave = O(Z /2S)
vwave = O/ T
vwave = O f
Mechanical Energy Conservation
83
Standing Wave
81
SHM to Wave Motion
Simple Harmonic Oscillator (SHO)
x(t) = R0 cos (Z t)
x = R0 cos T
where T = Z t
Kin. Equation of SHM
(a) FT if P = 40.0 g/m and f1 = 20.0 Hz?
(b) f2 and wavelength of second
harmonic?
(c) f2 and wavelength of second
overtone (or 3rd harmonic)?
[Q] How can you describe the
shape of the rope?
T = 2S / Z
[A]
A, O
f2 = 2 f1
f3 = 3 f1
Wave Motion
84
82
Q15.2
Example 2
Which of the following wave functions describe a wave
that moves in the –x-direction?
A transverse traveling wave on a cord is represented by
y(x, t) = 0.48 sin(0.56x + 84t)
where y and x are in meters and t in seconds. For this wave,
determine:
(a) the amplitude,
(b) wavelength, frequency, velocity (magnitude and direction),
(c) maximum and minimum speeds of particles of the cord, and
(d) maximum acceleration (magnitude) of the particles.
A. y(x,t) = A sin (–kx – Zt)
B. y(x,t) = A sin (kx + Zt)
C. y(x,t) = A cos (kx + Zt)
D. both B. and C.
E. all of A., B., and C.
[A] …
The wave function for a sinusoidal wave
moving in the +x-direction is
y(x, t) = A sin(Z t – k x),
where k = 2π/O , Z = 2Sf, O= v T …
Wave Motion
87
A15.2
85
Graphs
Which of the following wave functions describe a wave
that moves in the –x-direction?
A. y(x,t) = A sin (–kx – Zt) = A sin ( – (kx + Zt) )
B. y(x,t) = A sin (kx + Zt)
C. y(x,t) = A cos (kx + Zt)
D. both B. and C.
E. all of A., B., and C.
Oscillations
88
86
Understanding Problem
Example 3
A transverse wave pulse travels to the right along a string with
speed v = 2.0 m/s. At t = 0, the shape of the pulse is given by the
function y = 0.45 cos(3.0x) where y and x are in meters and t in
seconds. For this wave, determine:
(a) the wavelength, frequency, and amplitude,
(b) maximum and minimum speeds of particles of the string, and
(c) maximum and minimum accelerations (magnitudes) of the
particles.
Example 12.9
I1 ( r ) I 0 u 10 E1 / (10dB) at P1
I 2 ( r ) I 0 u 10 E 2 / (10dB) at P2
[A] …
Wave function is:
y(x, t) = 0.45 cos(3.0x 6.0t)
12.42: "By what factor must the sound intensity be increased
to increase the sound intensity level by 12.5 dB?”
Æ I2/I1 = ? = 10^{E 2 /10 – E1 /10 } = 10^{ (E2 – E1 ) / 10}
Æ 12.5 dB = E2 – E1
The wave function for a sinusoidal wave
moving in the +x-direction is
y(x, t) = A sin(Z t – k x),
where k = 2π/O , Z = 2Sf, O= v T …
Wave Motion
91
The Doppler Effect
89
Understanding Problem
™ Shifts in observed frequency can be caused by motion
motion o
of the
source, the listener, or both. Examples 12.10-12.13.
fL
v (0)
fS
v ( 30 m/s)
v (0)
fS
v ( 30 m/s)
L
fL
L
OL
v ( 30 m/s)
OS
v (0)
Example 12.9
fS = 300 Hz
v = 340 m/s
OL
I1 ( r ) I 0 u 10 E1 / (10dB) at P1
I 2 ( r ) I 0 u 10 E 2 / (10dB) at P2
v ( 30 m/s)
OS
v (0)
92
90
The Doppler Effect
™ Shifts in observed frequency can be caused by motion of the
source, the listener, or both. Examples 12.10-12.13 and P.1253,54,60
6
SA
L
95
How to Study Chap. 14
1) Heat transfer,
equilibrium, temperature:
P.14-5, 24, 27, 53, 56,
1
64, 74, 82
2) Thermal expansion: P.142
15, 16, 73
3) Phase change,
calorimetry: P.14-32,
44, 49
3
1
SB
' L D L0 ' T
Q mc ' T
H
dQ
dt
kA
THigh TLow
L
96
93
Thermal Expansion
Heat Capacity / Calorimetry
™ Substances have an ability to “hold heat”
that goes to the atomic level.
Q = m c 'T
[J] = [kg] [?] [K]
™ D: The expansion is proportional
to the original length and the
temperature change (for
reasonable 'T). (Table 14.1)
™ c = specific heat capacity [J / (kg * K)]
ƒ cwater = 4.19 x 103 J/(kg*K) vs. ccopper = 0.39 x 103 J/(kg*K)
™ What we see in life?
