1 chapter 6. the second law of thermodynamics. 2 ranque and his tube: no moving parts compressed air...
TRANSCRIPT
1
Chapter 6.
The second law of Thermodynamics
2
Ranque and his tube: No moving parts
compressed air 2 kmoles, 4 atm., 294K
COLD HOT1 kmole, 1 atm., 333K
1 kmole, 1 atm., 255K
hollow tube
Is this a crazy idea or what?
3
In earlier chapters we alluded to aspects of thermodynamics which require further discussion or analysis.
1. Many processes do not occur in nature even though they are not forbidden by the first law. Is there some other law stating that these processes cannot occur?
2. Is there some general way to decide if a process is reversible or irreversible?
3. Can we write the first law solely in terms of state variables?We examine these questions in this chapter. We can convert to a state variable by multiplying it by the integrating factor (1/P): where V is a state variabledV
Pr
Similarly if we multiply by an integrating factor which turns out to be (1/T) we obtain an exact differential. We write
dST
r S=entropy (reversible processes)
đWđW
đQ
đQ
4
The r in the above equations emphasizes that we are dealing with reversible processes.Hence dU= đQ- đW PdVTdSdUbecomes
The boxed equation is the first law in terms of state variables.It connects any two neighboring equilibrium states. Of course we must still prove that dS is an exact differential.
The boxed formula is probably the most important equation in thermodynamics.
5
Irreversible processes.A reversible process is one whose direction can be reversed
by an infinitesimal change in some property. It is a quasi-static process in which no dissipative forces are present. All reversible processes must be quasi-static, but not all quasi-static processes are reversible (e.g., slowly letting the air out of a tire). In a reversible process everything (including the surroundings) can be restored to its initial state.
All natural processes are, in actual fact, irreversible. How can we determine if a process is reversible or irreversible?
On the other hand any process involving dissipative work, such as rubbing two solids together, is irreversible.
6
Dissipative work is done on the system. As an example consider a system consisting of a stirrer and fluid. Work done by rotating the stirrer is converted to “heat” energy. To reverse the process, the same amount of heat would need to be extracted from the fluid to perform the original amount of work, with no other changes. This does not violate the first law. A new law, the second law of thermodynamics, will be introduced which forbids this reverse process.
Consider a gas in one portion of a chamber which is then allowed to undergo a free expansion into an evacuated section of the chamber. The gas will not compress itself back to the original volume. Again any process which attempts to restore the gas to its original volume, with no other changes, would violate the second law.
Consider heat flowing from a high-T body to a lower-T body in the absence of other effects. The reverse process does not take place. It is impossible to create a device to simply reverse the process. Again the second law would be violated.
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The impossibility of the above reverse processes occurring is contained in the second law of thermodynamics.Clausius statement: It is impossible to construct a device that
operates in a cycle and whose sole effect is to transferheat from a cooler body to a hotter body.
Kelvin-Planck statement: It is impossible to construct a devicethat operates in a cycle and produces no other effect thanthe performance of work and the exchange of heat with asingle reservoir.
Carnot’s theoremNo engine operating between two reservoirs can be more
efficient than a Carnot engine operating between the same two reservoirs.
Remember that a Carnot engine is an ideal reversible engine and hence can be run as a refrigerator.
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2QW
2Q
WQ 2WQ 2
High T
Low T
2T
1T
I R2
R2=Carnot refrigerator (so reversible)I= Engine (not an ideal reversible one)
In the drawing all quantities are magnitudes. The diagram is of a self-contained device operating between two reservoirs. One part of the device is a Carnot refrigerator and the other part is an engine. The system is adjusted so that the work output from I is used to drive R2.
Let us assume that the efficiency of I is greater than R2. We would then have and so
22 Q
W
Q
W
22 QQ
The heat extracted from the low temperature reservoir is
2222 )()( QQWQWQ By our assumption, this is positive.
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The heat deposited to the high-T reservoir is: 22 QQ
Therefore the net result of this cyclic system is to extract some heat from the low temperature reservoir and deposit it into the high-T reservoir, without any work being done by the surroundings. This violates the Clausius statement of the second law. Hence our original assumption is false and the efficiency of I must be less than, or at most equal to, the efficiency of the Carnot cycle.
