Calorimetric calculations:
dQ  mcdT or dQ  nCdT
Q  mL (latent heat )
( specific heat )
General Formulas applicable to ALL processes in an Ideal Gas:
PV  nRT
dU  dQ  dW
dW  PdV
dU  nCV dT
3
R (monoatomic)
2
CP  CV  R
CV 
Specific Processes:
Isothermal: T=constant  dT = 0  dU =0  dQ=dW
f
f
 Vf 
nRT
W   PdV  
dV nRT ln  
V
 Vi 
i
i
Adiabatic: dQ=0  dU= -dW
f
dW   dU
 W    nCV dT  nCV T f  Ti 
i

PV  const
TV  1  const
Isobaric: P=constant
f
 W   PdV P V f  Vi 
i
Isochroic: V=constant  dV=0  dW=0  dU=dQ
Entropy:
General Reversible Processes: dS=dQ/T
 S  nCV ln T f Ti   nR ln V f Vi 
Calorimetric Processes:
( specific heat ) : dQ  mcdT
or
dQ  nCdT
dT
dT
or  nC
T
T
T 
T 
 S  mc ln  f  or  nC ln  f 
 Ti 
 Ti 
(latent heat ) : Q  mL
 dS  mc
 S 
mL
T remains constant at phase change
T
Non Reversible Processes: (one cannot write dS=dQ/T for the process)
But one can use a surrogate reversible process with the same initial and final
states to calculate S .
2nd Law of Thermodynamics:
Stot  S system  Senvironment  0
Physics 262/266 - exam#1
1.
(25 pts) Answer the following questions. (Use the space provided below
and on the next page if needed.)
a). (9 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?
b). (8 pts) A Carnot engine operates between a high temperature reservoir at TH and a
low temperature reservoir at TL . One can increase the efficiency of the engine by
either increasing TH by T or decreasing TL by T . Which change TH  T  or
TC  T  will result in a larger efficiency for the engine?
c). (8 pts) i) An ideal gas is being taken
from an initial temperature T1 to a
final temperature T2 in two separate
processes: (a) isobaric and (b)
isochoric, as shown.
i)
ii)
Will the entropy of the system
decrease or increase in both
processes?
ii) Which process will result
in a larger change in the
system’s entropy?
P
b
a
T2
T1
i
V
SOLUTION:
a) i) 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 .
ii) Work done by the gas is the area under the curve. From the graph shown, we
have the following ranking Wa  Wb  Wc .
3/11
Physics 262/266 - exam#1
iii) 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 .
TL
so that both decreasing
TH
TL and/or increasing TH will gives a smaller TL TH and makes e closer to 1. As
T  T
for a fixed change in temperature, T , the ratio L
will generally be
TH
TL
so that e TL  TL  T   e TH  TH  T  .
smaller than
TH  T
b) For a Carnot engine, the efficiency is given by e  1 
To see that , we can consider the following ratio:
TL  T  TH  TLTH  TH  TL  T  T 2
TL TH  T 
TLTH
1
TH  TL
T 
T 2  1
TLTH
TLTH
Both blue terms are positive so that this ratio will always be less than 1 indicating
that,
 1
TL  T
TL
will always be smaller than
.
TH
TH  T
c) i) 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).
iii)
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  S v .
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.
4/11
Physics 262/266 - exam#1
2. (25 pts)
In a calorimetry experiment, a piece of hot iron initially at 500oC was placed in 1.50g of
water at 20.0oC. After the system has come to equilibrium, all water was vaporized and
the final temperature for the steam and hot iron was measured to be at 110oC? What was
the mass of the hot iron?
[ Lv  22.6  102 J g ; csteam  2.080 J g  C , cwater  4.186 J g  C , ciron  0.450 J g  C ]
Let the mass of the iron piece be mi
Q  0
gives
 heat released, 
 heat absorbed, 

  heat absorbed, 
hot
iron
latent
heat
absorbed







  water warming 
  steam warming   0





 cooling

 20 C  100 C   to vaporize at 100 C  100 C  110 C 







 
 500 C  110 C 
This gives
mi ci 110 C  500 C   mwcw (100 C  20 C )  mw Lv  mwcs 110 C  100 C   0
Rearranging terms, we then can solve for mi ,
mi ci  500 C  110 C   mwcw (100 C  20 C )  mw Lv  mwcs 110 C  100 C 
mi 
mi 
mi 
mw cw (80 C )  Lv  cs 10 C  
ci  390 C 
1.50 g   4.186 J
g  C  (80.0 C )   22.6 102 J g    2.08 J g  C 10.0 C  
1.50 g   4.186 J
g  C  (80.0 C )   22.6 102 J g    2.08 J g  C 10.0 C  
 0.450 J
g  C  390 C 
 0.450 J
g  C  390 C 
mi  22.4 g
22.4g of hot iron was initially used in the carlorimetry experiment.
6/11
Physics 262/266 - exam#1
3. (25 pts)
A
isothermal
Pressure
A Carnot cycle operated between two
temperature reservoirs at TH=500oC and
TL=200oC has one mole of an ideal monoatomic
gas as its working substance. You are given the
following parameters for the cycle: VA=10.0L,
VB=20.0L. Calculate a) the net work done by
the gas per cycle, b) the heat absorbed by the
gas per cycle, c) the heat expelled by the gas per
cycle. d) What is the efficiency of this cycle?
[1L  103 m3 , 1 atm= 1.013  105 Pa ]
B
adiabatic
adiabatic
D
C
isothermal
Volume
To get the volume at state C and D, we can use
the adiabatic processes:
TCVC 1  TBVB 1
1 (  1)
T 
VC  VB  H 
 TL 
Similarly,
TDVD 1  TAVA 1
1 (  1)
3
 500  273.15  2
 20 L 
  41.776 L
 200  273.15 
3
T 
 500  273.15  2
VD  VA  H 
 10 L 
  20.888L
 200  273.15 
 TL 
Note that since we really only need the ratio VD/VC, we can simply divide the above two
equations (without doing the numerical calculations) to get
VD VA

