Review for Final Exam - Physics and Astronomy at TAMUpeople.physics.tamu.edu/.../2013C/Review_phys201_Fi… · · 2013-12-04Review for Final Exam 1 ... [Fall 2013 Final Exam Schedule]

RReview for Final Exam

1

Problem 1 Understanding motion of objects in 2D (Chap. 3 + 2) Problem 2 Understanding Newton's laws (Chap.4 + 5 + 6) 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)

� BONUS ASSIGNMENT: Here (Deadlne: 12:30 pm, Dec. 3 (Tue)) � FINAL EXAM: Dec 6 (Fri) (3 pm - 5 pm) at MPHY 203

[Fall 2013 Final Exam Schedule] [Formular Sheet for Exam4 and Final]

� Old PHYS 218 Exam Solutions: 1. 2. 3. Final. � TENTATIVE PLAN for Final Exam as of Dec 1 (6 PM):

EExams 1, 2, 3, and 4

3

Resending Nov-9 Message *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) 40 or better, D (>=50)

Exams – returned during the classes and picked up in my office.

Resending Nov-9 Message

HHow to Understand Your Scores

RReview for Final Exam

4

5

6

7

8

1

Kinematics (2D)

I

III

UUnderstanding Projectile Motions

II

9

PProjectile Motion

10

– 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

PProjectile Motion

11

– 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)

FFurther Look at Projectile Motion

(2) ax = 0 ay = �g = �9.80 m/s2

(1) Choose an origin & an x-y coordinate system

vx = constant

(3) vy = 0

(4) y = 0

12

Projectile motion as horizontal motion with constant velocity (ax = 0) and vertical motion with constant acceleration (ay = �g).

II

[Quick Quiz 2] Is this a motion with a constant acceleration or with a varying acceleration?

13

TThe Same Problem? II

Kinematics (2D) Kinematics (2D)

y = 0

Vy = 0

y = �1.00m(1) Choose an origin & an x-y coordinate system

(2) ax = 0 ay = �g = �9.80 m/s2

(3) Use kinematic eqs.in x and y separately.

Vx = constant

14

II

SSame Problems?

15

R

R

y = 0 m

y = ? m

II

16

II

Kinematics (2D)

t = 7.6 s

Example 2: A projectile is launched from ground level to the 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 2sin�� cos� = sin2�� 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 �). b. Find a formula of tan� in terms of g, R, H and T.

17

A projectil launched from ground level tole is la too the t

The Same Problem?

II

Question: How fast must the motorcycle leave the cliff-top?

18

The Same Problem? Example 3.3. I

FFiring at a More Complex Target �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.

19

http://www.youtube.com/watch?v=cxvsHNRXLjw

III

Kinematics (2D)

A boy on a small hill aims his water-balloon 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.

20

A boy on a small hill aims his water-balloon 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.

III

Kinematics (2D) Kinematics (2D)

Same Concept

200 m, given � x?

d, given � H?

H?

?

21

H

III

I

222

http://link.brightcove.com/services/player/bcpid36804639001?bckey=AQ~~,AAAACIJPQzk~,qiwYyUrE_-dz5lglGrCClkfJDM1jW3zH&bclid=0&bctid=109459228001

Can you explain? Magic? or Physics?

Giancoli’s Textbook 3rd Ed.

We see “An apple in motion (x direction) tends to stay in motion (x direction).” � Motion with constant velocity in x direction.

III

23

2

Newton’s Laws of Motion

Kinematics (r, v, a) ��

Structure of Newtonian Mechanics

Inertial Reference Frame (Newton’s 1st Law)

Action-Reaction (Newton’s 3rd Law)

Mass (m)

The Nature of Force The Nature of Object The Nature of Motion

F = m a (Newton’s 2nd Law)

� �

Kinematics (r, v, a) � � �



24


The force on a hokey puck causes the acceleration

If the net force on a hokey puck is zero (equilibrium), the acceleration is zero.

0

0��

��

��

��

a

F

225

Force: Acceleration/Equilibrium Acceleration � Kinetic Equations (see Chap. 2 &3)


[A]

226

Quick Quiz

(b)

A hockey puck is sliding at a constant velocity across a flat horizontal ice surface. Which is the correct free-body diagram?


a�FN

Ff

Fg

mFa f�

� �

FN

Ff

Fg

Ignore the truck and two people!

Ff is the force on the box by the truck’s bed.