)R()I(
10
Suppose now that I is replaced by a Carnot engine R1. There is no reason that we cannot do this. The analysis proceeds as before and,based on the above analysis, )1()2R()1R(
The nice thing about our Carnot devices is that they are reversible.Hence we can reverse the engines so that R1 is now the refrigeratorand R2 is the heat engine. Again the analysis proceeds as before and we obtain )2()1R()2R(
Comparing equations (1) and (2) we conclude that
)2R()1R(
All Carnot engines (operating between the same temperatures) have the same efficiency.
In the above analysis no particular working substance was assumed.The efficiency of the Carnot cycle is independent of the working substance.
11
Breaking news!
Two devils have invaded our laboratory from their den at the centerof the Earth.
The red devil is capable of violating the Clausius Statement of the 2nd Law. That is, she can take energy from a low T reservoir and deposit it into a reservoir at higher T, without any other changes, in some cyclic process. (She argues that it’s OK because the 1st Law is not violated.)
The blue devil is capable of violating the Kelvin-Planck Statementof the 2nd Law by extracting energy from a single reservoir and performing some work in a cyclic process without depositing any energy into a reservoir at lower T. (He argues that it’s OK because the 1st Law is not violated.)
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1QW
2Q
High T
Low T
2T
1T
1Q 1Q engine
We wish to demonstrate that, if the Clausius Statement of the 2nd
Law is violated, then the Kelvin-Planck statement is also violated.We begin with an ordinary engine (blue). It takes some energy from the high T reservoir, does some work and then dumps some waste energy into the low T reservoir.
But the dreaded red devil (Violator of theClausius statement) sneaks into the system.The devil takes the energy that the engine deposited into the low T reservoir and puts it back into the high T reservoir.
This complete system then takes energy from the high T reservoirand performs some work with
no change in the low T reservoir.This violates the Kelvin-Planck statement of the 2nd Law!
13
In other words, the Kelvin-Planck statement says that it is impossibleto construct any cyclical system (no matter how complex!) whichtakes energy from a single reservoir and perform useful work. Butthis statement is violated if we permit a violation of the Clausius statement (with the aid of the red devil).
A violation of one statement of the 2nd Law leads to a violation of the other statement of the 2nd Law. Hence they are equivalent.
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We wish to demonstrate that, if the Kelvin-Plank Statement of the 2nd Law is violated, then the Clausius statement is also violated.
2Q
W
12 QQ
High T
Low T
2T
1T
R
1Q
A device (R) would like to extract some energy from the low T reservoir and deposit the same amount of energy into the high T reservoir. This does not violate the 1st Law. We know that this is impossible as it would cause Herr Clausius to be very unhappy. Enter the blue devil (which violates the Kelvin-Planck Statement.) He extracts some energy from the high T reservoir and uses it to perform an equal amount of work without depositing energy into the low T reservoir.
He uses this work to drive the refrigerator. The result of this system is to take a certain amount of energy from the low T reservoir and deposit the same amount into the high T reservoir, without any influence from, or change of, the environment external to the system.
1Q
15
We are now going to study entropy in some detail. In preparationfor this discussion we prove a theorem given on the next slide.
In other words, the Clausius Statement says that it is impossible to construct any cyclical system (no matter how complex!) which takes energy from a low T reservoir and transport it to a high Treservoir with no change in the system or envionment. But thisstatement is violated if we permit a violation of the Kelvin-Planck statement (with the aid of the blue devil).
Again a violation of one statement of the 2nd Law leads to a violation of the other statement of the 2nd Law. Hence they are equivalent.
To emphasize that our development is completely general we introduce a generalized force Y (which could be P) and a generalized displacement X (which could be V) and we consider some reversible process if as shown on the diagram.
ai
adiabatY
b
adiabat
X
f
reversible processisotherm
17
We now demonstrate the following: Any arbitrary reversible process, in which the temperature may change in any manner, can be replaced by two reversible adiabatic processes connected by a reversible isothermal process in such a way that the heat exchanged over the isothermal process equals that exchanged over the original arbitrary process.
