Problem 1.8-3. Work and heat
Problem 1.8-3. Work and heat
(A
(A→→B)B) W W == −−PP A A
((
V V B B −−V V A A))
== −−4
4
⋅⋅10
10
33[ [ ]]
J J( (
PPV V PP V V))
[ [ ]]
J J UU ==
2
2
..
5
5
B B B B −− A A A A ==10
10
44∆ ∆ QQ == ∆∆U U −−W W ==
1
1
..
4
4
⋅⋅10
10
44[ [ ]]
J J (B (B→→C)C) W W PP V V dV dV[ [ ]]
J J B B B B 3 310
10
7
7
))
((
== ⋅⋅ − − = =∫∫
3 3 ⋅⋅ − − − − −− −− ⋅⋅ 33 B B B B C C C C..
..
..
(C (C→→A)A) W W ==0
0
QQ U U( (
PP V V PP V V))
[ [ ]]
J J C C C C A A A A 3 310
10
5
5
..
7
7
5
5
..
2
2
−− == −− ⋅⋅ = = ∆ ∆ = =[ [ ]]
J J dV dV V V P P W W B B B B 3 310
10
67
67
..
2
2
))
((
== −− ⋅⋅ − − = =∫∫
[ [ ]]
J J W W U U Q Q == ∆∆ −− ==12
12
..
67
67
⋅⋅10
10
33 (A (A→→B, parabola)B, parabola)Problem 1.10-1. Shapes of S vs U
Problem 1.10-1. Shapes of S vs U
Most difficu
Most difficult: b, c, f lt: b, c, f and iand i
b) b) 3 3 / / 2 2 3 3 / / 1 1 2 2
= =V
V
NU
NU
R
R
S
S
θ
θ
S S U UInconsistent with the postulates because entropy does not scale linearly with the Inconsistent with the postulates because entropy does not scale linearly with the
Consistent: a, c, e, g, i Consistent: a, c, e, g, i Inconsisten Inconsistent: b, d, f, t: b, d, f, h, jh, j s s zze e o o tt e e ssyysstteemm 3 3 / / 2 2 3 3 / / 1 1 2 2
= =V
V
U
U
N
N
R
R
S
S
λ
λ
λ
λ
λ
λ
θ
θ
λ
λ
λ
λ
==λ
λ
22 / / 33 not true for not true for arbitraryarbitrary λλc) c) 2 2 / / 1 1 2 2 0 0 2 2 2 2 / / 1 1
+ +
= =v
v
V
V
R
R
NU
NU
R
R
S
S
θ
θ
θ
θ
S S U U Consistent with the postulates, zero of energyConsistent with the postulates, zero of energy is a
Problem 1.10-1. Shapes of S vs U
Problem 1.10-1. Shapes of S vs U
f) f)
= = 0 0 2 2ln
ln
v
v
R
R
N
N
UV
UV
NR
NR
S
S
θ
θ
S S U U Inconsistent with postulate IV,Inconsistent with postulate IV, S should be zero at S should be zero at ∂∂S/ S/ ∂∂UU→∞→∞ 2 2 i) i)
= =NR
NR
V
V
R
R
U
U
00exp
exp
S S U U S S U U First plot First plot U vs S U vs S Then rotate it Then rotate it around the U/S around the U/S bisector to
bisector to obtain S vs U obtain S vs U
Consistent with the postulates Consistent with the postulates
Problem 1.10-3. Entropy maximum for a
Problem 1.10-3. Entropy maximum for a
compos
compos
ite
ite
system
system
S S
For a
For a composite composite system (S=Ssystem (S=S11+S+S22))
( (
))
6610
10
27
27
⋅⋅ −− = = A A NV NV( (
))
6610
10
8
8
⋅⋅ −− = = A A NV NV80
80
= = + + = = B B AA U U
U U U U A A AA++ BB
( (
))
11 / / 33 3 3 / / 1 12
2
3
3
U U A A U U U U A A const const S S − − + + == -- need to need to maximaximize wmize with respeith respect to Uct to U
A A
( (
))
[ [
11 / / 33 11 / / 33]]
2
2
3
3
0
0
A A A AA A U U U U U U U U − − + + ∂ ∂ ∂ ∂ = = 22 / / 33
( (
))
22 / / 333
3
2
2
0
0
== −− −− −− −− A A AA U U U U
U U Solving for U Solving for UAA
( (
2
2
/
/
3
3
))
51
51
..
