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ERT 108 Physical Chemistry
The Second Law of Thermodynamics
by Miss Anis Atikah binti Ahmad
anisatikah@unimap.edu.my
Outline
•The Second Law of Thermodynamics•Heat Engines•Entropy•Calculation of entropy changes•Entropy, Reversibility and Irreversibility•The thermodynamics temperature scale•What is entropy?
The Second Law of Thermodynamics
•Kelvin-Plack formulation of the second law of thermodynamics:
It is impossible for a system to undergo cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings.
The Second Law of Thermodynamics
It is impossible to build a cyclic machine that converts 100% heat into work.
Does this system violate the first law?
The Second Law of Thermodynamics
Any heat engine must eject heat into the cold reservoir
Heat Engines
•Heat engine: a device that operates in a thermodynamic cycle and does a certain amount of net positive work as a result of heat transfer from a high-temperature body to a low-temperature body.
(eg: the internal-combustion engine and the gas turbine)
Heat Engines
•The efficiency of heat engine:
• The efficiency value is less than 1, qC has a negative value and qH has a positive value.
Work output per cycle
Energy input per cycle H
C
H
CH
H
cycle
q
q
q
q
w
1
Cold reservoir
Hot reservoir
Heat EnginesThe cycle for a reversible heat engine (Carnot cycle):
Heat, Work and ΔU for Reversible Carnot Cycle
Work flow in Carnot cycle
Carnot cycle
•For a complete cycle (assuming perfect gas);
•Dividing by T and integrating over Carnot cycle;
dwdqdU
PdVdq
dVVnRTdqdTCV
V
dVnR
T
dq
T
dTCV
First Law
Carnot cycle
•Thus;
V
dVnR
T
dq
T
dTCV
00
0Tdq
0 a
d
d
c
c
b
b
a T
dq
T
dq
T
dq
T
dq
T
dq
Differential of state function;independent of the path taken to reach final state.
Carnot cycle•Since bc and da are adiabatic; dq=0;
•Thus;
0 a
d
d
c
c
b
b
a T
dq
T
dq
T
dq
T
dq
T
dq0 0
0 d
c
b
a T
dq
T
dq
T
dq
0C
C
H
H
T
q
T
q
T
dq
Carnot cycleFor Carnot cycle the efficiency can be also
written as;
Where is the maximum possible efficiency for the conversion of heat to work.
rev Work output per cycle
Energy input per cycle
HCHC TTqq /Because
H
C
H
C
T
T
q
q 11
rev
Exercise 1
•Calculate the maximum work that can be done by reversible heat engine operating between 500 and 200 K if 1000 J is absorbed at 500 K
Solution•Calculate the maximum work that can be
done by reversible heat engine operating between 500 and 200 K if 1000 J is absorbed at 500 K
6.0500
200111
K
K
T
T
q
q
H
C
H
C
JJqw H 60010006.0
Entropy, S
T
dqdS rev
2
1
12 T
dqSSS rev
Closed sys, rev. process
Calculation of Entropy Changes
1. Cyclic process;
2. Reversible adiabatic process;
3. Reversible phase change at constant T & P
at constant P, , thus
0S
02
1
T
dqS rev
00S
T
qdq
TT
dqS rev
revrev
2
1
2
1
1
Hqq Prev T
S
Rev. phase change at const. T & P
Rev. adiab. proc.
Calculation of Entropy Changes
4. Reversible isothermal process
5. Constant-pressure heating with no phase change
If is constant over the temperature range, then
T
qdq
TT
dqS rev
revrev
2
1
2
1
1
dTCq PP
T
qS rev
2
1 T
dqS rev
2
1 T
dqP
dTT
CS
T
T
P2
1
PC 12ln TTCS P
Rev, isothermal proc.
Const. P, no phase change
Calculation of Entropy Changes
6. Reversible change of state of a perfect gas;
dwdUdqrev
Perfect gas
PdVdTCV
dVVnRTdTCV
2
1 T
dqS rev
2
1
2
1
dVVnRdTT
CV
2
1 1
2lnV
VnRdT
T
CV
Calculation of Entropy Changes
7. Irreversible change of state of a perfect gas;
8. Mixing of different inert perfect gases at constant T & P
Perfect gas 2
1 1
2lnV
VnRdT
T
CS V
21 SSS
bbaa VVRnVVRn lnln
bbaa xRnxRn lnln
PRTnnV ba
PRTnV aa
PRTn
PRTnnVV
a
baa
ax
1
Exercise 2 One mole of a perfect gas at 300 K is reversibly
and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath in the surroundings is very large, T remains essentially constant at 300 K during the process. Calculate ΔS of the system.
Exercise 2-Solution One mole of a perfect gas at 300 K is reversibly and
isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath in the surroundings is very large, T remains essentially constant at 300 K during the process. Calculate ΔS.
• This is an isothermal process, ΔT=0, thus ΔU=0 (for perfect gas, U depends only on T. ( )
dTCdU V
0 wqU
wqrev PdV 12ln VVnRTdVV
nRT
LLKKmolJmol 2510ln300314.81 11
J310285.2
13
62.7300
10285.2
KJK
J
T
qS rev
Exercise 3
Calculate ΔS for the melting of 5.0 g of ice (heat of fusion= 79.7 cal/g) at 0°C and 1 atm.