In the above diagram, adiabats are drawn passing through i and f.
18
Curve ab represents an isothermal process. It is selected so that the area under iabf is equal to the area under the original curve. This ensures that the work done in the two processes is equal.
Since the two paths connect the same states, the change in internal energy must be the same also. Therefore the heat exchanged must also be the same. However, for the curve iabf, the heat exchange takes place only during the isothermal part. This completes the proof.
19
Now we consider some reversible cycle as shown below.
adiabats
reversiblecycle
T2
T1
We draw a large number of adiabats, dividing the cycle into a large number of narrow strips, one of which is shown on the diagram. The isotherms (T1 and T2) are drawn as discussed above. We now have a Carnot cycle.
P
V
. i
f.
20
Heat energy Q2 enters the system during the isothermal process at T2
Heat energy Q1 leaves the system during the isothermal process at T1
We have previously shown, for a Carnot cycle,
Now we revert to a notation in which heat entering a system is positive and heat leaving the system is negative. Hence and we have
2
2
1
1
T
Q
T
Q
02
2
1
1 TQ
TQ
11 QQ
{Remember that “heat energy” means an energy transfer by the heating process.}
21
Clausius Theorem: The integral of around any reversible cycle equals zero.
đQ/T
0R
T
Continuing with the other Carnot cycles we have
04
4
3
3
2
2
1
1 TQ
TQ
TQ
TQ
The Q’s represent infinitesimal heat transfers. Taking the limit, the sum becomes an integral and we have
đQ
22
We can use the Clausius Theorem to show the following:
Consider 2 points (i, f) on any reversible cycle giving
2path1path0i
fT
f
iT
RT đQ đQ đQ
2path1pathf
iT
f
iT đQ đQ
is independent of the path taken. It depends onlyon the state of the system at the initial and final points. Hence we can introduce a state function S, called the entropy, by
đQ/T
TRdS đQ
Reversible process!
23
The main points so far in this chapter are:
First Law: PdVTdSdU
Second Law:
Clausius statement: It is impossible to construct a device that operates in a cycle and whose sole effect is to transferheat from a cooler body to a hotter body.
Kelvin-Planck statement: It is impossible to construct a devicethat operates in a cycle and produces no other effect thanthe performance of work and the exchange of heat with asingle reservoir.
A Carnot engine is the most efficient engine.2
1
T
T1
Clausius Theorem 0R
T TRdS đQ đQ
24
In preparation for the next point we review the efficiency.
2
1
2
1
2
12
2 T
T1
Q
Q1
Q
Q
W
Remember that is negative. Hence is negative. 1Q2
1
Q
Q
Hence the smaller (more negative) is, the smaller the efficiency. 2
1
Q
Q
Carnot !!
For example, if the efficiency is zero, while if ,
the efficiency is 0.5
1Q
Q
2
1 2
1
Q
Q
2
1
25
Now consider an irreversible cycle between the same two reservoirs. Since I < R this gives
2
1
2
1
2
1
TT
I
I or 02
2
1
1 TQ
TQ II
or, for an infinitesimal heat transfer, 02
I2I
I1TT đQ đQ
As before 0TI
đQ
In this expression /T is not an entropy.đQ
Combining our result with the expression for a reversible process we have
Clausius Inequality. 0T đQ
This is sometimes taken as a statement of the second law.Now we have a way to decide whether or not a process is reversible!
(!!!!!!!!!)
26
Now we wish to examine, again, the change in entropy of an irreversible process.
Consider going between two equilibrium states if by some irreversible process. We complete the cycle by fi by some reversible process. Because part of the cycle is irreversible
0i
fT
f
iTT
RI đQ đQđQ
S
0f
iT
f
iT
f
iT
f
iT
RI
RI
đQđQ
đQđQ f
iT
f
i
IdSS đQ
The change in entropy is greater than the integral going from the initial state to the final state in the irreversible process!
