Transcript Do Now (11/7/13):

Do Now (11/7/13):
• What are the major topics and concepts from
our unit on Fluids and Heat?
AP Physics Fluids and Thermo
Review
11/7/13
AP Guidelines
• No calculators during multiple choice
• Formula sheets during multiple choice and
free response
• At bell, multiple choice will be collected and
free response will be handed out
1) Which of the following could NOT be used to
indicate a temperature change? A change in:
A) color of a metal rod
B) length of a liquid column
C) pressure of a gas at constant volume
D) electrical resistance
E) mass of one mole of gas at constant pressure
2. One mole of gas is in a closed container of volume Vo and
pressure Po. The initial temperature of the gas is T0 The container
of gas is heated up to a temperature of 3To. Which of the
following statements is (are) true about gas after it has reached
this new temperature:
I. The pressure of the gas increases to 3Po.
II. The kinetic energy of the gas molecules increases by a factor of 3.
III. The root mean squared speed of the gas molecules increases by a
factor of 3.
(A) I only
(B) II only
(C) I and II only
(D) III only
(E) I, II, and III
3. At which of the points—A, B, C, or D—is the
gas at the greatest temperature?
(A) Point A
(B) Point B
(C) Point C
(D) Point D
(E) Gas is at the same temperature at all points.
4. In which of the following processes is the gas
undergoing a cooling phase?
I. AB
II. BC
III. CD
IV. DA
(A) II only
(B) IV only
(C) III and IV
(D) II, III, and IV only
(E) I only
5. In which process is no work done?
I. AB
II. BC
III. BD
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only
6. A sample of gas (assumed to be ideal) is
adiabatically compressed to have its volume
reduced to 20% of its initial volume. If the
internal energy of the gas is increased by
200 J, the work done on the gas must be
(a) 40 J
(b) 100 J
(c) 200 J
(d) 400 J
(e) zero
Answer:
6. Since the process is adiabatic, no heat energy
flows from the surroundings to the gas or
from the gas to the surroundings. So, the
entire work done in compressing the gas is
utilised in increasing the internal energy of the
gas. The work done on the gas is therefore
equal in value to the increase in internal
energy, which is C. 200 J.
7. The Greenland ice sheet can be one km thick.
Estimate the pressure underneath the ice.
(The density of ice is 918 kg/m3.)
a. 9.0 × 105 Pa (9 atm)
b. 2.5 × 106 Pa (25 atm)
c. 4.5 × 106 Pa (45 atm)
d. 9.0 × 106 Pa (90 atm)
e. 1.2 × 107 Pa (120 atm)
8. Water (density = 1 × 103 kg/m3) flows at 10 m/s
through a pipe with radius 0.030 m. The pipe
goes up to the second floor of the building, 2.0 m
higher, and the pressure remains unchanged.
What is the radius of the pipe on the second
floor?
a. 0.046 m
b. 0.034 m
c. 0.015 m
d. 0.012 m
e. 0.010 m
9. In a large tank of liquid, the
hydrostatic pressure at a given
depth is a function of:
a. depth.
b. surface area.
c. liquid density.
d. Choices a and c are both valid.
e. Choices a and b are both valid.
10. What volume of water is displaced
by a submerged 2.0-kg cylinder made
of solid aluminum?
(aluminum density = 2.7 × 103 kg/m3
and water density = 1.0 × 103 kg/m3)
a. 7.4 × l0−4 m3
b. 1.4 × 103 m3
c. 9.9 × 103 m3
d. 6.0 × 102 m3
e. 4.2 × 102 m3
11. Water is being sprayed from a nozzle at
the end of a garden hose of diameter 2.0
cm. If the nozzle has an opening of diameter
0.50 cm, and if the water leaves the nozzle
at a speed of 10 m/s, what is the speed of
the water inside the hose?
a. 0.63 m/s
b. 0.80 m/s
c. 2.5 m/s
d. also 10 m/s
e. 12 m/s
12. A fountain sends water to a height
of 100 m. What must be the
pressurization (above atmospheric) of
the underground water system? (1 atm
= 105 N/m2)
a. 1 atm
b. 4.2 atm
c. 7.2 atm
d. 9.8 atm
e. 12 atm
13. A heavily loaded boat is floating in a pond. The
boat starts to sink because of a leak but quick
action plugging the leak stops the boat from
going under although it is now deeper in the
water. What happens to the surface level of the
pond?
a. It stays the same.
b. It goes up.
c. It goes down.
d. More information is needed to reach a
conclusion.
