Entropy
 For a reversible ideal gas Carnot cycle:
Efficiency
=
q rev
T
W
 1  2rev  1  2
qrev
q1
T1

q1 q2
 0
T1 T2

dqrev
0
T
 The efficiency of any reversible engine has to be the same as the Carnot
cycle:
Since the engine is reversible, we can run it backwards. Use the work (-w’)
out of the Cannot engine as work input (w) to run the left engine backwards.

But

'
 
Total work out = 0 (-w’ = w > 0)
W ' W
W W W

 ' 

 q1   q1' since q1  0, q1'  0
'
q1
q1
q1
q1
q1
 (q1' +q1 )>0
nd
This contradicts the 2 law (Clausius). This says that we have a net flow of
heat into the hot reservoir, but no work is being done!
∴
The efficiency of any reversible engine is   1 
T2
T1
 We can approach arbitrarily closely to any cyclic process using a series
of only adiabats and isotherms.
∴ For any reversible cycle

dqrev
0
T
 This defines Entropy, a function of state
dS 
Note:
2 dq
dqrev
rev
 S  S 2  S1  
1
T
T
Entropy is a state function, but to calculate ΔS requires a
reversible path.
 An irreversible Carnot (or any other) cycle is less efficient than a
reversible one.
**
An irreversible isothermal expansion requires less heat
than a reversible one.
 irrev
q2rev
q2rev
 1  irrev  1  irrev   rev
q1
q1
dqirrev dqrev


also
T
T

(q2  0)
dqirrev
0
T
**
 Leads to Clausius inequality

dq
0
T
 dqrev
  T  0
contains 
 dqirrev 0
  T
The entropy of an isolated system never decreases
(A): The system is isolated and irreversibly (spontaneously)
changes from [1] to [2]
(B): The system is brought into contact with a heat reservoir and
reversibly brought back from [2] to [1]
Path (A): qirrev = 0

Clausius
(isolated)
2 dq
dq
irrev
0
1
T
T
 0!

1
2
dqrev
0
T
dqrev
 S1  S 2   S  0
2 T

∴
1
S  S2  S1  0
This gives the direction of spontaneous changel
S  S2  S1
But!
independent of path
ΔSsurroundings depends on whether the process is
reversible or irreversible
(a) Irreversible: Consider the universe as an isolated system containing our
initial system and its surroundings.
(b) Reversible:
Examples of a spontaneous process
Connect two metal blocks thermally
in an isolated system
(ΔU = 0)
Initially
T1  T2
dS  dS1  dS2 
dq1 dq2
(T  T )

 dq1 2 1
T1
T2
T1T2
(dq1  dq2 )
dS  0 for spontaneous process
 if
T2  T1  dq1  0

T2  T1  dq1  0 
in both cases heat flows
from hot to cold as expected
Joule expansion with an ideal gas
adiabatic
 mol gas (2V,T)
1 mol gas (V,T)
ΔU = 0 q = 0 w = 0
S  Sbackwards
Compress back isothermally and reversibly
qrev  0
1 mol gas (2V,T) = 1 mol gas (V,T)
Sbackwards  
v RdV
dqrev
dw
1
 

 R ln
2v V
T
T
2
∴ S  R ln2  0 spontaneous
Note that to calculate ΔS for the irreversible process, we needed to find
dqrev
dq
and
rev
a reversible path so we could determine
 T