1. No category

Physics 262/266 - exam#1
1. (20 pts) Answer the following questions. (Use the space provided below
and the next page if needed.)
a) (6pts) Systems A and B are in thermal equilibrium with each other. i) Do they
necessarily have the same internal energy? ii) Will they have the same temperature?
iii) If system A is also in thermal equilibrium with another system C, what can you
say about the temperature for system C as compare with system B?
b) (6 pts) An ideal gas is reversibly expanded
from an initial state i to the final state f
through three different thermodynamic
processes (a, b, c).
i)
ii)
iii)
Which process will result in the
largest U?
Which process will result in the
largest work done by the gas W?
Which process will absorb the largest
amount of heat Q?
c) (4 pts) An ideal gas initially at state i is
being taken from an initial temperature T1
to a final temperature T2 through two
separate processes: (a) isobaric and (b)
isochoric, as shown.
P
b
i)
ii)
Will the entropy of the system
decrease or increase in both
processes?
Which process will result in a
larger change in the system’s
entropy?
a
T2
T1
i
V
d) (4 pts) The same amount of heat is applied to 1.00 mol of an ideal monoatomic gas
and 1.00 mol of ideal diatomic gas. Which one will result in a larger increase in
temperature?
SOLUTION:
a)
i)
No. The fact that systems A and B are in thermal equilibrium simply
means they have the same temperature and this fact implies that
microscopic elements within the system will have the same average
translational KE per molecule but the amount of microscopic elements
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Physics 262/266 - exam#1
ii)
iii)
within each system might be different and their molecular structure might
be different (e.g. monoatomic vs. diatomic) such that the total internal
energy for the two systems are not necessary the same even if they are in
thermal equilibrium.
Yes. If not, there will be heat exchanges between the two systems.
System C must have the same temperature as system B. 0th Law.
b)
i)
ii)
iii)
U is a state variable. Since all three processes (a, b, & c) have the same
initial and final states, we have U a  U b  U c .
Work done by the gas is the area under the curve. From the graph shown,
we have the following ranking Wa  Wb  Wc .
From the 1st Law, we have Q  U  W . Since all U ’s are the same,
the ranking for Q will be the same as the ranking for W, i.e., Qa  Qb  Qc .
c)
i)
ii)
In both processes (isobaric and isochoric), heat enters the system as the
temperature increases. Since dS  dQ T , the change in entropy for both
processes will also be positive (an increase).
For this problem, the change in temperature for both processes is the same.
Since C p  Cv , we have dQ p  dQv and S p  Sv .
Tf
Tf

dQ p ,v
 T 
dT
 nC p ,v 
 nC p ,v ln  f  
 note : S p ,v  
T
T

 Ti  
Ti
Ti
Thus, the isobaric process will result in a larger entropy increase
d) An ideal diatomic gas has more degrees of freedom so that its molar specific heat
is larger (Cdiatomic>Cmonoatomic). Recall that Q=nCT so that for the same amount
of heat input to the system, the temperature raise for a diatomic gas would be
smaller than for a monoatomic gas.
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2. (25 pts) 12.0g of water initially at 15.0oC was poured into a mental cup with mass
100.g at 450.oC. After the system has come to a thermal equilibrium, all water was
vaporized and the final temperature for the steam and metal cup was measured to be
exactly at 100.oC. i) What is the specific heat for the metal cup? ii) Calculate the total
entropy change S for the water heating from 15.0oC to 100.oC and vaporizing at
100.oC?
[ Lv  22.6  102 J g ; csteam  2.080 J g  C , cwater  4.186 J g  C ]
a) Let the specific heat for the metal cup be cm .
Q  0
gives
 heat released, 

  heat absorbed, 
tal
hot
me

   water warming    latent heat absorbed   0
  to vaporize at 100 C 
 cooling
 






 15 C  100 C 


 450 C  100 C 
This gives
mmcm 100 C  450 C   mwcw (100 C  15 C )  mw Lv  0
Rearranging terms, we then can solve for cm ,
mm cm  450 C  100 C   mwcw (100 C  15 C )  mw Lv
cm 
cm 
mwcw (100 C  15 C )  mw Lv
mm  450 C  100 C 
12.0 g  4.186 J / g  C  (100 C  15 C )  22.6 102 J / g 
100 g   450 C  100 C 
cm  0.897 J / g o C
b) In heating liquid water slowly, for a small change in temperature dT, the infinitesimal
heat absorbed by the water is given by:
dQheating  mw cw dT
Then, the infinitesimal change in entropy is dS heating 
dQheating
T
. For the full range of
temperature change 15 C  100 C  , we have:
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Physics 262/266 - exam#1
T 
mw cw dT
 mw cw ln  f  (temp has to be in K)
T
T
 Ti 
Ti
Ti
 373.15 K 
S heating  12.0 g  4.186 J g  K  ln 
  12.985 J / K (heating is an entropy
 288.15 K 
increasing process)
Tf
S heating 

