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Entropy
Two reversible adiabatic can not intersect each other
CA
B
C
isotherm
RA
AB and CB are two reversible adiabatic and they intersect at B.
AC is a reversible isotherm. Through AC we can transfer heat to the cycle ABC which is
A close cyclic process. From the cycle ABC there is no heat loss but work out put. So
This violates 2nd law of thermodynamics. Hence two RA can not intersect each other.
p
v
i
f
a b
RA
Rev isotherm
Any rev path
Through i, f there is a
Rev process. We construct
two rev adiabatic through
i and f, then a rev isotherm
ab such that:
Area under iabf = area under if,
First law for if : , π‘žπ‘–π‘“ = 𝑒 𝑓 βˆ’ 𝑒𝑖 + 𝑀𝑖𝑓
Process iabf: π‘žπ‘–π‘Žπ‘π‘“ = 𝑒 𝑓 βˆ’ 𝑒𝑖 + π‘€π‘–π‘Žπ‘π‘“
𝑀𝑖𝑓 = π‘€π‘–π‘Žπ‘π‘“, π‘ π‘œ π‘žπ‘–π‘“ = π‘žπ‘–π‘Žπ‘π‘“
p
v
π‘žπ‘–π‘Žπ‘π‘“ = π‘žπ‘–π‘Ž + π‘ž π‘Žπ‘ + π‘ž 𝑏𝑓 = π‘ž π‘Žπ‘ =π‘žπ‘–π‘“
Thus any rev path can be substituted by a
reversible Zig-zag path, between the same end
states, consisting of a rev adiabatic + rev
isotherm+ another rev adiabatic.
Clausius theorem
p
v
Any rev cyclic process
RA
Rev isotherm
T1
T2
Lets us consider a smooth close curve which represents a rev cycle. Then we take two
Elemental parts from this cycle as shown in the picture (upper part and lower part). The
upper part cab represented by an isotherm+ two RA and similarly the lower part of the
curve. The upper isotherm is at a temp say T1 and the lower isotherm is at T2. Now this
elementary process can be thought of as a Carnot engine which receives heat dq1 at T1
and rejects heat dq2 at T2. So for this elementary cycle
π‘‘π‘ž1
𝑇1
=
π‘‘π‘ž2
𝑇2
π‘œπ‘Ÿ 𝑖𝑓 𝑀𝑒 π‘Žπ‘‘π‘‘ π‘€π‘–π‘‘β„Ž π‘‘β„Žπ‘’
Signs of heat transfer then we get
π‘‘π‘ž1
𝑇1
+
π‘‘π‘ž2
𝑇2
= 0. If we add all the elementary cycles such
that we cover the entire cyclic process then we get :
dq1
dq2
π‘‘π‘ž1
𝑇1
+
dq2
T2
+
dq3
T3
+ β‹― .
dqn
Tn
= 0 or
π‘‘π‘ž
𝑇
= 0 , hence
π‘‘π‘ž
𝑇
𝑖𝑠 π‘Ž π‘π‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘‘π‘¦ π‘™π‘–π‘˜π‘’ 𝑒, 𝑇, 𝑣 π‘Žπ‘›π‘‘ β„Ž π‘€β„Žπ‘–π‘β„Ž
is called entropy , ds.
R1
R2
a
b
π‘Ž
𝑏 π‘‘π‘ž
𝑇 𝑅1
+ 𝑏
π‘Ž π‘‘π‘ž
𝑇 𝑅2
= 0, οƒž π‘Ž
𝑏 π‘‘π‘ž
𝑇 𝑅1
= π‘Ž
𝑏 π‘‘π‘ž
𝑇 𝑅2
Hence entropy is a property, independent of path
So, 𝑑𝑠 =
π‘‘π‘ž π‘Ÿπ‘’π‘£
𝑇
or, π‘Ž
𝑏 π‘‘π‘ž
𝑇
= π‘Ž
𝑏
𝑑𝑠 = 𝑠 𝑏 βˆ’π‘  π‘Ž
Clausius theorem
𝑑𝑠 =
π‘‘π‘ž π‘Ÿπ‘’π‘£
𝑇
; If π‘‘π‘ž π‘Ÿπ‘’π‘£ = 0, then the process is called rev adiabatic and then ds=0
Hence s=Const;
T
S
A B
CD
π‘‘π‘ž
π‘‘π‘ž2
RA
AB is a general process
Either rev or irreversible,
All other processes are rev.
