The thermodynamic definition of entropy
Atkins, Chapter 4
∆S Sys
heat that would be transferred in a
reversible process (= maximum heat)
qrev
=
T
temperature at which heat
exchange takes place
Why is this a suitable definition?
q>0
q>0
dS Sys
dqrev
=
T
small change
Adding heat increases the entropy
(and disorder) of the system
The entropy (and disorder) of the system
increases in an isothermal expansion
Nils Walter: Chem 260
Entropy change in a system can be
calculated from the reversible heat
Irreversible
Irreversibleexpansion
expansion
p1 = 15 atm
V1 = 1 L
T = 298 K
n = 1 mol
p1 = 15 atm, V1 = 1 L
p2 = 1 atm
V2 = 15 L
T = 298 K
n = 1 mol
Pressure
p2
w = -1.42 kJ
w = -4.12 kJ
p2 = 1 atm, V2 = 15 L
Volume
Reversible
Reversibleexpansion
expansion
wrev
æ V2 ö
= − nRT lnçç ÷÷ = −4.12 kJ
è V1 ø
qrev = -wrev = + 4.12 kJ
Þ
The
Thesystem
systementropy
entropyincreases
increasesby
by
the
thesame
sameamount
amountin
inboth
both
expansions
expansions(state
(statefunction!)
function!)
qrev
4.12 kJ
∆S =
=Nils Walter: Chem
= 13260
.8 J K −1
T
298.15 K
How do classical and statistical
thermodynamic definitions of ∆SSys match?
Classical thermodynamic definition of S in the isothermal expansion of a perfect gas:
state 2
dqrev 1
∆S = S 2 − S1 = ò
=
T
T
state1
state 2
ò dq
state1
rev
æ V2 ö
qrev
=
; wrev = − nRT lnçç ÷÷ = − qrev
T
è V1 ø
isothermal Þ ∆U=
q + w = Cv ∆T = 0
∆
Þ
æ V2 ö
∆S = nR lnçç ÷÷
è V1 ø
classical
classical
and
entropy
statistical
change
entropy
in thechange in
isothermal
the isothermal
expansion
expansion
of a perfect
of a perfect
gas gas
Statistical thermodynamic definition of S in the isothermal expansion of a perfect gas:
V1
W1∝ (V1 )N
∆S = k ln W2 - k ln W1 = k ln (V2)N - k ln (V1)N
N
æ V2 ö
æ V2 ö
æ V2 ö
∆S = k lnçç ÷÷ = Nk lnçç ÷÷ = nN A k lnçç ÷÷
è V1 ø
è V1 ø
è V1 ø
N molecules of gas
V2
W2∝ (V2 )N
R
Þ
æ V2 ö
÷÷ Walter: Chem 260
∆S = nR lnçç Nils
è V1 ø
The second law of thermodynamics
∆∆SSUniverse
= ∆S System++∆∆SSSurroundings
≥0
Universe = ∆SSystem
Surroundings ≥ 0
The
The net
net entropy
entropy will
will increase
increase or
or stay
stay
the
the same.
same. It
It will
will never
never decrease.
decrease.
∆SSurr > 0
∆SSYS < 0
∆SSurr ≥
q
T
q
∆SUniv = 0 only for a reversible process
∆SUniv > 0 for all other processes
Atkins:
tends to
increase
Atkins: The
Theentropy
entropyof
ofthe
theuniverse
universe
increase
Nilstends
Walter:to
Chem
260
∆SSys is a state function, while ∆SSurr and
∆SUniv are pathway dependent
p2
Irreversible
Irreversibleexpansion
expansion
w = - p2 ∆V = -1.42 kJ
qSys = -w = 1.42 kJ
p2 = 1 atm
V2 = 15 L
T = 298 K
n = 1 mol
Reversible
Reversibleexpansion
expansion
wrev
∆∆SSSurr ==-4.8
JJKK-1-1
-4.8
Surr
æ V2 ö
= −nRT lnçç ÷÷ = −4.12 kJ = −q rev = −q Sys
è V1 ø
∆S Sys
qrev
=
T
∆S Surr = −
qSys
T
∆∆SSUniv ==99JJKK-1-1
Univ
p1 = 15 atm, V1 = 1 L
Pressure
p1 = 15 atm
V1 = 1 L
T = 298 K
n = 1 mol
∆∆SSSys == 13.8
JJKK-1-1
13.8
Sys
w = -1.42 kJ
w = -4.12 kJ
p2 = 1 atm, V2 = 15 L
Volume
∆∆SSSys == 13.8
JJKK-1-1
13.8
Sys
∆∆SSUniv ==00JJKK-1-1
Univ
NilsJWalter:
∆∆SSSurr ==-13.8
KK-1-1 Chem 260
-13.8
J
Surr
Isothermal compression
Irreversible
Irreversiblecompression
compression
p1 = 15 atm
V1 = 1 L
T = 298 K
n = 1 mol
w = - p1 ∆V = + 21.3 kJ
qSys = -w = - 21.3 kJ
∆∆SSSurr ==++71.4
JJKK-1-1
71.4
Surr
p2 = 1 atm
V2 = 15 L
T = 298 K
n = 1 mol
Reversible
Reversiblecompression
compression
wrev
∆∆SSSys ==--13.8
JJKK-1-1
13.8
Sys
æ V1 ö
= −nRT lnçç ÷÷ = +4.12 kJ = −q rev = −q Sys
è V2 ø
∆S Sys
qrev
=
T
∆S Surr = −
qSys
T
∆∆SSUniv ==57.6
57.6
Univ
JJKK-1-1
p1 = 15 atm, V1 = 1 L
Pressure
p1
p2 = 1 atm,
V2 = 15 L
Volume
∆∆SSSys ==--13.8
JJKK-1-1
13.8
Sys
∆∆SSUniv ==00JJKK-1-1
Univ
∆∆SSSurr ==++13.8
JJKK-1-1 Chem 260
13.8
Nils
Walter:
Surr