8
8
[[
]]
1
1
33 / / 22 J J U U UU A A == +
+ = =
Problem 2.3-4. Restrictions due to the Postulates
r m nN
V
AU
S
=Most common problem: forgetting to check if entropy is additive (extensive) When we increase the size of a thermodynamic system by a factor of λ, all
extensive variables should scale with λ: U→ λU, V→ λV, N→ λN, S→ λS.
Subjecting the fundamental relation to this scaling transformation:
( ) ( ) ( )
n m r n m r n m rN
V
AU
N
V
U
A
S
=λ
λ
λ
=λ
+ +λ
r m n+ + =λ
λ
n
+m
+r
=1
Problem 2.3-4. Restrictions due to the Postulates
r m nN
V
AU
S
= S U S must be increasing with U, so n>0∂
S
0<n<1,
∂
U
Additional condition: P should increase with U/V:
r m n
N
V
mAU
V
S
T
P
−1 = ∂ ∂ = n m rN
V
nAU
U
S
T
11
− = ∂ ∂ =V
U
n
m
V
S
T
P
= ∂ ∂ = >0
n
m
0
>m
Problem 2.6-3. Energy in Equilibrium
]
[
10
5
.
2
3 ) 2 ( ) 1 (J
U
U
+ = × (1) =?
U
2
) 1 ( =N
N
(2) =3
) 1 ( ) 1 ( ) 1 (2
3
1
U
N
R
T
= ) 2 ( ) 2 ( ) 2 (2
5
1
U
N
R
T
=The equations of state of the two systems are:
Since the two systems are in thermal equilibrium:
) 2 ( ) 1 (
T
T
= ) 2 ( ) 2 ( ) 1 ( ) 1 (2
5
2
3
U
N
U
N
= (2) (1)6
15
U
U
=]
[
10
5
.
2
3 ) 2 ( ) 1 (J
U
U
+ = ×2
.
5
10
[
]
6
15
(1) 3 ) 1 (J
U
U
+ = × (1)714
[
]
J
U
=Problem 2.7-3. The Indeterminate Problem of Thermodynamics
0
) 2 ( ) 1 ( + = ≡dU
dU
dU
Since the composite system is isolated a) Show that
P
(1) =P
(2) ) 2 , 1 ( ) 2 , 1 ( ) 2 , 1 (dV
P
dU
= −Since the piston is adiabatic, δ Q (1,2) = 0 and thus
0
) 2 ( ) 2 ( ) 1 ( ) 1 ( − = −P
dV
P
dV
Since the total volume of the composite system is constant
dV
(1) = −dV
(2)) 2 ( ) 1 ( , .
b) Show that thermodynamics does not tell us what the temperatures of the two systems are
0
) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 2 ( ) 1 ( + = + + + = ≡dV
T
P
T
dU
dV
T
P
T
dU
dS
dS
dS
The entropy maximum condition requires:
0
) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( = + + +T
dV
P
dU
T
dV
P
dU
The numerators in this expression vanish identically because of the adiabaticity of the wall, therefore the denominators (T(1)and T(2)) can be arbitrary and still satisfy the entropy
Problem 2.8-1. Semi-permeable partition
Two different gases N1 and N2 N1(1), N
2(1) N1(2), N2(2)
N1 can freely flow through the membrane while N2cannot.
) 2 ( )
1
( althou h one of the as com onents can freel
N1
penetrate the membrane
P is intensive parameter conjugate to volume, volume is constrained
initial
N
N
N
N
1(1) + 1(2) = 1(1) + 1(2)(
U
U
)
initialU
U
(1) + (2) = (1) + (2) Conservation conditions: Equilibrium conditions: ) 2 ( ) 1 (T
T
= ) 2 ( ) 2 ( 1 ) 1 ( ) 1 ( 1T
T
µ
µ
=Problem 2.8-1. Semi-permeable partition
−
−
+ =N
N
R
N
N
N
R
N
N
V
U
NR
NA
S
5 / 2 1 1 2 2 2 / 3ln
ln
ln
U
NR
U
S
T
2
3
1
= ∂ ∂ = 2 1N
N
N
= +2
3
NRT
U
=T
T
T
(1) = (2) = initialU
U
U
U
(1) + (2) = (1) + (2) Solve for T...
1 1 = ∂ ∂ =N
S
T
Here, when differentiating do not forget that N=N1+N2(
)
+ + − = 2 / 3 2 1 1 2 / 3 1ln
2
5
N
N
N
V
U
R
R
A
T
µ
After some algebra:
]
[
73
.
272
K
T
=Problem 2.8-1. Semi-permeable partition
T
T
) 2 ( 1 ) 1 ( 1µ
µ
=(
N
N
)
initialN
N
1(1) + 1(2) = 1(1) + 1(2) Solve for N10
.