Consider an elemental cycle, as shown in the picture. π‘‘π‘ž = heat supplied at 𝑇1,
π‘‘π‘ž2 = β„Žπ‘’π‘Žπ‘‘ π‘Ÿπ‘’π‘—π‘’π‘π‘‘π‘’π‘‘ π‘Žπ‘‘ 𝑇2
1 βˆ’
π‘‘π‘ž2
π‘‘π‘ž π‘Žπ‘›π‘¦
≀ 1 βˆ’
π‘‘π‘ž2
π‘‘π‘ž π‘Ÿπ‘’π‘£
π‘‘π‘ž2
π‘‘π‘ž π‘Žπ‘›π‘¦
β‰₯
π‘‘π‘ž2
π‘‘π‘ž π‘Ÿπ‘’π‘£
π‘‘π‘ž π‘Žπ‘›π‘¦
π‘‘π‘ž2
≀
π‘‘π‘ž π‘Ÿπ‘’π‘£
π‘‘π‘ž2
π‘‘π‘ž π‘Žπ‘›π‘¦
π‘‘π‘ž2
≀
𝑇
𝑇2
; 𝑠𝑖𝑛𝑐𝑒
π‘‘π‘ž π‘Ÿπ‘’π‘£
π‘‘π‘ž2
=
𝑇
𝑇2
Or,
Or,
Or,
Or,
π‘‘π‘ž π‘Žπ‘›π‘¦
𝑇
≀
π‘‘π‘ž2
𝑇2
οƒž
π‘‘π‘ž π‘Žπ‘›π‘¦
𝑇
≀ 𝑑𝑠, π‘“π‘œπ‘Ÿ π‘Žπ‘›π‘¦ π‘π‘Ÿπ‘œπ‘π‘’π‘ π‘ 
For a reversible process:
π‘‘π‘ž
𝑇
= 𝑑𝑠
For any other process : 𝑑𝑠 β‰₯
π‘‘π‘ž
𝑇
For any cyclic process:
π‘‘π‘ž
𝑇
≀ 𝑑𝑠
π‘‘π‘ž
𝑇
≀ 0
𝐼𝑓
π‘‘π‘ž
𝑇
= 0, cycle is reversible,
π‘‘π‘ž
𝑇
< 0, cycle is irreversible and feasible
π‘‘π‘ž
𝑇
> 0, cycle is impossible
Clausius Inequality
Entropy principle
For any process undergone by the system, 𝑑𝑠 β‰₯
π‘‘π‘ž
𝑇
, if the process is reversible then 𝑑𝑠 =
π‘‘π‘ž
𝑇
For isolated system, dq=0, so, 𝑑𝑠 β‰₯ 0, if isolated process is reversible, π‘‘β„Žπ‘’π‘› 𝑑𝑠 = 0, 𝑠 = 𝑐
For irreversible process, 𝑑𝑠 > 0, π‘ π‘œ π‘’π‘›π‘‘π‘Ÿπ‘œπ‘π‘¦ π‘Žπ‘™π‘€π‘Žπ‘¦π‘  π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘ 
Causes of entropy increase: for a closed system entropy increases due to
(a) External interaction, (b) internal irreversibility
𝑑𝑠 = 𝑑𝑠 𝑒π‘₯𝑑 + 𝑑𝑠𝑖𝑛𝑑
=
π‘‘π‘ž
𝑇
+ 𝑑𝑠𝑖𝑛𝑑; so, 𝑑𝑠 >
π‘‘π‘ž
𝑇
; π‘œπ‘›π‘™π‘¦ π‘“π‘œπ‘Ÿ π‘Ÿπ‘’π‘£π‘’π‘Ÿπ‘ π‘–π‘π‘™π‘’ π‘π‘Ÿπ‘œπ‘π‘’π‘ π‘  𝑑𝑠𝑖𝑛𝑑 = 0
𝑑𝑠 =
π‘‘π‘ž
𝑇
+ π‘‘π‘ π‘”π‘’π‘›οƒž 𝑠2 βˆ’ 𝑠1 = 1
2 π‘‘π‘ž
𝑇
+ 𝑠𝑔𝑒𝑛1βˆ’2
𝑑𝑠
𝑑𝑑
=
π‘ž 𝑐𝑣
𝑇
+ 𝑠 𝑔𝑒𝑛
Entropy change Entropy transfer Entropy generation
dq
T
𝑠 𝑔𝑒𝑛
𝑠2 βˆ’ 𝑠1
For an open system π‘šπ‘–
π‘ž
𝑠 𝑔𝑒𝑛
π‘š 𝑒
π‘šπ‘– 𝑠𝑖 +
π‘ž 𝑐𝑣
𝑇
+ 𝑠 𝑔𝑒𝑛 = π‘š 𝑒 𝑠 𝑒 +
𝑑𝑠 𝑐𝑣
𝑑𝑑