 0.5 .
VC VB
a) Net work done by the cycle:
Isothermal branches (AB, CD) ( U  0 ):
V
WAB  nRTH ln( B )
VA
(work done by)
 1mol (8.314 J / mol  K )(773.15 K ) ln(20 /10)  4.456kJ
V
WCD  nRTL ln( D )
VC
(work done on)
 1mol (8.314 J / mol  K )(473.15 K ) ln(0.5)  2.727 kJ
Adiabatic branches (BC, DA) (Q = 0):
3
WBC  U BC   nCV (TC  TB )  (1mol ) (8.314 J / mol  K )(473.15  773.15) K  3.741kJ
2
3
WDA  U DA  nCV (TA  TD )  (1mol ) (8.314 J / mol  K )(773.15  473.15) K  3.741kJ
2
Thus,
7/11
Physics 262/266 - exam#1
Wnet  4.456kJ  2.727 kJ  3.741kJ  3.741kJ  1.729kJ  1.73kJ
(net work is done BY gas).
b) Now we calculate the heat transfers:
AB: isothermal. QAB  WAB  4.456kJ
BC: adiabatic. QBC  0
CD: isothermal. QCD  WCD  2.727kJ
DA: adiabatic. QDA  0
Thus,
Qabosorbed  QAB  4.456kJ  4.46kJ
And the system releases heat during only the process CD so that
Qrelease  QCD  2.727kJ  2.73kJ
d) The efficiency of this heat engine is
e
Wnet
1.729kJ

 0.388  38.8%
4.456kJ
Qabsorbed
Shorter Method:
This is a Carnot cycle and its efficiency is given by:
e  1
TL
473.15 K
 1
 38.8% (same as before)
TH
773.15 K
Now, we need to calculate the net work done and it is given by the two isothermal
branches:
V 
V 
Wnet  WAB  WCD  nRTH ln  B   nRTL ln  D 
 VA 
 VC 
(Work done by the adiabatic branches (BC and DA) cancel out each other.)
From the adiabatic relation, TV  1  const , we can write:
TAVA 1  TDVD 1 and TBVB 1  TCVC 1 . Dividing these two equations, we have
TBVB 1 TCVC 1

. Since TA  TB  TH and TC  TD  TL , this ratios simplify to:
TAVA 1 TDVD 1
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Physics 262/266 - exam#1
VB VC

 2 . Inserting these ratios back into the work equations, we have,
VA VD
Wnet  nRTH ln  2   nRTL ln  0.5  1mol  8.314 J mol  K   773.15K  473.15K  ln 2
Wnet  1.729kJ
Then, we can use the definition of efficiency to get the absorbed heat:
e
Wnet
QH
 QH 
Wnet 1.729kJ

 4.456kJ
e
0.388
 QH  4.456kJ
Finally, to calculate heat related, we use the relation:
Wnet  QH  QL

QL  QH  Wnet  4.456kJ  1.729kJ  2.727kJ
 QL  2.727kJ
All these values match with the previous direct but longer calculations.
9/11
Physics 262/266 - exam#1
4. (25 pts)
a) One mole of a monoatomic gas is being compressed reversibly from an initial state
V1  5.00 L, P1  5.00 105 Pa  to a final state V2  3.00 L, P2  3.00 105 Pa  . Calculate
the entropy change S for this reversible process.
b) What will the entropy change S be for the same ideal gas if the process is done
irreversibly but with the same initial and final states as in part a)?
c) 10.0 g of water is heated slowly from 20 C to its boiling point at 100 C and all of it is
converted into steam at 100 C . What is the total entropy change S for this heating and
vaporization process? [ Lv  22.6  102 J g , cwater  4.186 J g  K ]
a) For any reversible process, the change in entropy is given by:
T
Sa  nCv ln  f
 Ti

 Vf 
  nR ln  

 Vi 
We have the volume ratio and to get the temperature ratio, we can use the ideal gas law,
T f Pf V f
PV
we have
 const 

T
Ti
PV
i i
Thus, we have ( Cv  3R / 2 for a monoatomic ideal gas)
 3   PV 
 V 
S a  (1mole)  8.314 J / mole  K    ln  2 2   ln  2  
 V1  
 2   P1 V1 
 3   9 
 3 
S a   8.314 J / K    ln    ln     17.0 J / K
 5 
 2   25 
The change of entropy is negative since the process is a compression.
b) Since entropy is a state variable and the initial and final states are the same as in part
a), the change in entropy for a nonreversible process between the same state 1 to 2 is the
same as in part a):
Sb  Sa  17.0 J / K
c) In heating liquid water slowly for a small change in temperature dT, the infinitesimal
heat absorbed by the water is given by:
dQheating  mcwater dT
10/11
Physics 262/266 - exam#1
Then, the infinitesimal change in entropy is dSheating 
dQheating
T
. For the full range of
temperature change  20 C  100 C  , we have:
T 
mcwater dT
 mcwater ln  f  (temp has to be in K)
T
T
 Ti 
Ti
Ti
 373.15 K 
S heating  10.0 g  4.186 J g  K  ln 
  10.10 J / K (heating is an entropy
 293.15 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 
mLv
10.0 g  22.6 102 J / g  60.57 J / K

Tboiling
 373.15K 
So, the TOTAL entropy increase is the sum of the two,
Stot  10.10 J / K  60.57 J / K  70.7 J / K
11/11