27
Combining with the reversible case:TdS đQ
For isentropic processes: dS = 0 and so 0T đQ
For an adiabatic process đQ = 0 and so
Now consider an isolated system. Then đQ =0 and so 0dS
The entropy of an isolated system increases in any irreversible process and is unaltered in any reversible process.
It is the system’s entropy that governs the direction of spontaneous change.
0dS
Let us again consider an isolated system. Suppose we start at some initial state which has an entropy Si .Now we begin some process. As the process proceeds, the entropy can only increase or, at best, remain unaltered. If now, at some stage, we reverse the direction of the process, the entropy must continue to increase, or remain the same. If the entropy continues to increase, the system can never regain the initial state with entropy Si . Hence it is obvious that , for a process to be reversible, the entropy cannot change.
We now can state when a process is irreversible:In an isolated system, an irreversible process is one in
which the entropy increases.
29
Now suppose the isolated system is in equilibrium, so no changes are taking place in the system. Then
0dS system 0systemdS
If , this implies a change towards equilibrium. Since the entropy of the system can only increase, this means that the equilibrium state is the state of maximum possible entropy. This is true regardless of the type of process taking place in the system.
Thus, at equilibrium, the entropy of an isolated system has its maximum value with respect to all possible variations and the condition for this maximum is that 0systemdS
An example of an isolated system is a universe. We must then have
0 universeS
0 gssurroundinsystemuniverse SSS
NOTE: One of these two terms can be negative!!!
30
This principle of increasing entropy gives a direction for the sequence of natural events. The laws of mechanics are second order equations in time and are unaltered by the transformation t -t. According to mechanics all physical processes can run backwards. This is not so. The process can only go in a direction which increases the entropy. Entropy provides the “arrow of time” for physical processes.
As will be discussed later, increasing entropy is associated with increasing disorder.
31
Entropy and available energy.What is the significance of the statement that entropy is
created in an irreversible process? Of course, it tells us whether or not a process is possible. For example a chemist might wish to know if two substances will react under certain conditions. If the reaction would decrease the entropy of the universe, it is not possible.
Something about the energy is irretrievably lost during an irreversible process, even though the energy is conserved. This is often stated as the “availability of the energy to do useful work”.
Consider a simple situation:- the transfer of Q from a high-T reservoir to a low-T reservoir through a conductor. This reduces the capacity of Q to be converted to do useful work. Let us consider this in more detail.Suppose an even lower temperature reservoir exists.
Consider a Carnot engine operating between and and extracts energy Q from the high-T reservoir.
2T 1T
0T
2T 0T
2
01T
T
Q
W
32
Increasing T of reservoirs
2T
1T
0T
C
C
Q
Q
Q
W
W
same
33
Now suppose that the heat Q had been transferred to the reservoir at and we now operate the same Carnot engine between and extracting energy Q from the T1 reservoir, then
1T1T 0T
1
01T
T
Q
W
The difference is the loss in work that Q can do as a result of it being transferred from the reservoir at to the reservoir at the lower temperature .
WW
2T1T
WWSTT
Q
T
QT
T
T
T
TQWW 0
21
0
2
0
1
0
S is the increase in entropy which has resulted from the energy transfer.
Although we have chosen to discuss a simple case, the result is true in general. Although the energy is not lost, it has been degraded in that it is less available do useful work.
34
To summarize:All natural processes are irreversibleEvery irreversible process makes energy less available to
do external work.All natural processes continually make energy less available
to do useful work.
35
EXAMPLE: We consider an irreversible engine.
2T
1T
I
2Q
1Q
W
Because the process is irreversible, we cannot draw the cycle. We consider one complete cycle.Since S is a state variable: 0)( IS
1
1
2
2)(T
Q
T
QreservoirsS
)()()( reservoirsSsystemSuniverseS Must have 0)( universeS
2
1
2
1
2
2
1
1
1
1
2
2 0T
T
Q
Q
T
Q
T
Q
T
Q
T
Q
2
1
2
12
2
1Q
Q
Q
Q
WI
Since RIRI T
T
T
T
Q
Q
2
1
2
1
2
1 1
36
EXAMPLE: A 0.5kg sample of water (the system) at 90oC cools to room temperature (surroundings) at 293K. Calculate the entropy changes.
