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Unformatted text preview: Second Law Of Thermodynamics MOTIVATION: By now we are fully aware that if two. systems are at different
temperatures and there is a conducting wall in between they will exchange heat DQ until
equilibrium is attained. It is found that in this process the higher temperature reduces and DQ the lower temperature rises [Re call min?“ sign in the equation 3 = Jutlg]. The question we need to answer is: why does exchange of heat invariably involve transfer
from high T to low T, or, why will heat not “ﬂow” spontaneously from low T to high T?
The Second Law of Thermodynamics was formulated to provide a succinct answer to
their question. It will help to deﬁne a property which gives DIRECTION to a
thermodynamic process. . The starting point comes as follows: we know that 4.18J of mechanical work will mimic
the effects of l cal of Heat 4.18JofDW=>l cal ofDQT A 7' A and this can be done as often as we like using a cycle. At its simplest level the second law
asserts that it is impossible to construct an engine which operates cyclically and whose
only effect 13 to pick up heat DQ from a reservoir and deliver DW= DQ. That 15, the
following schema: , Is IMPOSSIBLEH So already you begin to discern a “Direction” 4.18 J of DW:> l cal of DQ but the
Reverse cannot be done on a repeating basis. So then, what is possible? All existing engines produce useful work DW by picking up DQH but to do so they must
reject DQC as shown below: HOT Rama/om TH>TC F 61.0.1.3) .IQE’S’cI‘wél/u/R Tc] Pick up DQH from Hot Reservoir } Produce(DQH + DQC) Dump DQC into Cold Reservoir output per cycle To fulﬁll the promise that the second law provides a basis for the direction of all
thermodynamic processes we must use it to identity a unidirectional property. NOT
SURPRISING THAT IT WILL BE CALLED
ENTROPY [FROM GREEK WORD gvrponoyoyl]. To do so we discuss a cyclic process proposed by Carnot Carnot Cycle. The working substance in our “engine” is going to be an ideal gas so that we can use the
equations we wrote for thermodynamic processes. The cycle is shown in PV— Diagram We start at P], V1 and carry out 4 processes: . (l) Isothermal Expansion at TH, we pick up DQH from hot reservoir. dU=0 W=DQH= nRTHlnYI:l —>(1)
l (2) Adiabatic Expansion [T V7 ”' = ConstJ. Gas is allowed to cool until temperature is Tc. DQ = 0, TC Var—l = TH V27_I “>(2) (3) Isothermal Contraction at TC. Discard DQC into Cold Reservoir dU= 0 DW = DQC = nR TC 111%— —> (3) 3 Note: DQC is a negative quantity.
Pt. P4, V4 is chosen judiciously so that we can carry out, (4) Adiabatic Contraction and Gas is allowed to warm up until temperature is back at
TH. \
TH V4H = Tc V4 H ’9 (4) Next, analyze Eqs. (1) through (4) Combine (1) and (3)
DQH + DQC = ”R ln£+ lnﬂ
TH TC V1 V3
= nR 1n VZV"
V1 V3
= nR lnl
Because from (2) and (4)
1
n _ [1]? _ 5
V3 TH 4
That is, > (A) This has two very fundamental consequences: L Efﬁciency of our engine.
77 _ Output _ DQH +DQC _1_ TC
Input DQH T1,, .4 :> 77 is determined SOLELY by ratio of TEMPElgTURES! dS=DQ I_L If we deﬁne a new property by the equation T w The change in S over a closed loop is ZERO” S is unique. It is appropriate to
use (1. Change in S independent of Path. S is called ENTROPY.
R
deﬁnes dS =DTQ 7 (B) change of Entropy in a REVERSIBLE PROCESS.
CARNOT CYCLE IN ST DIAGRAM. Isotherm Adiabatie Is the Quantity deﬁnedrin (B) unidirectional? The answer is yes. To prove it we proceed
as follows: ' Imagine that we carry out an IRREVERSIBLE ADIABATIC PROCESS ~ DQ = 0, but
only initial and ﬁnal state is in equilibrium so only they are represented on the PV_
diagram. P L 9mm WEN/2w ' l “WWIWMWWWMWJWW V Since iand f are" equilibrium states, we can assign values Si and Sf to them.
Next, carry out three reversible processes. f —> A reversible adiabatic, DQ = O and by
i Eq.(B) d3 = 0, SO SA = Sf ,
A —> B reversible isotherm where system exchanges DQ with reservoir.
Choose B carefully so that B —>A reversible adiabatic [(DQ) = 0] takes us back to 1'. Again, AS = 0 , so 83 = 8,. Altogether, we have an irreversible cycle i —>f——>A—>B—>z‘ in which the only heat
exchange is with the single reservoir at T. The second law forbids one taking DQ firm
the reservoir and converting it to DW. It only allows us to discard DQ into the reservoir. dSAB=%=SB—~SA <0 SO 83 < SA and indeed Sf> Si That is, in an irreversible adiabatic DQ = 0, but
dS > 0 So indeed change of S is unidirectional. In an adiabatic process S can only increase
except in limiting case when process is reversible when d8 = 0.
To summarize, in terms of entropy, 2nd law says: in any adiabatic process. QEVEﬂle'gLE: . fﬂglﬁ“VE$/eoﬁ’fﬁég
FED {#ng ”Wt; #51) I‘M/9 ”7???
2.
mmmmmmmmwmwn ‘ W p NOTE: FINAL STEP IN ARGUMENT IS THAT ANY PROCESS CAN BE
RENDERED ADIABATIC BY PUTTING THE INSULATING BOUNDARY
OUTSIDE ALL OF THE SYSTEMS INVOLVED IN THE PROCESS. ...
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 Spring '10
 Shawhan
 Physics, Thermodynamics

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