Second Law
description
Transcript of Second Law
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The second law of thermodynamics
A cup coffee does not get hotter in a cooler room
HOT COFFEE
Heat
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The first law of thermodynamics places no restriction on the direction of a process, but satisfying the 1st law alone does not ensure that a process will actually take place. This inadequacy is remedied by introducing the
second law of thermodynamics.
The 1st law is concerned with the quantity of energy and the transformations of energy from one form to another with no regard to its quality.
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The second law helps us to:
determine the quality as well as the degree of degradation of energy during a process. determine the theoretical limits for the
performance of engineering systems such as power plants, heat engines and refrigerators. predict the degree of completion of chemical
reactions.
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Statement of the second law:Lord Kelvin and Planck(1) No apparatus can operate in such a way that its only effect (in system and surroundings) is to convert heat absorbed by the system completely into work done by the system.
[It is impossible by a cyclic process to convert the heat absorbed by a system completely into work done by the system]
[It is impossible to build a heat engine that produces a net work output by exchanging heat with a single
fixed-temperature region]
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Statement of the second law (Cont.)
(2) No process is possible that consists solely in the transfer of heat from one temperature level to a higher one.
[It is impossible to build a device (a refrigerator or heat pump) that has as its only effect the transfer of heat from a low- to a
high-temperature region]
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For reversible process:
CHC
H
TTTT CH
C
H QQ QQ ==
If both QH and QC refer to the heat engine itself (rather than to heat reservoirs) then QH is positive and QC is negative.
)9.5(..........0QQ
QQ
CH
CH
=+
=
CH
CH
TT
TT
For a complete cycle of a Carnot engine, the working fluid periodically returns to its initial state; i.e. its T, P, and U return to their initial values; i.e. the summation of changes in the properties of the working fluid is zero for any complete cycle.
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Equation (5.9) suggests that a new property (Q/T) should exist!
When isothermal steps becomes infinitesimal, then:
only cycles reversiblefor 0 T
dQ
0T
dQ T
dQ
rev
C
C
H
H
=
=+
i.e. sum of changes of dQrev/T for a cyclic process = 0
dQrev/T must be a property ( like T, P, U)
This property is called entropy (S)
(5.12) trev TdSdQ =
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The entropy change of a heat reservoir is always Q/T for heat source or sink whether transfer is reversible or irreversible. For an adiabatic and reversible process dQrev = 0 and dSt = 0i.e. the entropy is constant during an isentropic process.
Summary
(1) The change in entropy of any system undergoing a reversible process is
(2) For an irreversible (real) process applied to an arbitrarily reversible process that achieves the same change of state as the actual process.Entropy is a state function entropy changes for reversible and irreversible processes are identical.
(3) Entropy is a property derived from the second law just as U was derived from first law.
= TdQS revt = TdQS revt
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Entropy changes of an ideal gas
First law of thermodynamics dU = dQ + dWReversible Process dU = dQrev PdVBut H = U + PV dH = dU + PdV + VdP
dH =(dQrev PdV) + PdV + VdPor dQrev = dH - VdP
Ideal gas dH = CPdT and V = RT / PdQrev = CPdT RT dP/P
This is a general equation for calculation of entropy changes of an ideal gas, since it relates properties only, i.e. independent of the process causing the changes (reversible/irreversible) [T1, P1T2, P2]
1
2
T
rev
lnS
TdQ
2
1PPR
TdTCgIntegratin
PdPR
TdTCdSor
T
P
P
=
==
-
=
+=+=
==
CH
CHtotal
CH
tC
tHtotal
C
tC
H
tH
TTTTQS
TQ
TQ
SSS
TQ
STQ
S and
If heat Q is transferred between two reservoirs TH and TC
Since TH > TC Stotal is positive For this irreversible process
reversible becomes process when i.e. T when 0 )(T as
H
H
Ctotal
Ctotal
TSTS
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Stotal is always positive for an irreversible(actual) process, and approaches zero as the process becomes reversible.
What happens if Q=0 (Adiabatic process)?
Consider an irreversible adiabatic expansion from A to B.Suppose fluid is restored to its initial state by a reversible process. If the irreversible results in an entropy change of fluid, there must be heat transfer during the reversible process such that:
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Both processes constitute a cycle dU = 0, and
== AB
revtB
tA
t
TdQSSS
=== revrevrevirr dQQWWWSince impossible by cyclic process to convert heat completely into work Qrev cannot directed into system
is negative is also negative
0)&( = tAtBtotal SSSSStB
tA SS revdQ
tA
tB SS
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Stotal is always positive, approaching zero as a limit when process becomes reversible. True for any process.
Stotal 0Consider our cyclic heat engine working between TH and T
No net changes in its properties Total entropy change of process (S&S) is the sum of
entropy changes of reservoirs
)1(W
gives Q elminating , C
H
CHtotalC
CH
C
C
H
Htotal
TTQST
QQWTQ
TQ
S
+==
+=
-
This is the general equation for work done by heat engines working between TH and TC.
Minimum work output = 0 equivalent to irreversible heat transfer between two reservoirs
C
C
H
Htotal T
QTQ
S +=
Maximum work output obtained when engine is reversible
revH
total
QS
==
W and 0
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dVPdW A=
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TdSdQ =
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The Second Law can be stated for a processas follows:
"It is impossible to have a process in which the sum of the change in the total entropy of the system and the change in the total entropy of the surroundings is negative."
That is,
(S2 - S1)system + (S2 - S1)surroundings > = 0