Estimate ΔS for the reverse process
Exercise 3- Solution Calculate ΔS for the melting of 5.0 g of ice (heat
of fusion= 79.7 cal/g) at 0°C and 1 atm. Estimate ΔS for the reverse process.
• Identify type of process:▫ Phase change at constant T & P▫ At constant P, q= ΔH▫ Thus,
• Calculate ΔS;
TS
KJKcalK
ggcal
TS 1.646.1
15.273
57.79
Exercise 3- Solution Calculate ΔS for the melting of 5.0 g of ice (heat
of fusion= 79.7 cal/g) at 0°C and 1 atm. Estimate ΔS for the reverse process.
• ΔS for reverse process (freezing of 5g liquid water );
KJS 1.6
Exercise 4The specific heat capacity cP of water is nearly
constant at 100 cal/g K in the temperature range of 25°C to 50°C at 1 atm.
(a) Calculate ΔS when 100 g of water is reversibly heated from 25°C to 50°C at 1 atm.
(b) Without doing a calculation, state whether ΔS for heating 100g of water from 50°C to 75°C at 1 atm will be greater, equal to or less than ΔS for the 25°C to 50°C heating.
Exercise 4- SolutionThe specific heat capacity cP of water is nearly constant at
100 cal/g K in the temperature range of 25°C to 50°C at 1 atm.
(a) Calculate ΔS when 100 g of water is reversibly heated from 25°C to 50°C at 1 atm.
2
1 T
dqS rev
2
1 T
dqP
K
KKcal
T
TCdT
T
CP
T
T
P
298
323ln100ln
1
22
1
KJ7.33
KgcalgmcC PP 00.1100
Kcal100
Exercise 4 - SolutionThe specific heat capacity cP of water is nearly constant at 100 cal/g K
in the temperature range of 25°C to 50°C at 1 atm.
(b) Without doing a calculation, state whether ΔS for heating 100g of water from 50°C to 75°C at 1 atm will be greater, equal to or less than ΔS for the 25°C to 50°C heating.
Thus, ↑T, ↓ΔS ΔS for heating 100g of water from 50°C to 75°C at 1 atm will be
smaller than ΔS for the 25°C to 50°C heating.
TS 1
2
1 T
dqS rev
Entropy, Reversibility and Irreversibility
•Reversible Process, ΔSuniv = 0
surrsystemuniv dSdSdS
surr
rev
sys
rev
T
dq
T
dq
sys
rev
sys
rev
T
dq
T
dq
In reversible process, any heat flow btween
system & surroundings must occur with no finite
temperature difference
0
Entropy, Reversibility and Irreversibility
•Irreversible Process, ΔSuniv > 0
revrev dwdqdwdqdU
dwdwrev
When energy leaves the system as work,
dwdwrev
revrev dwdwdqdq
0 revdwdw
More work is done when a change is reversible than when it is irreversible;
Recall first law;rearranging
rearranging
•Irreversible Process, ΔSuniv > 0
Substituting into
Entropy, Reversibility and Irreversibility
revrev dwdwdqdq 0 revdwdw
0 dqdqrev
dqdqrev
T
dq
T
dqrev
T
dqdS
Clausius inequality
Dividing by T;
•Irreversible Process, ΔSuniv > 0
Suppose that the system is isolated from its surroundings, thus dq=0
Entropy, Reversibility and Irreversibility
T
dqdS 0dS
0 univsurrsys SS
•Entropy & Equilibrium
Thermodynamic equilibrium in an isolated system is reached when the system’sentropy is maximized.
Entropy, Reversibility and Irreversibility
S
Time
Equilibrium reach
S=Smax
This expression enabled Kelvin to define thermodynamic
temperature scale
The thermodynamics temperature scale
H
C
T
T1 HC TT 1
Kelvin scale is defined by using water at its triple point as the notional of hot source and defining that temperature as 273.16 K
If it is found that the efficiency of heat engine equal to 0.2, then the temperatureof cold sink is (0.8) x 273.16 K =220 K, regardless of the working substance of theengine.
rearranging
-a scale that is independent of the choice of a particular thermometric substance.
What is entropy?
•Entropy is a measure of the probability, p of the thermodynamic state
a
a a
a b b
bb
a
a a
a b b
bb
a
aa
ab
b
b
b
Partition removed
System proceed to equilibriumIrreversible mixing of perfect gas at
constant T & P
The probability that all a molecules will be in the left half & all b molecules in right half is extremely small.The most probable distribution has a and b molecules equally distributed.
What is entropy?
•Entropy is a measure of molecular disorder of a state.
Eg: In mixing two gases, the disordered (mixed state) is far more probable than the ordered (unmixed) state.
What is entropy?
• Entropy is related to the distribution or spread of energy among the available molecular energy levels.
• The greater the number of energy levels, the larger the entropy is.
• Increasing the system’s energy (eg:by heating) will increase its entropy because it allows higher energy levels to be significantly occupied
• Increasing the volume of a system at constant energy also allows more energy level to be occupied.
What is entropy?
•Boltzmann made link btween distribution of molecules over energy levels and the entropy;
Where k= 1.381 x 10-23 JK-1
W= probability/ways in which the molecules of a system can be
arranged while keeping the energy constant
WkS ln
Exercise
•True or false?▫ΔSuni for a reversible process in a
closed system must be zero▫ΔS for a reversible process in a closed
system must be zero▫For a closed system, equilibrium has
been reached when S has been maximized.
•What is ΔSuni for each steps of a Carnot cycle?