ERT 108 Physical Chemistry
The Second Law of Thermodynamics
by Miss Anis Atikah binti Ahmad
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?