Kkg
JcP
4180
Since the entropy is a state variable, the change in entropy is the same as if the water were cooled reversibly from KTtoKT 293363 12
2
1P
T
T
P T
Tlnmc
T
dTmc
T
Q)system(S
1
2
P is constant, so
K
JsystemS
Kkg
JkgsystemS 448)(
363293
ln4180)5.0()(
The heat enters the room (surroundings), which is at constant T and so its entropy increases (the entropy change must be positive).
The energy by the heating process that leaves the water is
)TT(mcdTmcQ)system(Q 21P
T
T
P
1
2
đ
đ
37
K
JgssurroundinS 499)(
)()()( gssurroundinSsystemSuniverseS
K
J51
K
J499
K
J448)universe(S
051)( K
JuniverseS
K
J499
T
TTmc)gssurroundin(S
1
12P
The energy that enters the surroundings is )TT(mc 12P
38
EXAMPLE: Two reservoirs are separated by a diathermic wall. We consider an interval of time over which some heat is exchanged, the amount being sufficiently small so that the
KTH 500 KTL 200
JQ 5104
adiabatic walls
diathermic wall
LresHressystemuniverse SSSS )()()()(
K
J
K
J
T
Q
T
QS
L
L
H
Huniverse 200
104500104
)(55
K
JS universe 1200)(
Can the heat flow be in the opposite direction? Why not?
temperatures of the reservoirs do not change appreciably.
39
EXAMPLE (Problem 6.5):A thermally insulated 50 ohm resistor carries a current of 1A for 1 sec. The initial temperature of the resistor is 100C, its mass is 5 grams and the specific heat capacity at constant pressure is 850 J/(kgK).(a) What is the entropy change of the resistor? (b) What is the entropy change of the universe?
(a)dT
Rdtimcmc
dT
QRdtiQRiP
2
PP
P
22
K
KkgJ
kg
sAT
mc
RdtidT
P
8.11)850)(105(
)1)(50()1(
3
22
KTKT fi 8.294283
T
dTmcdsdTmcQbut
T
QdS P
P
283
8.294ln850)105(ln 3
Kkg
Jkg
T
Tmc
T
dTmcS
i
fP
T
T
P
f
i
đđ
đđ
40
K
JresistorS 174.0)(
(b) Surroundings isolated, so 0)( gssurroundinS
K
JuniverseS 174.0)(
41
Remember:T
QdS (reversible process)
Q into system, the entropy increases
Q out of system, the entropy decreases
For any closed cycle (because U and S are state variables):
0SU
0T
Q
R
đ
đ
đ
đ
42
EXAMPLE:
C85.274 C85.1370
C85.548C85.0
1
3109
V
2
31020
V
1P
2P b
a
c
d
Consider one cycle of this reversible process. The heat capacities are:
K
JC
K
JC PV 108
(a) Q=?
J2192Q)K274(K
J8dTCQ ab
b
a
V
P(Pa)
)( 3mV
J10960Q)K1096(K
J10dTCQ bc
c
b
P J6576Q)K822(
K
J8dTCQ cd
d
c
V J5480Q)K548(
K
J10dTCQ da
a
d
P JQcycle 1096
For the cycle JWU 10960 (b) ?P W=area enclosed
)()())((
12121212 VV
WPPPPVVW
PaPPm
JP 4
12331096.9
1011
1096
đ
đ
đ
đ
43
(c)
i
f
T
TlnCS
T
dTC
T
QS?
T
Q
K
JS
K
J
T
TCS ab
a
bVab 545.5
274548
ln8ln
K
JS
K
J
T
TCS bc
b
cPbc 10.11
5481644
ln10ln
K
JS
K
J
T
TCS cd
c
dVcd 545.5
1644822
ln8ln
K
JS
K
J
T
TCS da
d
aPda 10.11
822274
ln10ln
0T
Q because S is a state variable (like U)
đđ
đ
EXAMPLE (Problem 6.1): A Carnot engine operates on 1 kg of methane which we will consider to be an ideal gas. The value of the ratio of specific heats is 1.35. If the ratio of the maximum to the minimum volume is 4 and the efficiency is 0.25, find the entropy increase of the methane during the isothermal expansion.