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:

27


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?

a = 0.20 g

FG

FN

Ff

where Ff = �s FN = �s (mg) Thus, �s (mg) = m (0.20g) So, �s = 0.20

Ff

Which one is correct?

Newton’s 2nd law: F = m a � Ff = m (0.20g)

x

28

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 c d eTaken from Fig. 5-38 (Giancoli)

29


Draw a Diagram � FBD � Newton’s Laws D Di

30


Draw a Diagram � FBD � Newton’s Laws

31

332

Weight on Planet

� 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?


Example 2: A block (mass m1) is placed on a smooth horizontal 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.

FN

Fg1

FT

FT

Fg2

FT = m1 a Fg2 – FT = m2 a

m2 g – m1 a = m2 a

m2 g = m1 a + m2 a

a = m2 g / (m1 + m2 ) a

a

x 33

Circular Motion

The quantity v2/R is not a force - it doesn’t belong in the free-body (force) diagram

Which one is correct?

F�

Rvm

2

F�

F�F

�

INCORRECT CORRECT

F.B.D. of Uniform Circular Motion

34

Thinking …

Circular Motion

Swinging a ball on the end of string … [Quick Quiz] Which one is correct?

(a) (b)

335

� Model airplane on a string

336

Example Problem 1

Rotation � Center-seeking Force � FT

r̂r

vm amF

2

T

rad ��

�

One revolution every 4.00 seconds yy

Tr

timecetandis v 2�

��

Circular Motion

EExample 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?

37

r̂rvm am

F

2

f

rad ��

�

mgF �N

�smin

= Ff/FN

Circular Motion

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?

338

Example Problem 2

No Skidding on Banked Curve

r̂r

vm am2

rad ��

mgF N

�s = Ff/FN

UUnderstanding Satellite Motion

39

r̂r

vm am2

rad ��

r̂r

mMGF 2��

• G = 6.67 x 10�11 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

A tetherball problem – Example 6.2

440

41

3

Work and Energy

A 50.0-kg crate is pulled 40.0 m by a constant force exerted (FP = 100 N and � = 37.0o) by a person. Assume a coefficient of friction force �k = 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.

WWork Energy Theorem

42

Wnet = Wi = 1302 [J] (> 0)

Wnet = Kf – Ki

= (1/2) m vf2 � 0

Energy Method Equations of Motion

43

Energy Conservation

Find h Using Energy Conservation

=

ISEE ISEE

44

EEEEEnnnnnnnnnnnneeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeerrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooonnnnnnnnnnnnnnnnnnnnnnnnnservation

45

46

447

Momentum Conservation nnnnn

(A) M

omen

tum

Con

serv

atio

n

(B) Energy Conservation

1

2

Same Concept with Ballistic Pendulum 8-85

8-44, 8-83

48

Momentum Conservation

(A) M

omen

tum

Con

serv

atio

n

(B) Energy Conservation

1

2

Same Concept with Ballistic Pendulum

A

B

C

Skeet+Pellet

49


50

Similar Problem

51

52

4

Rotational Motion

NNewton’s Laws for Rotation

1st part [s–2]

3rd part [N m]

2nd part [kg m2]

Krot = (1/2) I �2 (�K = (1/2) m v2)

53

Chap. 10

Angular Momentum Conservation

RRotational Energy and Inertia

m1

m2

m3 54

Mass x Distance^2

Rotational Motion

RRotational Inertia

55

(factor) (Mass x Distance^2)

MMechanical Energy Conservation

K = Km + KM

Krot = (1/2) I �2 (�K = (1/2) m v2)

Rotational Motion

Solid disk (M, R0)

m

57

0

Mechanical Energy Conservation

RRace of the objects on a ramp �This is a classic multiple-choice question from

MCAT-style standardized tests. �Refer to Figure 9.23.

2cm

2

21

21 �IMMgH �� v

Icm = Large � v = Small

59

Can you explain why?

Rotational Motion

Analyze the rolling sphere in terms of forces and torques: find the magnitudes of the velocity v and ….

60

Can you find v?

Rotational Motion

61

Ktrans + Krot = Krolling

amF �� net

�� I�net

62

Rotational Motion

663

Angular Momentum Conservation P10-30 (MP)

EExample 2

64

�i�i��f�f

P10-33 P10-34 (MP)

65

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.

Q10.11

666

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.

A10.11

667

The four forces shown all have the same magnitude: F1 = F2 = F3 = F4.

Which force produces the greatest torque about the point O (marked by the blue dot)?

