PES 2130
Exam 1/page 1
PES 2130 - Physics 3
Exam 1
Name:_____SOLUTIONS_______
Score:
/ 100
Instructions
Time allowed for this is exam is 1 hours 15 minutes
10 written problems
For written problems:
Write all answers in the spaces provided. An extra sheet is included at the back if you
need more space for your calculations. Answers to questions involving calculations
should be evaluated to appropriate significant figures and given in decimal form.
Despite an incorrect final result on written problems, credit may be obtained for method
and working, provided these are clearly and legibly set out.
Only required equipment may be placed on the desk.
No talking permitted.
No questions. If you are unsure of what is being asked, take your best guess and
state your reasoning.
Cell phones must be turned off.
Equipment
This Question/Answer booklet
Equation sheet (need to be turned in with Question/Answer booklet)
Pens, pencil, eraser, ruler, water bottle
Graphing or scientific calculator (cell phones may not be used as a calculator)
PES 2130
page intentionally left blank
Exam 1/page 2
PES 2130
Exam 1/page 3
Question 1 (10 points)
a) (2 points) A solid steel object has a hole in it. Which of these illustrations (#1 or #2 or
neither) more correctly shows how the size of the object and the hole change as the
temperature increases?
2 points: Illustration #1
b) (8 points) A hole in a steel plate (coefficient of linear expansion = 12.0 x 10 -6 K-1) has
a diameter of 20.0 cm at 0.00oC. What is the radius of the hole when the steel plate is
heated red hot, 700.oC ?
2 points
1 point
New diameter:
L = 20.168 cm
2 points
2 points
Radius = diameter/2
R = 10.084 cm = 10.1 cm
0.5 points
0.5 points
-0.5 units
-0.5 sig figs
-0.5 math mistake
- 1 did not calculate radius
PES 2130
Exam 1/page 4
Question 2 (10 points)
10.0g of ice at 0.00oC is mixed with 100 g water at 10.0oC in a thermos. The Specific
heat of water is cw = 4187 J kg-1 K-1 and for ice ci = 2108 J kg-1 K-1. The latent heat of
fusion to melt ice is 333 kJ/kg.
a) (2 points) What is the composition of this mixture at thermal equilibrium? Choose
from ice, water, or ice-water.
The final composition is all water.
(2 points)
b) (8 points) What is the final temperature of the composite?
The ice will first melt and then the water from the ice will get warmer as the initial water
will get colder.
Q  0  Q  Q
i
w
0  mi L  mi cw Ti  mw cw Tw
1 point
1 point +1 point +1 point
0  mi L  mi cw (T  Tii )  mw cw (T  Tiw )
0  mi L  mi cwT  mi cwTii  mw cwT  mw cwTiw
T
mi cwTii  mw cwTiw  mi L
mi cw  mw cw
rearrange 2 points
Tii  0
T
mw cwTiw  mi L
mi cw  mw cw
(0.1)(4187)(10)  (0.01)(333 103 )
857
T
K
K
(0.01)(4187)  (0.1)(4187)
460.57
T  1.86  C
Final answer 2 points
- 0.5 sig figs
- 0.5 units
- 0.5 wrong numerical result
else correct
PES 2130
Exam 1/page 5
Question 3 (10 points)
2.0 L of an ideal gas is at 10.0oC. It is heated at constant pressure to 40.0oC.
a) (3 points) Draw the corresponding p-V diagram and label the graph appropriately.
2 points shape
1 points correct labels
b) (7 points) Calculate the final volume of the gas.
pressure is constant hence:
pV = nRT
(2 points)
Vi nR V f