75
) 2 ( 1 ) 1 ( 1 =N
=N
To find P:V
NR
T
P
V
S
= = ∂ ∂(
)
) 1 ( ) 1 ( 2 ) 1 ( 1 ) 1 (V
RT
N
N
P
= +]
[
680
) 1 (kPa
P
= Similarly:P
(2) =567
[
kPa
]
Numerical Problem
First calculate P and T by differentiating the fundamental relation. Then use the parametric plot of P(S) and T(S) to plot the P(T) dependence.
Nu=1;V =0.01;U0=1; T@S_ D:=U0S V J2 + S NuNExpB S NuF; P@S_ D:=U0S 2 V 2ExpB S NuF;
Param etricPlot@ T@SD,P@SD<, S,0,0.8<,Fram e->True,TextStyle-> FontFam ily->"Tim es",FontW eight->"Bold",FontSize->14<, Fram eLabel-> Tem perature""@KD,Pressure" "@PaD<,PlotRange-> AllD
14000 0 100 200 300 400 500 Temperature @KD 0 2000 4000 6000 8000 10000 12000 e r u s s e r P @ a P D ÜGraphics Ü
Problem 3.3-1. Fundamental relation from the equations of state
v
As
T
23
= Equations of state: 2 3v
As
P
=a) To find the chemical potential, let us use the Gibbs-Duhem relation. Since the independent variables are s and v, it is convenient to use the Gibbs-Duhem relation in the energy representation:
− 3 23
As
As
−
v
v
2
− =v
As
d
d
3µ
ds
v
As
dv
v
As
ds
v
As
dv
v
As
d
2 2 3 2 2 33
2
6
3
+ − − =µ
ds
v
As
dv
v
As
d
2 2 33
− =µ
const
v
As
+ − = 3µ
Problem 3.3-1. Fundamental relation from the equations of state
Molar form of the Euler equation of state in the energy representation is:
+ −
=
Ts
Pv
u
Substituting the equations of state into this equation:
const
v
As
const
v
As
v
v
As
s
v
As
u
= − − + = + 3 3 2 3 23
const
v
As
u
= + 3Problem 3.3-1. Fundamental relation from the equations of state
Now let us directly integrate the molar form of the equations of state:
Pdv
Tds
du
= −dv
v
As
ds
v
As
du
2 3 23
− =
= − =v
As
d
dv
v
As
ds
v
As
du
3 2 3 23
const
v
As
u
= + 3Problem 3.5-1. Fundamental relation from the equations of state
c) Equations of state:buv
a
buv
c
v
u
P
+ + =buv
a
u
T
+ =Equations of state are functions of u and v, so convenient to work in the entropy representation in molar form:
P
s
s
=
∂
∂ =1
Are these equations of state compatible?
T
T
v
u
v
∂
u
∂bv
u
a
T
+ =1
bu
v
c
T
P
+ = where
∂ ∂ ∂ =
∂ ∂ ∂u
v
s
v
u
s
2 2v
T
u
T
P
∂
∂ = ∂
∂1
b
b
= YesProblem 3.5-1. Fundamental relation from the equations of state
dv
T
P
du
T
ds
=1
+bu
dv
v
c
du
bv
u
a
ds
+ +
+ =( )
dv
bu
( )
dv
ad
( ) ( )
u
bv
du
cd
( ) ( )
v
bu
dv
v
c
du
bv
du
u
a
ds
= + + + =ln
+ +ln
+( )
uv
d
vdu
= - This is wrong because v is a variable, not a constant( ) ( )
( ) ( )
( )
( )
( ) ( )
( )
ln
( )
(
)
ln
)
(
ln
ln
ln
ln
uv
bd
v
cd
u
ad
du
v
dv
u
b
v
cd
u
ad
dv
bu
v
cd
du
bv
u
ad
ds
+ + = + + + = + + + =( )
u
c
( )
v
buv
const
a
s
=ln
+ln
+ + (same problem in 3.3-1)Problem 3.5-6. VdW and ideal gas in equilibrium
2v
a
b
v
RT
P
− − =v
RT
P
=In this problem, temperature, pressure and molar volumes of the van der Waals fluid and the ideal gas are the same in equilibrium
VdW: Ideal:
P
RT
v
=Substituting this molar volume into the equation of state for VdW:
( )
2 2RT
aP
b
P
RT
RT
P
− − =Here P is the only unknown, so need to solve for P
After simple algebra:
[ ]
Pa
a
RT
b
RT
P
1
=3
.