ƒ One of the best reasons to spray water on a fire is that it
suffocates combustion. But, another reason is that water has a
huge heat capacity. Stated differently, it has immense thermal
inertia. In plain terms, it’s good at cooling things off because
it’s good at holding heat.
ƒ Taking a copper frying pan off the stove with your bare hands is
an awful idea because metals have small heat capacity. In plain
terms, metals give heat away as fast as they can.
™ E: Volume expansion (Table 14.2; Example 14.4)
™ Examples 14.6 and 14.7 ; Examples 14.8 and 14.9
99
Stress on a Spacer
™ Consider a aluminum spacer
(L0=10 cm) at 17,2 oC.
™ Thermal Expansion
97
Phase Changes
™ The steam contains the energy (heat of vaporization) that it took to
become a gas. This is 2.3 MILLION joules per kg of water.
Example 14.5
™ The ice needs to absorb the latent heat of fusion to become a liquid.
' L D L0 ' T
? ' L / L0 D ' T
™ Stress (Chap. 11)
Y ˜ ' L / L0
? ' L / L0
FT / A
FT / ( A ˜ Y )
™ Thermal Stress
D ' T FT / ( A ˜ Y )
? FT / A D ' T ˜ Y
D (a lum inum )
Y (a lm inum )
2 . 4 u 10 5 K 1
Q/m= Lf = 3.34 x 105 J/kg
Q/m= Lv = 2.26 x 106 J/kg
0 . 70 u 10 1 1 P a
™ Example: Road Expansion and
Contraction
100
98
Understanding Problem
Understanding Problem
Phase Change
Q/m= Lv = 2.26 x 106 J/kg
Q ste p 1 Q ste p 2
'L
D L0'T
'd
d d0
m iro n c iro n ' T
At oC
At 78 oC
103
Chap. 15 : pV = nRT
D d0'T
101
Understanding Problem
Q water
dQwater
dt
104
m ice L f Ÿ
dQiron
dt
dQwater m ice L f
dt
600 sec
THig h TLow
kA
L
102
Molecular Heat Capacities
Key Numbers and Equations
n CV 'T = (3/2) n R 'T
CV = (3/2) R
107
105
The first law of thermodynamics
Q
'U
Details of Kinetic Property (II)
™ In simple terms, “the
energy added to a system
will be distributed between
heat and work”.
™ “Work” is defined
differently than we did in
earlier chapters, here it
refers to a p'v (a pressure
increasing a volume).
Q = 'U + W
'U
Container
Kav = (1/2) m v2
V, N, T, p
(3/2) nR T = N Kav = Ktrans
o N = Number of particles in
volume V
o Number of particles per unit
volume is N/V
Kav = (3/2) k T
o pV = n R T l pV = N k T
vi,x = -30 m/s
vi,x = 30 m/s
Q = n CV 'T + p (V2 – V1)
Ktrans = N Kav = (3/2) n R T
108
Chap. 8
'p = pf,i – pi,x
= (040 kg)(+30 m/s) – (0.40 kg)(-30 m/s)
=2 (Mass) x |vx|
106
Thermodynamic processes
™ A process can be adiabatic and have no heat transfer
in or out of the system
™ A process can be isochoric and have no volume change.
™ A process can be isobaric and have no volume change.
™ A process can be isothermal and have no temperature
change.
You heat a sample of air to twice its original
temperature in a constant-volume container. The
average translational kinetic energy of the molecules is
C. twice the original value.
111
109
You compress a sample of air slowly to half its original
volume, keeping its temperature constant. The internal
energy of the gas
You heat a sample of air to twice its original
temperature in a constant-volume container. The
average translational kinetic energy of the molecules is
A. decreases to half its original value.
B. remains unchanged.
C. increases to twice its original value.
A.
B.
C.
D.
half the original value.
unchanged.
twice the original value.
four times the original value.
Q = 'U + W
Q = 'U + W
Q = n CV 'T + p (V2 – V1)
Q = n CV 'T + p (V2 – V1)
Ktrans = N Kav = (3/2) n R T
Ktrans = N Kav = (3/2) n R T
112
110
You compress a sample of air slowly to half its original
volume, keeping its temperature constant. The internal
energy of the gas
B. remains unchanged.
113
You expand a sample of air adiabatically (Q = 0) from
its initial volume of 1.42 m3 (p = 1.38 x 105 Pa) to 2.27
m3 (p = 2.29 x 104 Pa). Compute the work done by air.
Cv of air is 20.8 J/(mol*K).
Q = 'U + W
0 = n CV 'T + W
'T = Tf – Ti
= pfVf/nR – piVi/nR
'U = Cv (pfVf - piVi)/R
114