A. F1

B. F2

C. F3

D. F4

E. not enough information given to decide

Q10.1 F1

F2

F3

F4

O

68

The four forces shown all have the same magnitude: F1 = F2 = F3 = F4.

Which force produces the greatest torque about the point O (marked by the blue dot)?

A. F1

B. F2

C. F3

D. F4

E. not enough information given to decide

A10.1 F1

F2

F3

F4

O

69

Which of the four forces shown here produces a torque about O that is directed out of the plane of the drawing?

A. F1

B. F2

C. F3

D. F4

E. more than one of these

Q10.2 F1

F2

F3

F4

O

70

Which of the four forces shown here produces a torque about O that is directed out of the plane of the drawing?

A. F1

B. F2

C. F3

D. F4

E. more than one of these

A10.2 F1

F2

F3

F4

O

71

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. (0.80 m)(900 N)sin 19°

B. (0.80 m)(900 N)cos 19°

C. (0.80 m)(900 N)tan 19°

D. none of the above

Q10.3

772

A. (0.80 m)(900 N)sin 19°

B. (0.80 m)(900 N)cos 19°

C. (0.80 m)(900 N)tan 19°

D. none of the above

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?

773

A. m2g = T2 = T1

B. m2g > T2 = T1

C. m2g > T2 > T1

D. m2g = T2 > T1

E. none of the above

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?

774

A. m2g = T2 = T1

B. m2g > T2 = T1

C. m2g > T2 > T1

D. m2g = T2 > T1

E. none of the above

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?

775

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. T = w

B. T > w

C. T < w

D. not enough information given to decide

Q10.6

776

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. T = w

B. T > w

C. T < w

D. not enough information given to decide

A10.6

777

A solid bowling ball rolls down a ramp.

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

D. more than one of the above

E. The answer depends on whether the ball rolls without slipping.

Q10.7

778

A solid bowling ball rolls down a ramp.

Which of the following forces exerts a torque on the bowling ball about its center?

A10.7

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.

779

80

5

81

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.

Visualizing SHM

[Bonus (10 pts)] If you have time, sketch x-t and v-t graphs of this motion.

xmka x ��xkF x ��

vmax, ax=0

Mechanical Energy Conservation

SSHM to Wave Motion

82

[Q] How can you describe the shape of the rope? [A]

x = R0 cos ��where �� = � t

x(t) = R0 cos (� t)

Kin. Equation of SHM

Simple Harmonic Oscillator (SHO)

T = 2� / �

A, ��

Figure 15.4

and TIME dependence…

83

vwave = ��/ T�

vwave = ��(� /2�)�

vwave = �� f�

SStanding Wave

Wave Motion

84

(a) FT if � = 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)? f2 = 2 f1 f3 = 3 f1

Wave Motion

Example 2 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] …

85

The wave function for a sinusoidal wave moving in the +x-direction is y(x, t) = A sin(� t – k x), where k = 2π/� , � = 2�f, ��= v T …

Oscillations

GGraphs

86

Which of the following wave functions describe a wave that moves in the –x-direction?

A. y(x,t) = A sin (–kx – �t)

B. y(x,t) = A sin (kx + �t)

C. y(x,t) = A cos (kx + �t)

D. both B. and C.

E. all of A., B., and C.

Q15.2

87

Which of the following wave functions describe a wave that moves in the –x-direction?

A. y(x,t) = A sin (–kx – �t) = A sin ( – (kx + �t) )

B. y(x,t) = A sin (kx + �t)

C. y(x,t) = A cos (kx + �t)

D. both B. and C.

E. all of A., B., and C.

A15.2

88

Wave Motion

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. [A] … Wave function is: y(x, t) = 0.45 cos(3.0x �� 6.0t)

Example 3

89

The wave function for a sinusoidal wave moving in the +x-direction is y(x, t) = A sin(� t – k x), where k = 2π/� , � = 2�f, ��= v T …

UUnderstanding Problem

90

2dB)10(

02

1dB)10(

01

Pat 10)(

Pat 10)(2

1

/

/

IrI

IrI�

�

��

��

Example 12.9


91

12.42: "By what factor must the sound intensity be increased to increase the sound intensity level by 12.5 dB?”

� I2/I1 = ? = 10^{�2 /10 – �1 /10 } = 10^{ (�2 – �1 ) / 10} � 12.5 dB = �2 – �1

Example 12.9

2dB)10(

02

1dB)10(

01

Pat 10)(

Pat 10)(2

1

/

/

IrI

IrI�

�

��

��

� Shifts in observed frequency can be caused by motion of the source, the listener, or both. Examples 12.10-12.13.