Ti
p Tf
V f  Vi
Tf
(2 points)
Ti
V f  (2)
40  273
L  2.2 L
10  273
- 0.5 sig figs
- 0.5 units
- 0.5 wrong numerical result else correct
-1 temperature not converted into Kelvin
not in Kelvin:
V f  (2)
40
L  8.0 L
10
(3 points)
PES 2130
Exam 1/page 6
Question 4 (10 points)
a) (4 points) State the "Equipartition Theorem" for 1 mole of gas.
Every "degree of freedom" (2 points)
that a microscopic object has is associated with energy of
- 1 point if answered
1
RT per mole ( 2 points)
2
1
kB T per molecule
2
b) (6 points) You have 1.00 mol of an ideal monatomic gas and 1.00 mol of an ideal
diatomic gas whose molecules can rotate. Initially both gases are at room temperature. If
the same amount of heat flows into each gas, which gas will undergo the greatest increase
in temperature and why? Use equations to explain.
Answer: the monatomic gas
(2 points)
detailed explanation (4 points)
(-2 points if answer was: KEtr depends on temperature only)
3
1 mole of monatomic gas:  KEtr av  RT
2
5
1 mole of diatomic gas:  KEtr av  RT
2
Assume W = 0:
1st law: Eint  Q  W
Eint  Q    KEtr av
1 mole of monatomic gas:
3
3
Q  RT  R  T fm  T 
2
2
2Q
 T  T fm
3R
1 mole of diatomic gas:
5
5
Q  RT  R  T fm  T 
2
2
2Q
 T  T fd
5R
Hence Tfm > Tfd
PES 2130
Exam 1/page 7
Question 5 (10 points)
A straight-line process is done on a monatomic ideal gas, as shown. The temperature of
the gas at point A is 500 K and the temperature at point B is 300 K.
a) (2 points) For the straight-line process shown, is work done ON the gas or BY the gas?
For full points you need to explain your reasoning without doing a calculation.
answer:
Work is done BY the gas.
b/c gas expands
1 point
1 point
b) (6 points) What is the magnitude of work done?
Work = are underneath p-V diagram
2 points
Work = area of bottom rectangle + area of top triangle
2 points
W = (2.0 m3)(1.0 Pa)+ (0.5) (2.0 m3)(4.0 Pa)
W = 6.0 J
2 points
- 0.5 sig figs
- 0.5 units
- 0.5 wrong numerical result
c) (2 points) Is heat added to the gas or removed from the gas? For full points you need to
explain your reasoning.
answer: Heat is added
1 point
Explanation:
1 point
Work is done by the system and therefore energy must be added into the system in form
of heat.
Calculation (was not required)
Q=∆Eint + W = 3/2 nR∆T +W
3 pV 
3  5 1 
Q   1 1   T2  T1   W  
  300  500  J  6 J  3 J  6 J  3 J
2  T1 
2  500 
PES 2130
Exam 1/page 8
Question 6 (10 points)
The figure below shows a flow diagram for a heat engine.
TL=300K
a) (6 points) Does the heat engine violate the first law of thermodynamics? You must
show a calculation to explain your answer.
first law: Eint  Q  W
for cycle: 0  Q  W
Q = QH+ QL = 30J - 20J = 10J
Q=W=10J
2 points
1 point
2 points
First law OK
1 points
b) (4 points) What is the efficiency (in %) of this engine?
heat engine efficiency:  

Wout 10 J 1

  0.33
QH
30 J 3
The efficiency is 33%
Wout
QH
2 points
1 point
1 point
- 0.5 answer not in %
- 1 used TH and TL instead of QH and QL in part a)
PES 2130
Exam 1/page 9
Question 7 (10 points)
Two 250g potatoes at 295K are thrown into a 2.00 L pot of water at 375K. The final
equilibrium temperature of the potato-water system is 363 K. Assume that this system is
isolated from the surroundings.
a) (2 points) Will the change in entropy of this system increase or decrease? For full
points you need to explain your reasoning without doing a calculation.
(1 points) The entropy of the system will increase
(1 points)
- The second law of thermodynamics can be stated in terms of entropy: No process is
possible in which the total entropy of an isolated system decreases.
- Because heat does not flow spontaneously from cold to hot.
b) (8 points) Calculate the change in entropy of this system using specific heat of potato,
cp = 3430 J kg-1 K-1, and specific heat of water, cw = 4187 J kg-1 K-1. You will also need
dx
 x  ln( x) .
Stotal  S p  S w
f
S  
i
dQ
T
dQ  mcdT
f
S  mc 
i
2 points
2 points
1 point
 Tf 
dT
 mc ln  
T
 Ti 
 Tf
Stotal  m p c p ln 
T
 ip
1 point

