# The Second and Third Laws of Thermodynamics

Document Sample

```					CHEM 332 – Physical Chemistry II                                               February 23, 2005

The Second and Third Laws of Thermodynamics

The Second Law of Thermodynamics

Entropy S is a state function which is a measure of the disorder of a system: the greater the
disorder, the greater the entropy. The Second Law of Thermodynamics tells us the direction of
spontaneous (natural) change in an isolated system. The law states that for any spontaneous
change in an isolated system

Stotol  S  Ssur  0                                  (1)

where S is the entropy change of the system and Ssur is the entropy change for the
surroundings. The “=” applies for a change carried out reversibly, while the “>” holds for an

irreversible (spontaneous) change. A process for which Stotal  0 is impossible (much like going
backwards in time)! Since the universe is an isolated system, the Second Law says that the
entropy of the universe is always increasing.


Another statement of the second law of thermodynamics comes from the study of the efficiency
of heat engines: No process is possible in which the sole result is the absorption of heat from a
reservoir and its complete conversion into work.

To calculate Stotal we need to find the entropy change of the system S and of the surroundings
Ssur. The thermodynamic definition of entropy is

dqrev
dS                                                 (2)
T

We use this expression to calculate S for the system whether or not the change was actually
carried our reversibly. Thus     

dqrev
S         T
(3)

No matter how the change was actually carried out, we calculate S for the system via Eq(3): we
do this by constructing a reversible path from the initial state to the final state.

Arguments used to find Ssur are more subtle. The surroundings consist of everything in the
universe except for the system: they are enormous! The surroundings are such a huge reservoir
that they can be assumed to have a constant volume and constant temperature. These
assumptions lead to the approximations

Page 1 of 6
CHEM 332 – Physical Chemistry II                                                    February 23, 2005

dqrev dqrev,V dqsur    dq
dSsur                                                          (4)
T      T      T       T

The argument leading to the second to last equality is that for a change carried out at constant
volume, dw  pextdV  0 , so dUsur  dqrev,V  dqsur . Since U is a state function dU cannot

depend on the path by which the heat is transferred. The last equality assumes that the heat
gained (or lost) by the system comes from (or goes to) the surroundings so dqsur  dq .
Integration of Eq(4) gives
                    
dqsur    1          q
Ssur     dS sur          T

T
 dq 

T
(5)

The Clausius Inequality

The Clausius inequality is

dq
dS               or dq  TdS                               (6)
T

where dq is the actual heat change of the system which may be irreversible or reversible. As in
Eq(1), the “=” applies for a change carried out reversibly (since then dq  dqrev ), while the “>”


holds for an irreversible (spontaneous) change. The inequality follows directly from the
differential form of Eq(1), dStotal  dS  dSsur  0 . If we use Eq(4) to substitute for dSsur and then
solve for dS, we obtain Eq(6).


Calculation of  the System
S for

S for the system is found by imagining a reversible path from the initial state to the final state
and then using Eq(3). Note that phase changes are reversible processes carried out at a constant p
and T. Notes on how to calculate dqrev for common cases follow:

   Constant pressure case:
dqrev      dq             dH     C dT           T 
(1) Change from Ti to Tf: S        T
  rev,p  
T             T
 p
T
 nC p ln f 
Ti 
dq         dqrev, p     dH n H trans
(2) Phase change at T = Ttrans: S   rev                       
T           T           T     Ttrans
   Adiabatic case: 
(1) Reversible change: dqrev = 0 so S = 0.
(2) Irreversible change: dq = 0, but dqrev may not equal zero. You need to construct a

reversible path from the initial state to the final state and calculate S for that path.

Page 2 of 6
CHEM 332 – Physical Chemistry II                                                                   February 23, 2005

Isothermal case (and not phase transition):
(1) Reversible change for an ideal gas: dU  dqrev  dwrev  CV dT  0 so dqrev  dwrev .
dq       dwrev (pextdV) pdV nRdV                             V 
Thus dS  rev                                          and S  nRln f .
T        T           T          T        V                    Vi 
(2) Irreversible change for an gas: dU  dq  dw  CV dT  0 , so dq  dw . However
ideal                              
dqrev
since dS         , you need to construct a reversible path from the initial state to the final
               T                                      
state and calculate S for that path. This path is identical to the reversible path above and so
                               
the same S is obtained. The value of Ssur is different from for the irreversible case.
 Constant volume case:

dqrev    dq         dU   C dT           T 
(1) Change from Ti to Tf: S      T
  rev,V  
T         T
 V
T
 nCV ln f 
Ti 
(2) Phase change at T = Ttrans: Find U. Then H  U  (pV) . To calculate H, see the
constant pressure case above.

   Other cases: a path between the initial and final state may be constructed by summing the
cases above.                         

The Third Law of Thermodynamics

The Third Law of Thermodynamics states that the entropy of all perfect crystalline substances is
zero at T = 0 K. We therefore assign the value of zero entropy to perfect crystals at absolute zero.
All non-perfect crystals and all substances (including pure elements) at temperatures above 0 K
have positive entropy. Entropy increases with temperature, since disorder increases when the
temperature is raised.

Note that entropy values are absolute since they are based on the third law assignment of S = 0 to
all perfect crystals at 0 K. In contrast the zero of enthalpy and zero of free energy are undefined.