dQheating
Tf


Now, the water vaporizes into steam at the boiling point, the heat of vaporization is
absorbed at a fixed temperature. The increase in entropy is then given by:
Svaporize 
12.0 g  22.6 102 J / g  72.685 J / K
mLv

Tboiling
 373.15K 
So, the TOTAL entropy increase is the sum of the two,
Stot  12.985 J / K  72.685 J / K  85.66 J / K  85.7 J / K
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3. (25 pts) One mole of an ideal
A
Pressure
monoatomic gas starting at state A (PA =
is
isothermally
200.kPa;
VA=10.0L)
expanded to a volume of 20.0L at state B.
Then, the gas is taken to state C by an
isochoric process. To complete the cycle, the
gas is returned to state A through a reversible
adiabatic compression. a) Calculate the
temperatures, TA , TB , TC at state A, B, and C.
b) Find the net work done by the gas per
cycle. c) What is efficiency of the cycle?
[1Pa = 1N/m2; 1atm  1.013 105 Pa ;1L  103 m3 ]
isothermal
adiabatic
B
C
Volume
a) First, we need to use the ideal gas law to find the temperatures at state A, B, and C.
PAVA  nRTA
(2.0  105 N / m 2 )(10 103 m3 )
TA 
8.315 J / mol  K
TA  240.53K
TB  TA  240.53K (AB is isothermal)
TCVC 1  TAVA 1 (C  A adiabatic)
 1
5
1
V 
 10  3
TC  TA  A   240.53   K  151.52 K
 20 
 VC 
b) AB: isothermal expansion
V
WAB  nRTA ln( f )
Vi
(work done by gas)
 1mol (8.315 J / mol  K )(240.53K ) ln(20 /10)  1.386kJ
BC: isochoric (no work done)
WBC  0
CA: adiabatic:
WCA  U CA  nCV (TA  TC )
(work done on gas)
3
  R (240.53K  151.52 K )  1.110kJ
2
Thus, Wnet  1.386kJ  1.110kJ  0.276kJ (net work is done BY gas).
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Physics 262/266 - exam#1
c) Heat is absorbed during the isothermal expansion and is given by,
QAB  WAB  1.386kJ (since U AB  0 in an isothermal process)
Then, from conservation of energy, we have
Wnet  QAB  QBC
gas)
 QBC  Wnet  QAB  0.276kJ  1.386kJ  1.11kJ (“-“ released by
(Note that QCA  0 since it is adiabatic by definition.)
Now, with heat absorbed found, we can calculate the efficiency of the heat engine:
e
Wnet 0.276kJ

 0.199  19.9%
QAB 1.386kJ
.
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Physics 262/266 - exam#1
4. (25 pts) For very low temperatures, the molar specific heat of a solid can be
3
T 
approximated by a cubic function of the absolute temperature, C (T )     , where  =
 T0 
1940 J/mol K is a constant and T0 is a parameter depending on the material. For this
problem, you are given 2.50 mol of potassium salt (KCl). T0 for this substance is given as
230.K. a) Calculate the heat needed to raise the temperature of this substance from 15.0K
to 20.0K. b) What is the entropy change for this heating process? [dQ=nC(T)dT]
a) From the definition of specific heat, we have dQ  nC (T )dT and
 n
Q   nC (T )dT   3
T
i
 0
f
f 3
 n  T f4  Ti 4 
  T dT   3 

T 
4
i
 0 

 2.5mol (1944 J / mol  K ) 
4
4

  20 K   15 K 
3
4(230 K )


 10.9 J

b) (Note: For entropy calculations, temperature needs be in Kelvin.)
f
f
dQ
.
For calorimetric changes, we have
In general, we have S   dS  
T
i
i
dQ  nC (T )dT . Thus, we have
f
S  
i
f
 n  T f3  Ti 3 
nC (T )dT  n  2
  3   T dT   3 
T 
T 
T
3
 0 i
 0 

 2.5mol (1944 J / mol  K ) 
3
3

  20 K   15 K 
3
3(230 K )


 0.616 J / K

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