 Tf 
  mw cw ln  
 Tiw 

 363 
 363 
Stotal  (0.5)(3430)ln 
  (2)(4187) ln 

 295 
 375 
Stotal   355.738  272.349  J / K
Stotal  83.39 J / K  83.4 J / K
- 1 sig figs
- 1 units
- 1 wrong numerical result else correct
2 points
PES 2130
Exam 1/page 10
Question 8 (10 points)
a) (2 points) How did we define temperature of a substance?
Temperature is a measure of the average (translational) kinetic energy of the molecules,
characterized by the Maxwell-Boltzmann distribution.
give full points if "characterized by the Maxwell-Boltzmann distribution" part is missing.
b) (5 points) Calculate the translational root-mean-square (rms) velocity of water vapor
molecules at room temperature, i.e. at T = 293 K. The molar mass of a water vapor
molecule is 18.0 g/mol.
vrms 
3RT
M
2 points
M = molar mass (needs to be in kg/mol)
M = 18.0 x 10-3 kg/mol
vrms 
3RT

M
3(8.31)(293)
m / s  637m / s
18.0 103
1 point
2 points
- 1 sig figs
- 1 units
wrong answer if M not converted into kg/mol
3RT
3(8.31)(293)
vrms 

m / s  20.1m / s
M
18.0
c) (3 points) If a water molecule at room temperature is traveling on average at 587 m/s,
briefly explain why a puddle of water will eventually evaporate at room temperature.
Temperature is a measure of the average kinetic energy of particles, characterized by a
Maxwell-Boltzmann distribution. At room temperature, a large percentage of its particles
have a temperature close to 25°C, but some have a temperature farther away. When some
of those particles that happen to have a temperature of 100°C are at the surface of the
water, they have enough energy to break away and evaporate. This cools the rest of the
puddle down, but the puddle is constantly in interaction with other substances around it
(ground, air) which put that heat back into the puddle. So, the puddle maintains a constant
temperature (roughly) and shrinks smaller and smaller as particles near the surface get
enough kinetic energy to evaporate.
PES 2130
Exam 1/page 11
Question 9 (10 points)
The figure below shows several isotherms (lines of equal temperature) of an ideal gas on
a p-V diagram.
Starting at point a clearly indicate (draw path with label) in the figure the following
thermodynamics processes:
a) (2 points) constant pressure volume expansion
b) (2 points) constant volume change in pressure
c) (2 points) constant temperature volume expansion
d) (2 points) adiabatic volume expansion
a)
b)
c)
d)
e) (2 points) In which process of these four processes is the work done by the gas zero? Is
it process a), b), c) or d) and why?
Answer:
process b)
area below p-V diagram is zero: W = 0
1 point
1 point
PES 2130
Exam 1/page 12
Question 10 (10 points)
(Do this problem last.)
An ideal gas is expanded adiabatically from state 1 (p1, V1, T1) to state 2 (p2, V2, T2).
Staring with an infinitesimal change in the system
 nR  dT    p  dV 
(Equation 1)
proof by showing a detailed derivation that
T1 V11/    T2 V21/   ,
(Equation 2)
where α is a constant.
[Hint: Which state variable (p,V, or T) in Equation 1 does not appear in Equation 2?
dx
You will need   ln( x ) .]
x
no pressure p in Eq 2, hence use ideal gas law to sub in p:
2 points
pV = nRT
p = nRT/V
2 points
nRT
 dV 
V
dT
dV
rearrange and simplify: 

T
V
sub into Eq. 1:  nR  dT   
2 points
T2
integrate both sides from state 1 to state 2:
V
2
dT
dV

 
T
V
T1
V1
2 points
solve evaluate integrals, take exponential and rearrange:
T 
V 
 ln  2    ln  2 
 T1 
 V1 

1
 T2 
 V2 
V1
    
V2
 T1 
 V1 
1/ 
  T  
 2  
  T1  


1/ 
V 
 1 
 V2 
2 points
1/ 
T2  V1 
 
T1  V2 
T2V21/   T1V11/ 
______________________
END OF EXAM
____________________________
PES 2130
______
Exam 1/page 13
EXTRA SPACE BELOW FOR CALCULATIONS
_________________
PES 2130
Exam 1/page 14