2T
1T a
b
c
d
adiabat adiabat
P
V
kmolen
kmolekg
kgnCHkg 0635.0
161
1 4
25.0V4V35.1 bd
)1......(34
43
25.011
12
2
1
2
1
2
1
2
1
TT
T
T
Q
Q
T
T
Q
Q
isothermal expansion:
b
cc
b V
VnRTPdVWQsoU ln0 22
)2.....(V
VlnnR
T
Q)expansion isothermal(S
b
c
2
2
45
adiabatic expansion 1d1
1c2
1 VTVTconstantTV
Using (1) 35.01
11
2
1
1
43
43
43
d
c
d
c
V
V
T
T
V
V
76.1)4(43 35.0
1
b
c
b
d
d
c
b
c
V
V
V
V
V
V
V
V
From (2)
)76.1ln()Kkmole
J10314.8)(kmole0625.0(exp)iso(S 3
4CHforK
J294)expansionisothermal(S
Energy enters the engine during this part of the cycle so the entropy increases.
46
EXAMPLE (Problem 5.11): A reversible engine is connected to 3 reservoirs as shown below. During some integral number of complete cycles the engine absorbs 1200J from the reservoir at 400K and performs 200J of mechanical work. (a) Find the quantities of energy exchanged with the other reservoirs, and state whether the reservoir gives up or receives energy(b) Find the change in entropy of each reservoir.
R
W=200J
KT 4001 KT 3002 KT 2003 2Q
3Q
JQ 12001
(a) 0
U
WQU
after n cycles
WQQQWQ )( 321
)1.....(1000)(
)(2001200
32
32
JQQ
QQJJ
For n complete reversible cycles, the entropy change of the engine=0
K
J
T
Q
T
Q
T
Q
T
Q
T
Q
T
Q
4001200
01
1
3
3
2
2
3
3
2
2
1
1 (see slide 20)
47
)2.....(33
3
2
2
K
J
T
Q
T
Q
From (1) JQQ 100023 Placing in (2)
3222
3
3
2
2
2 3100031000
TK
JJQQ
T
T
K
J
T
JQ
T
Q
JQJJQQ 1200600100032
222
From (1) JQJQQ 2001000 323
(b) For the reservoirs where Q is for the engine T
QS
KJS
K
J
T
QS 3
4001200
11
1 KJS
K
J
T
QS 4
3001200
22
2
KJS
K
J
T
QS 1
200200
23
3 ΔS1+ ΔS2 + ΔS3=0 (Of course)
(Out of engine)
(Into engine)
It is the requirement that the entropy change be zero that heat istransferred into the reservoir which has the middle temperature.
48
EXAMPLE: A mass of liquid at a temperature T1 is mixed with an equal mass of the same liquid at a temperature T2. The system is thermally isolated. Calculate the entropy change of the universe in terms of cp and the temperatures and show that it is necessarily positive. (Assume reversible process connecting states.)
2T
0)()(0)2()1( 21 TTmcTTmcQQ fPfP
)1).....((21
2 212121 TTTTTTTTTT ffff
2
P
1
P2121 T
dTmc
T
dTmc
T
Q
T
QdSdSdS
21
lnln1 2
T
T
T
Tmc
T
dT
T
dTmcS ff
P
T
T
T
T
P
f f
2
2121
lnlnTT
Tmc
T
T
T
TmcS f
Pff
P
21
21
2ln2
TT
TTmcS P
(universe)
đđ
49
One way to show that this is an increase in entropy is to show that
12 21
21
TT
TT
x
xxfx
T
TLet
2
1)(
1
2
1041
2
12/3
xdx
df
x
x
xdx
df
041
18
)1(321
2
2
2/52/32
2
dx
fdxat
x
x
xdx
fd
Hence the minimum value of x=1 (temperatures the same) and so the minimum change in entropy is zero
A second way is as follows:
12
2)1(
2
221
2
1 2
x
xx
x
xxx
x
xf
50
EXAMPLE: The value for for some substances can be reasonably represented by
PcbTacP
(a) Find the heat absorbed and the increase in entropy of a mass m of the substance when its temperature is increased at constant pressure from T1 to T2.