SL fv

vfm/s) 30()0(

��

�

L

SL fv

vfm/s) 30()0(

��

�

L

fS = 300 Hzv = 340 m/s

observed frequency can be caused by mmotion omotionThe Doppler Effect

SL vv

��)0(m/s) 30(

��

�SL v

v��

)0(m/s) 30(

��

�

92

TThe Doppler Effect

L SA SB

� Shifts in observed frequency can be caused by motion of the source, the listener, or both. Examples 12.10-12.13 and P.12-53,54,60

93 95

6

HHow to Study Chap. 14 1) Heat transfer,

equilibrium, temperature: P.14-5, 24, 27, 53, 56, 64, 74, 82

2) Thermal expansion: P.14-15, 16, 73

3) Phase change, calorimetry: P.14-32, 44, 49

96

LTT

k AdtdQH

TmcQTLL

LowHigh ��

��

�� 0

1

1

2

3

HHeat Capacity / Calorimetry � Substances have an ability to “hold heat”

that goes to the atomic level. Q = m c �T [J] = [kg] [?] [K]

97

� c = specific heat capacity [J / (kg * K)] � cwater = 4.19 x 103 J/(kg*K) vs. ccopper = 0.39 x 103 J/(kg*K)

� What we see in life? � 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.

� 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.

� Examples 14.6 and 14.7 ; Examples 14.8 and 14.9

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.

Q/m= Lv = 2.26 x 106 J/kg Q/m= Lf = 3.34 x 105 J/kg

98

� The ice needs to absorb the latent heat of fusion to become a liquid.

TThermal Expansion � : The expansion is proportional

to the original length and the temperature change (for reasonable �T). (Table 14.1)

99

� �: Volume expansion (Table 14.2; Example 14.4)

SStress on a Spacer � Consider a aluminum spacer

(L0=10 cm) at 17,2 oC. � Thermal Expansion

� Stress (Chap. 11)

� Thermal Stress

� Example: Road Expansion and

Contraction

100

)(0

0

YA/FL/LA/FL/LY

T

T

��

��

TL/LTLL

� ��

�

0

0

YTA/FYA/FT

T

T

��

� � )(

Example 14.5

P a 10700(a lm inum )

K 1042(a lum inum )1 1

15

��

��

.Y

.


101

TLL � � 0�

Tdddd � � 00 ��

At �78 oC At �� oC


102

LTT

k Adt

dQdt

dQ

Lmdt

dQLmQ

LowHighironwater

ficewaterficewater

��

��sec 600


103

Q/m= Lv = 2.26 x 106 J/kg

TcmQQ i r o ni r o ns t e ps t e p �� 21

Phase Change

CChap. 15 : pV = nRT

104

Key Numbers and Equations

105

Details of Kinetic Property (II)

106

V, N, T, p

o N = Number of particles in volume V

o Number of particles per unit volume is N/V

o pV = n R T � pV = N k T

�p = pf,i – pi,x = (040 kg)(+30 m/s) – (0.40 kg)(-30 m/s) =2 (Mass) x |vx|

vi,x = 30 m/s

vi,x = -30 m/s

Chap. 8

Container

Kav = (3/2) k T

(3/2) nR T = N Kav = Ktrans

Kav = (1/2) m v2

107

Molecular Heat Capacities

CV = (3/2) R

n CV ��T = (3/2) n R �T

108

� 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).

The first law of thermodynamics

Q = n CV ��T + p (V2 – V1) �U

�U

Q = �U + W

Q

Ktrans = N Kav = (3/2) n R T

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.

109

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. half the original value. B. unchanged. C. twice the original value. D. four times the original value.

110

Q = n CV ��T + p (V2 – V1)

Q = �U + W


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

You compress a sample of air slowly to half its original volume, keeping its temperature constant. The internal energy of the gas

A. decreases to half its original value. B. remains unchanged. C. increases to twice its original value.

112

Q = n CV ��T + p (V2 – V1)

Q = �U + W


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).

114

0 = n CV ��T + W

Q = �U + W

�T = Tf – Ti = pfVf/nR – piVi/nR

�U = Cv (pfVf - piVi)/R

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Review for Final Exam - Physics and Astronomy at TAMUpeople.physics.tamu.edu/.../2013C/Review_phys201_Fi… · · 2013-12-04Review for Final Exam 1 ... [Fall 2013 Final Exam Schedule]