, for steady state
𝑑𝑠 𝑐𝑣
𝑑𝑑
= 0
π‘šπ‘– 𝑠𝑖 +
π‘ž 𝑐𝑣
𝑇
+ 𝑠 𝑔𝑒𝑛 = π‘š 𝑒 𝑠 𝑒 + π‘š2 𝑠2 βˆ’ π‘š1 𝑠1
Integral form of the equation
Change of entropy
within the CV
system
𝑻 𝒅𝒔 relation
First law and second law combined:
π‘‘π‘ž = 𝑑𝑒 + 𝑝𝑑𝑣 οƒž 𝑇𝑑𝑠 = 𝑑𝑒 + 𝑝𝑑𝑣
𝑑𝑠 =
𝑑𝑒
𝑇
+
𝑝
𝑇
𝑑𝑣 𝑑𝑠 =
𝑑𝑒
𝑇
+
𝑅
𝑣
𝑑𝑣
π‘‘π‘ž π‘Ÿπ‘’π‘£ = 𝑇𝑑𝑠
𝑠2 βˆ’ 𝑠1 =
𝑐 𝑣 𝑑𝑇
𝑇
+
𝑅
𝑣
𝑑𝑣
𝑠2 βˆ’ 𝑠1 = 𝑐 𝑣 ln
𝑇2
𝑇1
+ 𝑅 ln
𝑣2
𝑣1
1st Tds relation
Only for ideal gases
β„Ž = 𝑒 + 𝑝𝑣
π‘‘β„Ž = 𝑑𝑒 + 𝑝𝑑𝑣 + 𝑣𝑑𝑝
π‘‘β„Ž = 𝑇𝑑𝑠 + 𝑣𝑑𝑝
T𝑑𝑠 = π‘‘β„Ž βˆ’ 𝑣𝑑𝑝 2nd Tds relation
𝑑𝑠 =
π‘‘β„Ž
𝑇
βˆ’
𝑣
𝑇
𝑑𝑝 𝑠2 βˆ’ 𝑠1 =
𝑐 𝑝 𝑑𝑇
𝑇
βˆ’
𝑅
𝑝
𝑑𝑝 οƒž 𝑐 𝑝 ln
𝑇2
𝑇1
βˆ’ 𝑅 ln
𝑝2
𝑝1
Only for ideal gases

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Lect 6

  • 1. Entropy Two reversible adiabatic can not intersect each other CA B C isotherm RA AB and CB are two reversible adiabatic and they intersect at B. AC is a reversible isotherm. Through AC we can transfer heat to the cycle ABC which is A close cyclic process. From the cycle ABC there is no heat loss but work out put. So This violates 2nd law of thermodynamics. Hence two RA can not intersect each other. p v i f a b RA Rev isotherm Any rev path Through i, f there is a Rev process. We construct two rev adiabatic through i and f, then a rev isotherm ab such that: Area under iabf = area under if, First law for if : , π‘žπ‘–π‘“ = 𝑒 𝑓 βˆ’ 𝑒𝑖 + 𝑀𝑖𝑓 Process iabf: π‘žπ‘–π‘Žπ‘π‘“ = 𝑒 𝑓 βˆ’ 𝑒𝑖 + π‘€π‘–π‘Žπ‘π‘“ 𝑀𝑖𝑓 = π‘€π‘–π‘Žπ‘π‘“, π‘ π‘œ π‘žπ‘–π‘“ = π‘žπ‘–π‘Žπ‘π‘“ p v π‘žπ‘–π‘Žπ‘π‘“ = π‘žπ‘–π‘Ž + π‘ž π‘Žπ‘ + π‘ž 𝑏𝑓 = π‘ž π‘Žπ‘ =π‘žπ‘–π‘“ Thus any rev path can be substituted by a reversible Zig-zag path, between the same end states, consisting of a rev adiabatic + rev isotherm+ another rev adiabatic.