(b) From a plot for the specific heat for Cu (see Figure 4.1) find the change in the molal specific entropy for Cu, when the temperature is increased at constant pressure, from 500 K to 1200 K.
(a) 2
1
T
T
PP dT)bTa(mdTcmQdTmcQ
)(
2)( 2
12212 TT
bTTamQ
dTT
)bTa(mS
T
dT)bTa(m
T
dTmc
T
Qds
2
1
T
T
P
)(ln 12
1
2 TTbT
TamS
đ
đ
51
(b) From the graph, at 400K at 1000K
Kkmole
JcP
31075.29
Kkmole
JcP
3105.25
This gives Kkmole
Ja
Kkmole
Jb
4
21027.208.7
KKkmole
J
Kkmole
Js )5001200(08.7
5001200
ln1027.22
4
Kkmole
Js
41048.2
52
Example: A system is taken reversibly around the cycle adcba
(a) Calculate the heat transferred in each process.(b) Does the cycle operate as an engine or a refrigerator?(c) Find the efficiency of this cycle operating as an engine.(d) What is the coefficient of performance of this cycle
operating as a refrigerator.
a d
b c
4
R
4
3R
)(KT
K
JS
500
200
53
TdSQ
(a)
RRR
QRRR
QQQ dabccdab 1004
3
4)200(250
44
3)500(0
RQRQQQ dabccdab 1002500
(b) Q is positive for the cycle and there is no change in internal energy so the work is positive and hence it operates as a heat engine.
(c) 60.0R250
R150
R250
R100R250
Q
Q
W
bc
dabc
bc
(d) 67.0150
100
W
outheat COP
R
RCOP
WdUQ đ đ đ
54
compressed air 2 kmoles, 4 atm., 294K
COLD HOT1 kmole, 1 atm., 333K
1 kmole, 1 atm., 255K
hollow tube
Kkmole
JaircP
41091.2)(
PdVTdSdTCWQdU V
nRdTVdPTdSdTCnRdTVdPPdV V
dPP
nRTdTncVdPnRdTdTCTdS PV
i
f
i
fPP P
PlnnR
T
TlnncS
P
dPnR
T
dTncdS
(Treat as an ideal gas)
đ đ
55
Consider 1kmole, 4 atm, 294K 1kmole, 1atm, 255K
K
kJS 38.7
Consider 1kmole, 4 atm, 294K 1kmole, 1atm, 333K
K
kJS 15.15
The total change in entropy is then,
05.22 K
kJS
The scheme does not violate any law of thermodynamics!
56
Example: Construct a reversible process to show explicitly that the entropy increases during the free expansion of an ideal gas.
For the free expansion of an ideal gas we have:0000 TUQW
To calculate the change in entropy we will consider a reversible isothermal process. (Quasistatic expansion to final V)
V
nRdV
T
PdV
T
PdVdU
T
QdS R
i
V
V i
f VV
VnR
V
dVnRS
f
i
fVandln
increases entropy0S
Ways to increase entropy: (1) add energy (2) increase volume
đ
57
Recapitulation.
In this chapter we have introduced the extremely important concept of entropy. Entropy and energy are very different.Although there is a conservation law for energy, there is no conservation law for entropy. (S is conserved only in a reversible process.)
For example when hot and cold water are mixed, the energy flow out of the hot water equals the energy flow into the cold water. The total energy has not changed. However the entropy increase of the cold water is greater than the entropy decrease of the hot water. Hence the entropy of the system after the transfer is greater than the entropy before the transfer. Where did the additional entropy come from? Some entropy is created in the mixing process and once created it can never be destroyed.
1st Law: Energy can neither be created nor destroyed.2nd Law: Entropy can be created but it can never be destroyed.
58
Where do we go from here? We will apply the 2nd Law to several situations.We will also discuss the TdS equations. These useful equations for entropy are written in terms of the measurable quantities
. and,c,c VP