Download Physics 103 Hour Exam #2 Solution Point values are given for each

Physics 103 Hour Exam #2
Solution
Point values are given for each problem. With a problem, the points are not
necessarily evenly divided between the parts. The total is 60 points.
Equations:
K = 12 mv 2
E=
kc q
r2
K = 23 kB T
V =
kc q
r
P V = nm RT
0.499mhvi
η=√
2π(2rm )2
Constants and conversion factors:
m
g = 9.8 2
s
P V = N kb T
`= √
1
2π(2rm )2 n
R = 8.314
J
K
kb = 1.38 × 10−23
1 mole = 6.02 × 1023
N m2
C2
mass of electron: me = 9.11 × 10−30 kg
(incorrect! should be 9.11 × 10−31 kg)
kc = 9.0 × 109
mass of proton: mp = 1.67 × 10−27 kg
J
K
e = 1.6 × 10−19 C
charge of electron: −e
charge of proton: +e
1 eV = 1.6 × 10−19 J
1. [5 points] The electric field of a small charged particle is measured to be 100 N/C
at a distance of 4 mm away from the particle. What is the electric field 2 mm
away from the particle?
From E = krc2q , we know that E is inversely proportional to r2 . Reducing r by a
factor of 2 (from 4 mm to 2 mm) reduces r2 by a factor of 4 and hence increases E
by a factor of 4. Thus the field is (4)(100 N/C) = 400 N/C.
1
2. [10 points] A defibrillator sends approximately 0.1 C of charge through a
patient’s chest in about 2 ms=0.002 s. The average voltage during such a
discharge is 3000 V.
(a) How much energy is involved in this discharge?
E =Vq =
(3000 V) (0.1 C)
J
VC
= 300 J
(b) What is the power output of the defibrillator during the discharge?
P =
E
=
t
300 J
0.002 s
Ws
J
= 150, 000 Watt
3. [5 points] The atmosphere of Mars is composed almost entirely of carbon dioxide,
CO2 . The atomic masses of carbon and oxygen are 12.0 and 16.0, respectively.
Consider some place near the Martian surface where the pressure is 800 N/m2
and the temperature is 230 K. What is the rms speed of the CO2 molecules?
The molecular mass of CO2 is 12+(2)(16)=44, so the mass of one molecule is
44 g
mole
1 kg
m=
= 7.31 × 10−26 kg.
mole
6.02 × 1023
1000 g
Equating the kinetic energy of an individual ideal gas particle to 21 mv 2 gives
2
1
2 mv
=
3
2 kB T
r
v
=
3kB T
=
m
s
J
3 1.38 × 10−23 K
(230 K) kg m2
m
= 360
2
(7.31 × 10−26 kg)
s
Js
2
4. [15 points] In a conventional television set, electrons are accelerated through a
voltage of 25,000 V. They then travel the length of the television tube and hit
the television screen on the other side.
A
B
path of electrons
electron
source
25000 V
Television
Screen
(a) The two metal plates in the diagram have a 25,000 V difference in voltage
between them. Electrons are generated (with essentially no energy) behind
metal plate A. In order to accelerate electrons toward the television screen
on the right, which plate must be at a higher voltage, A or B?
Electrons are negatively charged, so they lose potential energy (and hence
gain potential energy) by moving to a place with higher voltage. Hence B
must be at a higher voltage.
(b) How much energy do the electrons have after they pass through the small
hole in plate B? (Give your answer in whatever unit you deem convenient.)
A particle with charge e accelerated through 25000 V has energy 25000 eV.
J
) = 4 × 10−15 J, but notice that
This is equal to (25000 eV)(1.6 × 10−19 eV
it is much simpler to get the answer in eV.
(c) What is the speed of the electrons as they emerge from the hole in plate
B?