5
⋅10
7
− =Problem 3.6-1. Microwave Background Radiation
4 / 1 4 / 3 4 / 13
4
V
U
b
S
= Fundamental relation 4bVT
U
= Equation of stateSubstituting U into expression for S: 3
3
4
bVT
S
=Isentropic process: Sinitial=Sfinal=S
3 3 3
3
4
2
3
4
3
4
f i f f i iT
bV
T
bV
T
bV
= = f iV
V
=2
3 32
f iT
T
=T
fT
i 32
1
=Note that energy of the background radiation is not conserved, it constantly decreases! Where does it go?
Problem 3.7-2. Rubber Band
T
cL
U
= 0Equation of state of the rubber band is:
Since T is constant, U is also constant:
dT
=0
dU
=0
From the conservation of energy:
dU
=W
+Q
=0
W
Q
= −Calculating work from the second equation of state:
dL
L
L
L
L
bT
W
0 1 0 − − =δ
0 1 0L
L
L
L
bT
− − = Γ We obtain:dL
L
L
L
L
bT
Q
0 1 0 − − − =δ
Problem 3.8-1. Paramagnet
The fundamental relation of a simple paramagnetic system is:
+ = 2 0 2 2 0exp
I
N
I
NR
S
NRT
U
Equations of state:
+ =
∂ = 2ex
S
I
T
U
T
∂ 0 ,NR
N
I
S
I N
+ =
∂ ∂ = 2 0 2 2 2 0 0 ,exp
2
I
N
I
NR
S
NI
IRT
I
U
B
N S e
+
− − =
∂ ∂ = 2 0 2 2 2 0 2 2 0 ,exp
2
1
I
N
I
NR
S
I
N
I
NR
S
RT
N
U
I Sµ
Problem 3.8-1. Paramagnet
N
V
B
TS
U
= + e +The Euler equation:
Substituting the equations of state into the Euler equation:
− − + +
+ = 2 2 2 02
1
2
exp
S
I
S
I
S
I
RT
U
0 0 0
+ = 2 0 2 2 0exp
I
N
I
NR
S
RT
U
Problem 3.9-6. Model of a solid insulator
(
)
+ − =R
s
v
v
b
As
u
3
exp
0 2 3 / 4 Fundamental relation: a) Show that at s→0, T→0(
)
+
+ − = ∂ ∂ =R
s
s
R
s
v
v
b
A
s
u
T
3
3
4
3
exp
3 / 4 3 / 1 2 0Problem 3.9-6. Model of a solid insulator
Problem 3.9-6. Model of a solid insulator
b) Show that c b) Show that cvv~ T~ T33 at Tat T→→00
( (
))
+ +
+ + − − = =R
R
ss
ss
R
R
ss
vv
vv
b
b
A
A
T
T
3
3
3
3
4
4
3
3
exp
exp
3 3 / / 4 4 3 3 / / 1 1 2 2 0 0( (
))
+ + − − = =R
R
ss
vv
vv
b
b
As
As
u
u
3
3
exp
exp
00 22 3 3 / / 4 4Let us calculate T/u: Let us calculate T/u:
+ + = =T
T
4
4
1
1
+ + = =u
u
T
T
4
4
1
1
Since sSince s→→0 at T0 at T→→0, we can simplify the expression for T at low temperature:0, we can simplify the expression for T at low temperature:
ss
u
u
T
T
~
~
On the other hand, from the fundamental relation at low T, we obtain:
On the other hand, from the fundamental relation at low T, we obtain:
u
u
~
~ ss
44 / / 33 Substituting into the expression for T:Substituting into the expression for T:
T
T
~
~ ss
11 / / 33ss
~
~ T
T
33Since Since
ss
u
u
T
T
~
~
u
u
~
~
Ts
Ts
~
~
T
T
44 33 4 4~
~
~
~
T
T
T
T
T
T
T
T
u
u
cc
vv vv ∂ ∂ ∂ ∂
∂ ∂ ∂ ∂ = = This isProblem 3.9-6. Model of a solid insulator
Problem 3.9-6. Model of a solid insulator
c) Show that c c) Show that cvv~ R at T~ R at T→∞→∞
R
R
u
u
R
R
ss
u
u
T
T
3
3
3
3
1
1
3
3
4
4
≈ ≈
+ + = =( (
))
R
R
T
T
RT
RT
T
T
u
u
cc
vv vv3
3
3
3
≈ ≈ ∂ ∂ ∂ ∂ ≈ ≈
∂ ