kB = 1.38 10-23 J/deg
8/21/14
R = 8.314 J/(molK)
Giancoli, 4th ed. Internal energy: Secs. 18-1, 19-2, 19-8.
T-2. Kinetic Theory of Gases.
1. Discussion. Gas particles in continuous random motion are
t
8/21/14
Giancoli, 4th ed. Pressure: Sections 13-1 through 13-6. Ideal Gas Law: Sections 17-6 through 17-9.
R = 8.314 J/(molK).
Standard conditions: 1 atm = 1.01325 105 Pa, T = 273.15 K. 1 liter = 10 c
Physics 7B
Page 1
TN T1 (rev. 2.0)
Why doesnt the temperature change in a free expansion?
A free expansion occurs when a gas is allowed to expand freely into a vacuum. In the usual setup,
there is an
p. 1
T-2. Notes: Energy Contained in an
Jason Zimba
Ideal Gas
At the level of this course, temperature is best understood in terms of the Equipartition Theorem,
which states that KEparticle = (d/2)kBT
NAME _
PHYSICS 4B Exam 1 P. Tsai
You may use a bound printed dictionary with no marks,
the formula sheet provided, and a calculator.
Turn off all cell phones and communication devices, and place them
8/31/15
Giancoli, Sec. 19-1 to 19-5, 19-8.
p. 1
T-3. Calorimetry
calorie =
specific heat, c =
latent heat, L =
Mechanical equivalent of heat. James Prescott Joule, c. 1845.
Power =
With signed heat fl
9/9/15
Giancoli, 4th ed. Sec. 19-6 through 19-9.
Reversible processes: Gas is always described by its equation of state because it is always in
equilibrium.
T-4.
p. 1
Heat, Work, & The First Law of
Th
T-4
T-4. Notes
Jason Zimba
In the following, you may assume that all of the processes described are reversible.
3. When an ideal gas undergoes adiabatic expansion, the temperature
_ goes up
X
goes dow
from the floor to
a uniform magnetic field pointing straight up,
y—direction be
be upward, the xadirection be eastward, and the
A room is filled with
the ceiling. Let the 2-direction
northward.
(a) A
165
Chapter 39
(d)
culates clockwise or counterclockwise. To
answer this, apply the righthand rule to the cross oint your right fingers in the
direction of v. Initially, that's the ans-direction. Then
Chapter 39 167
° rt J ' : ~Q and mass or gets shot rightward at initial speed so between the two
plates of a parallel~plate capacitor. The plates, fixed a distance 5 apart,
I have length l. The part
Chapter 39 1 71
Before moving on to part (c), let me make a theoretical point. As demonstrated in this
problem, when a particle feels both a magnetic field and an electric field, the total force is
Fa
Chapter 39
(b)
169
N
moo S
= S—-———
Q30 '
where in the last step, I used our expression for r from subproblem 1. _
lntuitiveiy, the particle doesn’t deflect if the electric force cancels the magnetic
Chapter 39 175
where I denotes the length and A denotes the cross-sectional area. Intuitively, a wider tunnel allows
more “car-current” to flow through, and therefore puts up less resistance. Also, if
Chapter 39 17‘s
OK, here goes. The infinitesimal force on a tiny piece of the rod containing charge dq is
dFmag = dqv x B
qugxﬁ [sincev=dl/dt]
=dqglth
=Idle. [sincev=dl/dtl
You’ll use this crucial for
Chapter 39 17?“
Let’s begin by ﬁnding 1'2, the torque produced by F2. In this
diagram, I’ll drawn the relevant radius vector, from the center of
mass to the contact point where F: acts. I’ve also brok
Chapter 39 i 79
(b)
To speed up an object, you must apply a force. But not just any force. At least a component
of the force must be directed along the object's direction of motion. For instance, cons
.
ems : t ’f"
Ampere’s Law and Blot—Saved
(Before attempting this problem, please work through your textbook’s sample problems that use the Biot—
Saoart law to calculate a magneticﬁeld. This prob
Chapter 39 I 181
These two conﬂicting intuitions "cancel out.” Our part (b) answer, B = 21rﬁn/ q, does not
depend on R. In words, the magnetic field needed to keep the particle circling the cyclotron
Chapter 40 185
Speciﬁcally, you can choose an arbitrary piece of the hexagon, point your fingers in the
direction of current flow (d1), and then curl your ﬁngertips in the direction of it,
towards the
Chapter 40 187
(c)
undeflected, along an imaginary axis through the center of the hexagon. Let me call that axis the
"central axis.”
Look at your textbook’s drawing of the magnetic ﬁeld produced b
Chapter 40 191
depends only on where the objects are located, not on how fast they’re moving. Therefore, in parts
(b) and (c), the particle feels the same electric force it felt in part (a). If the pa
Chapter 40 189
So, the particle feels an electrostatic repulsion, and nothing else. Since Page = ngdue ,0 rod, the
meat of this problem is ﬁnding the rod’s electric ﬁeld. As you know, Gauss’ law enabl
Chapter 40 1 93
and hence
B _, #_01 : ﬂoquo
2m 22ml. '
where in the last step I used my part (a) expression for the current. By the Way, we just derived the
textbook equation for the field generat
Chapter 40 195
Fortunately, the superposition principle applies to all kinds of ﬁelds, not just electric ones.
If 31 denotes the field created by semicircle 1, 132 denotes the field generated by wir
197
Chapter 40
Two Very long cylindrical wires of length L, each of radius R, run
right next to each other, as pictured here. In the end View, the left wire carries
current I“ into the page, While t
Chapter 40 199
I’ll now rederive this same expression by thinking in terms of current density, the current per
cross—sectional area. Since the current density is uniform, I = Io/Awire z 10/ (1:12)." T