The electron mass given on the first page of the exam was
m = 9.11 × 10−30 kg. This is wrong. The actual mass is 10 times smaller
than this. Oops!! We will use the incorrect value of m here, along with the
energy calculated in part (b):
2
1
2 mv
= 25000 eV
s
m
(2)(25000 eV)
1.6 × 10−19 J
kg m2
v =
= 3 × 107
2
−30
s
9.11 × 10
kg
1 eV
Js
3
5. [15 points] The pressure in a canister of nitrogen gas is 2 × 105 N/m2 . The mean
free path of nitrogen molecules within the canister is ` = 100 nm = 1 × 10−7 m,
and the mean time between collisions is 0.2 ns = 2 × 10−10 s.
A pump removes some of the nitrogen molecules from the canister, reducing
the pressure by half, to 1 × 105 N/m2 . The temperature and volume are not
changed.
(a) What is the mean free path now?
From the ideal gas law, P V = N kB T , the number density of the gas is:
n=
N
P
.
=
V
kB T
Thus n is proportional to P . Reducing P by half reduces n by half. The
1
mean free path, ` = √2π(2r
, is inversely proportional to n, so if n is
)2 n
m
reduced by half, ` is doubled. Hence it is now `=(2)(100 nm)= 200 nm.
(b) How does the speed of the nitrogen molecules now compare with their
original speed?
The temperature has not changed, so the energy per molecule, 32 kB T
has not changed. Thus there is no change in the speed of the nitrogen
molecules.
(c) What is the mean time between collisions now?
Speed is v = distance
time . In this situation, the distance is ` and the time is
t, the mean time between collisions. Hence v = `t and hence t = v` . If `
has doubled while v remains the same, t must also have doubled, giving
t=(2)(0.2 ns)= 0.4 ns.
4
6. [10 points] You have two types of molecule, call them A and B. You have a 1 m3
cube box which is separated into two halves (the right half and the left half)
by a wall in the middle. You fill the left half with gas made up of A molecules
and the right half with gas made up of B molecules. The two types of molecules
have the same sizes and the same masses.
If you remove the wall between the two halves, the gases of A and B molecules
will diffuse into one another. This means that, over a long period of time, the
gases will combine to form a uniform mixture of A and B molecules.
There are various ways of measuring the diffusion rate. For example, you could
measure how much time it takes for 25% of the A molecules to be in the right
side, with only 75% remaining on the left side. The details aren’t important.
The point is that you could come up with ways of measuring how fast diffusion
happens.
We haven’t studied diffusion in class. However, with your understanding of how
molecules in gases behave, you should be able to answer the following questions.
(a) You do two experiments in which you measure the time for gases A and
B to diffuse into each other. Each of the two experiments uses the same
type of molecules, the same number of molecules, and a box of the same
volume. The only difference is that the first experiment is done at a lower
temperature and the second experiment is done at a higher temperature.
In which experiment does diffusion happen faster, the cooler experiment
of the hotter experiment? (Briefly explain.)
1
, is the same in the two experiments,
The mean path length, ` = √2π(2r
2
m) n
but the speed of the particles is higher in the hotter experiment (since
energy is proportional to temperature and speed is proportional to the
square root of energy). Thus the particles do the same type of “bouncing
around” (same `) in each experiment, but they do it faster in the hotter
experiment. Thus diffusion is faster there.
(b) You do two more experiments, both at the same temperature, same volume,
and with the same number of molecules. In the first experiment, you use
molecules A and B. In the second experiment, you use molecules C and
D instead. Molecules C and D have the same size as A and B, but they
have larger molecular masses. In which experiment do you expect diffusion
to happen faster, the one with lighter molecules or the one with heavier
molecules? (Briefly explain.)
As in part (a), both experiments have the same path length `. Since the
temperatures are the same, the energy per molecule is the same in both
experiments; however, since energy is 12 mv 2 , in the experiment with lighter
molecules the molecular speeds are higher. Thus diffusion happens faster
in the experiment with lighter molecules.
5