14. A large stone is resting on the bottom of
the swimming pool. The normal force of the
bottom of the pool on the stone is equal to
the:
a. weight of the stone.
b. weight of the water displaced.
c. sum of the weight of the stone and the
weight of the displaced water.
d. difference between the weight of the stone
and the weight of the displaced water.
e. weight of the water in the swimming pool.
Free Response
• White Boards
Free Response #1 A
A small container of gas undergoes a
thermodynamic cycle. The gas begins
at room temperature. First, the gas
expands isobarically until its volume
has doubled. Second, the gas expands
adiabatically. Third, the gas is cooled
isobarically; finally, the gas is
compressed adiabatically until it
returns to its original state.The initial
state of the gas is indicated on
the PV diagram below. Sketch this
process on the graph.
Answer:
Free Response #1 B
A small container of gas undergoes a thermodynamic
cycle. The gas begins at room temperature. First,
the gas expands isobarically until its volume has
doubled. Second, the gas expands adiabatically.
Third, the gas is cooled isobarically; finally, the gas
is compressed adiabatically until it returns to its
original state. The initial state of the gas is
indicated on the PV diagram below.
Is the temperature of the gas greater right before or
right after the adiabatic expansion? Justify your
answer.
Answer:
The temperature is greater right before the
expansion. By definition, in an adiabatic process,
no heat is added or removed. But because the
gas expanded, work was done by the gas,
meaning the W term in the first law of
thermodynamics is negative. So, by ΔU = Q + W, Q
= 0 and W is negative; so ΔU is negative as well.
Internal energy is directly related to temperature.
Therefore, because internal energy decreases, so
does temperature.
Free Response #1 C
A small container of gas undergoes a
thermodynamic cycle. The gas begins at room
temperature. First, the gas expands isobarically
until its volume has doubled. Second, the gas
expands adiabatically. Third, the gas is cooled
isobarically; finally, the gas is compressed
adiabatically until it returns to its original state.
The initial state of the gas is indicated on
the PV diagram below.
Is heat added to or removed from the gas in one
cycle?
Answer:
In a full cycle, the gas begins and ends at the same
state. So the total change in internal energy is zero.
Now consider the total work done on or by the gas.
When the gas expands in the first and second
process, there's more area under the graph than
when the gas compresses in the second and third
processes. So, the gas does more work expanding
than compressing; the net work is thus done by the
gas, and is negative. To get no change in internal
energy, the Q term in the first law of
thermodynamics must be positive; heat must be
added to the gas.
Free Response #1 D
A small container of gas undergoes a
thermodynamic cycle. The gas begins at room
temperature. First, the gas expands isobarically
until its volume has doubled. Second, the gas
expands adiabatically. Third, the gas is cooled
isobarically; finally, the gas is compressed
adiabatically until it returns to its original state.
The initial state of the gas is indicated on
the PV diagram below.
Does this gas act as a refrigerator or a heat
engine?
Answer:
This is a heat engine. Heat is added to the gas,
and net work is done by the gas; that's what a
heat engine does. (In a refrigerator, net work
is done on the gas, and heat is removed from
the gas.)
Free Response:
Answer the free response sheet with your
group.
A.
a.
b.
c.
d.
240 K
720 K
2160 K
720 K
B.
• W = 8000 J
C.
• 29, 900 J
D.
• Efficiency – 33%
Good luck!!!