We follow the convention of setting the heat of formation H f and free energy of formation
o

G o of a pure element in its standard state equal to zero.
f

The entropy change for a reaction at standard conditions is

S 
o
n
products
prod   S prod 
o
n
rea cta nts
react   S react 
o
(7)

Note that none of the entropies in the equation will be zero at standard conditions since we are
way above 0 K!

Page 3 of 6
CHEM 332 – Physical Chemistry II                                                   February 23, 2005

More Convenient Criteria for Spontaneity.

So far we have learned that a process is spontaneous if dStot  0 . We now use the second form of
the Clausius inequality given in Eq(6) to derive more convenient tests for spontaneity which
depend only on the system. Throughout we assume only pressure-volume (pV) work. The most
useful criterion is that it the one involving the Gibbs function G, another function of state.

dU Criterion
By definition, dU  dq dw  dq  pextdV . For a process carried out at constant volume,
dUV  dq . From Eq(6), dq  TdS . Thus dUV  TdS . If the entropy is also constant,
dU S,V  0 . As always, the “=” applies for a reversible process. For a spontaneous process, the
internal energy change dU at constant S and V must be less than zero.

                                     
dH Criterion
   Since H = U + pV, dH  dU  d(pV)  dq  pextdV  pdV  Vdp. For a process carried out at
constant pressure, pext  p  constantand dp = 0 so dH p  dq . From Eq(6), dq  TdS . Thus
dH p  TdS . If the entropy is also constant, dH S, p  0 . For a spontaneous process, the
enthalpy 
change dH at constant S and p must be less than zero.
                                                           
                                        
The last two criteria for spontaneity involve two additional functions of state:

A = U – TS = Helmholtz Free Energy (or Work function)                       (8)
G = H – TS = Gibbs Free Energy (or Gibbs function)                        (9)

dA Criterion
Since A = U – TS, dA  dU  d(TS)  dU TdS  SdT . At constant T, dAT  dU  TdS . From
above we have dU V  TdS . Thus if the volume is also constant, dAT ,V  0 . For a spontaneous
process, the Helmholtz free energy change dA at constant T and p must be less than zero.
                                                 
dG Criterion
                                                  
Since G = H – TS, dG  dH  d(TS)  dH TdS  SdT . At constant T, dGT  dH  TdS . From
above we have dH p  TdS . Thus if the pressure is also constant, dGT , p  0 . For a
spontaneous process, the Gibbs free energy change dG at constant T and p must be less than
                                                  variables to control, the dG
zero. Since pressure and temperature are usually the easiest 
criterion proves the most useful.
                                                

Page 4 of 6
CHEM 332 – Physical Chemistry II                                                February 23, 2005

Relation of dA and dG to Work

It is easy to show that dA  dwrevfor an isothermal system doing only pressure-volume (pV)
work. Thus AT  w rev . Since a reversible system does the maximum work, calculation of A
for an isothermal change is equivalent to finding the maximum work. The derivation begins with
the definition of A. We then note that dU is independent of path, and choose a reversible path.

We also use the definition of dS from Eq(2) to give TdS  dqrev . The steps are as follows:

A  U  TS

dA  dU  d(TS)

dA  dU  TdS  SdT

dA  dqrev  dwrev  dqrev  SdT

dAT    dwrev

We can also show that dGp,T  dwrev, non pV for a system doing both pV and non-pV work. Thus
the maximum non-pV work which can be done by a system is Gp,T  w rev, non pV . We begin

with the definition of G; choose a reversible path for dU; use the definition of dS from Eq(2) to
give TdS  dqrev ; and note that dwrev  pextdV  pdV :


G  H  TS
                         dG  dH  d(TS)

dG  dH  TdS  SdT

dG  dU  pdV  Vdp TdS  SdT
dG  dqrev  dwrev  pdV  Vdp dq rev  SdT

dG  dwrev, pV  dwrev, non pV  pdV  Vdp SdT

dG   pdV  dwrev, non pV  pdV  Vdp  SdT

dG  dwrev, non pV  Vdp SdT

dGp,T    dwrev, non pV



Page 5 of 6
CHEM 332 – Physical Chemistry II                                              February 23, 2005

Additional Thermodynamic Expressions Combining the First and Second Laws

The First Law tells us dU  dq  dw  dqrev  dwrev (since we can choose any path to calculate
the change in a state function). From the Second Law we have the thermodynamic definition of
entropy of Eq(2) which gives TdS  dqrev . If we combine the definitions of U, H, A, and G with
these relationships, we obtain additional thermodynamic expressions using “natural variables.”

In the derivations below we assume only pV work.


U=U(S,V)
dU  dq  dw  dqrev  dwrev
dU  TdS  pdV                                      (10)


H=H(S,p)                      
H  U  pV

dH  dU  pdV  Vdp

dH  TdS  pdV  pdV  Vdp

dH  TdS  Vdp
(11)


A=A(V,T)
A  U  TS

dA  dU  TdS  SdT

dA  TdS  pdV  TdS  SdT

dA   pdV  SdT
(12)


G=G(p,T)
G  H  TS

dG  dH  TdS  SdT

dG  TdS  Vdp TdS  SdT

dG  Vdp SdT
(13)



Page 6 of 6

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 24 posted: 2/11/2010 language: English pages: 6