Transcript pchemc3-1.ppt

Chapter 3
The Second Law
Unit 1 The second law of thermodynamics and Entropy
Spring 2009
Spontaneous Change
The purpose of this chapter is to explain the origin of
the spontaneity of physical and chemical change.
Spontaneous change does not require work to be done
to bring it about.
Examples of spontaneous change:
• A gas expands to fill the available volume,
• a hot body cools to the temperature of its surroundings
• a chemical reaction runs in one direction rather than another.
The second law of thermodynamics
Kelvin’s statement
No process is possible in which the sole result is
the absorption of heat from a reservoir and its
complete conversion into work.
The second law of thermodynamics
• Engine:
heat is drawn
from a hot reservoir and
converted into work.
• All real heat engines have
both a hot source and a cold
sink; some energy is always
discarded into the cold sink
as heat and not converted
into work.
The second law of thermodynamics
In tems of entropy
The entropy of an isolated system increases in
the course of a spontaneous change: ∆S total > 0.
(Stotal is the total entropy of the system and its surroundings)
Entropy
• The definition of entropy instructs us to find the energy supplied as heat
for a reversible path between the stated initial and final states regardless
of the actual manner in which the process takes place.
• when the energy transferred as heat is expressed in joules and the
temperature is in kelvins, the units of entropy are joules per kelvin (J K−1).
• Entropy is an extensive property.
• The molar entropy is an intensive property.
Molar entropy, the entropy divided by the amount of substance, is
expressed in joules per kelvin per mole (J K−1 mol−1).
Example 3.1
Calculating the entropy change for the isothermal expansion of a
perfect gas
Calculate the entropy change of a sample of perfect gas
when it expands isothermally from a volume Vi to a
volume Vf.
Answer:
1. For isothermal process DU=0,
Vf
nRT ln 
Vi
qrev

DS 

T
T


  nR ln  V f
V
 i



2. According to the first law of thermodynamics DU=q + w,so q = -w
3. For a reversible change wrev= - nRT ln (Vf / Vi)
4.
qrev= nRT ln (Vf / Vi)
qrev
DS 

T
nRT ln(
T
Vf
Vi
)
Vf
 nR ln 
 Vi



Self Test 3.1
Calculate the change in entropy when the pressure of a
perfect gas is changed isothermally from pi to pf.
∆S = nR ln(pi/pf)
entropy of the surroundings, ∆Ssur
Self Test 3.2
Calculate the entropy change in the surroundings when
1.00 mol N2O4(g) is formed from 2.00 mol NO2(g) under
standard conditions at 298 K.
Entropy changes in expansion
• Isothermal Expansion
the change in entropy of a perfect gas that expands isothermally
from Vi to Vf is
The logarithmic increase in entropy of a
perfect gas as it expands isothermally.
Entropy changes in expansion
• Isothermal Expansion
∆Stot = 0
∆Stot = 0,which is what we should expect for a reversible process.
Entropy changes in expansion
• If the isothermal expansion occurs freely (w = 0) and
irreversibly, then
For system
q = 0 (because ∆U = 0); ∆Ssys = nRln(Vf/Vi)
For surrounding
∆Ssur = 0,
the total entropy change is
∆Stot > 0, as we expect for an irreversible process
Entropy in phase transition
DS phasetransition 
D transitionH
Ttransition
• If the phase transition is exothermic (∆trsH < 0,as in freezing or condensing),
then the entropy change is negative. This decrease in entropy is consistent
with localization of matter and energy that accompanies the formation of a
solid from a liquid or a liquid from a gas.
• If the transition is endothermic (∆trsH > 0, as in melting and vaporization), then
the entropy change is positive, which is consistent with dispersal of energy
and matter in the system.
Trouton’s rule
All liquids can be expected to have similar standard entropies of vaporizat ion
DSvaporization  85 J K -1 mol -1
a wide range of liquids give approximately the same standard
entropy of vaporization (about 85 J K−1 mol−1): this empirical
observation is called Trouton’s rule.
Restriction: no hydrogen bond or metallic bond
Application: predict the standard enthalpy of vaporization of a
liquid.
Illustration 3.3 Using Trouton’s rule
There is no hydrogen bonding in liquid bromine and Br2 is a
heavy molecule that is unlikely to display unusual behaviour in
the gas phase, so it is probably safe to use Trouton’s rule. To
predict the standard molar enthalpy of vaporization of bromine
given that it boils at 59.2°C, we use the rule in the form
Substitution of the data then gives
The experimental value is +29.45 kJ mol−1.
Self Test 3.3
Predict the enthalpy of vaporization of ethane
from its boiling point, −88.6°C.
Entropy when temperature change
Temperature change from
Ti →Tf
At constant pressure
If Cp is constant in the
temperature range Ti ,Tf
At constant volume
S (Tf )  S (Ti )  
Tf
Ti
CV dT
T
Example 3.2 Calculating the entropy change
Calculate the entropy change when argon at 25°C and
1.00 bar in a container of volume 0.500dm3 is allowed
to expand to 1.000 dm3 and is simultaneously heated
to 100°C
Step 1 : Vi → Vf at constant T
Step 2 : Ti → Tf at constant V
Step 1 + Step 2
Self Test 3.4
Calculate the entropy change when the same initial
sample is compressed to 0.0500 dm3 and cooled to
−25°C.
Calculating Changes in Entropy
1.
Reversible adiabatic Process qreversible=0
2.
For any cyclic process
3.
DS 
 dq

DS    reversible   0
T


 dqreversible   0
 
T 

For the reversible isothermal expansion or compression
of an ideal gas Vi , Ti  V f , Ti because DU=0
qreversible   wreversible  nRT ln
4.
Vf
Vi
Vf
dqreversible qreversible
DS  

 nR ln
T
T
Vi
For a reversible change in T at constant V or P
Vi , Ti  Vi , T f dqreversible =CV dT
Tf
nCV ,m dT
dqreversible
DS  

 nCV ,m ln
T
T
Ti
5.4 Calculating Changes in Entropy
Any reversible and irreversible process for an ideal gas
Vi , Ti  V f , T f
Pi , Ti  Pf , T f
DS  nR ln
Vf
Vi
DS  nR ln
 nCV ,m ln
Pf
Pi
Tf
Ti
 nCP ,m ln
Tf
Ti