  • 2. Clausius theorem p v Any rev cyclic process RA Rev isotherm T1 T2 Lets us consider a smooth close curve which represents a rev cycle. Then we take two Elemental parts from this cycle as shown in the picture (upper part and lower part). The upper part cab represented by an isotherm+ two RA and similarly the lower part of the curve. The upper isotherm is at a temp say T1 and the lower isotherm is at T2. Now this elementary process can be thought of as a Carnot engine which receives heat dq1 at T1 and rejects heat dq2 at T2. So for this elementary cycle π‘‘π‘ž1 𝑇1 = π‘‘π‘ž2 𝑇2 π‘œπ‘Ÿ 𝑖𝑓 𝑀𝑒 π‘Žπ‘‘π‘‘ π‘€π‘–π‘‘β„Ž π‘‘β„Žπ‘’ Signs of heat transfer then we get π‘‘π‘ž1 𝑇1 + π‘‘π‘ž2 𝑇2 = 0. If we add all the elementary cycles such that we cover the entire cyclic process then we get : dq1 dq2 π‘‘π‘ž1 𝑇1 + dq2 T2 + dq3 T3 + β‹― . dqn Tn = 0 or π‘‘π‘ž 𝑇 = 0 , hence π‘‘π‘ž 𝑇 𝑖𝑠 π‘Ž π‘π‘Ÿπ‘œπ‘π‘’π‘Ÿπ‘‘π‘¦ π‘™π‘–π‘˜π‘’ 𝑒, 𝑇, 𝑣 π‘Žπ‘›π‘‘ β„Ž π‘€β„Žπ‘–π‘β„Ž is called entropy , ds. R1 R2 a b π‘Ž 𝑏 π‘‘π‘ž 𝑇 𝑅1 + 𝑏 π‘Ž π‘‘π‘ž 𝑇 𝑅2 = 0, οƒž π‘Ž 𝑏 π‘‘π‘ž 𝑇 𝑅1 = π‘Ž 𝑏 π‘‘π‘ž 𝑇 𝑅2 Hence entropy is a property, independent of path So, 𝑑𝑠 = π‘‘π‘ž π‘Ÿπ‘’π‘£ 𝑇 or, π‘Ž 𝑏 π‘‘π‘ž 𝑇 = π‘Ž 𝑏 𝑑𝑠 = 𝑠 𝑏 βˆ’π‘  π‘Ž Clausius theorem
  • 3. 𝑑𝑠 = π‘‘π‘ž π‘Ÿπ‘’π‘£ 𝑇 ; If π‘‘π‘ž π‘Ÿπ‘’π‘£ = 0, then the process is called rev adiabatic and then ds=0 Hence s=Const; T S A B CD π‘‘π‘ž π‘‘π‘ž2 RA AB is a general process Either rev or irreversible, All other processes are rev. Consider an elemental cycle, as shown in the picture. π‘‘π‘ž = heat supplied at 𝑇1, π‘‘π‘ž2 = β„Žπ‘’π‘Žπ‘‘ π‘Ÿπ‘’π‘—π‘’π‘π‘‘π‘’π‘‘ π‘Žπ‘‘ 𝑇2 1 βˆ’ π‘‘π‘ž2 π‘‘π‘ž π‘Žπ‘›π‘¦ ≀ 1 βˆ’ π‘‘π‘ž2 π‘‘π‘ž π‘Ÿπ‘’π‘£ π‘‘π‘ž2 π‘‘π‘ž π‘Žπ‘›π‘¦ β‰₯ π‘‘π‘ž2 π‘‘π‘ž π‘Ÿπ‘’π‘£ π‘‘π‘ž π‘Žπ‘›π‘¦ π‘‘π‘ž2 ≀ π‘‘π‘ž π‘Ÿπ‘’π‘£ π‘‘π‘ž2 π‘‘π‘ž π‘Žπ‘›π‘¦ π‘‘π‘ž2 ≀ 𝑇 𝑇2 ; 𝑠𝑖𝑛𝑐𝑒 π‘‘π‘ž π‘Ÿπ‘’π‘£ π‘‘π‘ž2 = 𝑇 𝑇2 Or, Or, Or, Or, π‘‘π‘ž π‘Žπ‘›π‘¦ 𝑇 ≀ π‘‘π‘ž2 𝑇2 οƒž π‘‘π‘ž π‘Žπ‘›π‘¦ 𝑇 ≀ 𝑑𝑠, π‘“π‘œπ‘Ÿ π‘Žπ‘›π‘¦ π‘π‘Ÿπ‘œπ‘π‘’π‘ π‘  For a reversible process: π‘‘π‘ž 𝑇 = 𝑑𝑠 For any other process : 𝑑𝑠 β‰₯ π‘‘π‘ž 𝑇 For any cyclic process: π‘‘π‘ž 𝑇 ≀ 𝑑𝑠 π‘‘π‘ž 𝑇 ≀ 0 𝐼𝑓 π‘‘π‘ž 𝑇 = 0, cycle is reversible, π‘‘π‘ž 𝑇 < 0, cycle is irreversible and feasible π‘‘π‘ž 𝑇 > 0, cycle is impossible Clausius Inequality
  • 4. Entropy principle For any process undergone by the system, 𝑑𝑠 β‰₯ π‘‘π‘ž 𝑇 , if the process is reversible then 𝑑𝑠 = π‘‘π‘ž 𝑇 For isolated system, dq=0, so, 𝑑𝑠 β‰₯ 0, if isolated process is reversible, π‘‘β„Žπ‘’π‘› 𝑑𝑠 = 0, 𝑠 = 𝑐 For irreversible process, 𝑑𝑠 > 0, π‘ π‘œ π‘’π‘›π‘‘π‘Ÿπ‘œπ‘π‘¦ π‘Žπ‘™π‘€π‘Žπ‘¦π‘  π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘  Causes of entropy increase: for a closed system entropy increases due to (a) External interaction, (b) internal irreversibility 𝑑𝑠 = 𝑑𝑠 𝑒π‘₯𝑑 + 𝑑𝑠𝑖𝑛𝑑 = π‘‘π‘ž 𝑇 + 𝑑𝑠𝑖𝑛𝑑; so, 𝑑𝑠 > π‘‘π‘ž 𝑇 ; π‘œπ‘›π‘™π‘¦ π‘“π‘œπ‘Ÿ π‘Ÿπ‘’π‘£π‘’π‘Ÿπ‘ π‘–π‘π‘™π‘’ π‘π‘Ÿπ‘œπ‘π‘’π‘ π‘  𝑑𝑠𝑖𝑛𝑑 = 0 𝑑𝑠 = π‘‘π‘ž 𝑇 + π‘‘π‘ π‘”π‘’π‘›οƒž 𝑠2 βˆ’ 𝑠1 = 1 2 π‘‘π‘ž 𝑇 + 𝑠𝑔𝑒𝑛1βˆ’2 𝑑𝑠 𝑑𝑑 = π‘ž 𝑐𝑣 𝑇 + 𝑠 𝑔𝑒𝑛 Entropy change Entropy transfer Entropy generation dq T 𝑠 𝑔𝑒𝑛 𝑠2 βˆ’ 𝑠1 For an open system π‘šπ‘– π‘ž 𝑠 𝑔𝑒𝑛 π‘š 𝑒 π‘šπ‘– 𝑠𝑖 + π‘ž 𝑐𝑣 𝑇 + 𝑠 𝑔𝑒𝑛 = π‘š 𝑒 𝑠 𝑒 + 𝑑𝑠 𝑐𝑣 𝑑𝑑 , for steady state 𝑑𝑠 𝑐𝑣 𝑑𝑑 = 0 π‘šπ‘– 𝑠𝑖 + π‘ž 𝑐𝑣 𝑇 + 𝑠 𝑔𝑒𝑛 = π‘š 𝑒 𝑠 𝑒 + π‘š2 𝑠2 βˆ’ π‘š1 𝑠1 Integral form of the equation Change of entropy within the CV system
  • 5. 𝑻 𝒅𝒔 relation First law and second law combined: π‘‘π‘ž = 𝑑𝑒 + 𝑝𝑑𝑣 οƒž 𝑇𝑑𝑠 = 𝑑𝑒 + 𝑝𝑑𝑣 𝑑𝑠 = 𝑑𝑒 𝑇 + 𝑝 𝑇 𝑑𝑣 𝑑𝑠 = 𝑑𝑒 𝑇 + 𝑅 𝑣 𝑑𝑣 π‘‘π‘ž π‘Ÿπ‘’π‘£ = 𝑇𝑑𝑠 𝑠2 βˆ’ 𝑠1 = 𝑐 𝑣 𝑑𝑇 𝑇 + 𝑅 𝑣 𝑑𝑣 𝑠2 βˆ’ 𝑠1 = 𝑐 𝑣 ln 𝑇2 𝑇1 + 𝑅 ln 𝑣2 𝑣1 1st Tds relation Only for ideal gases β„Ž = 𝑒 + 𝑝𝑣 π‘‘β„Ž = 𝑑𝑒 + 𝑝𝑑𝑣 + 𝑣𝑑𝑝 π‘‘β„Ž = 𝑇𝑑𝑠 + 𝑣𝑑𝑝 T𝑑𝑠 = π‘‘β„Ž βˆ’ 𝑣𝑑𝑝 2nd Tds relation 𝑑𝑠 = π‘‘β„Ž 𝑇 βˆ’ 𝑣 𝑇 𝑑𝑝 𝑠2 βˆ’ 𝑠1 = 𝑐 𝑝 𝑑𝑇 𝑇 βˆ’ 𝑅 𝑝 𝑑𝑝 οƒž 𝑐 𝑝 ln 𝑇2 𝑇1 βˆ’ 𝑅 ln 𝑝2 𝑝1 Only for ideal gases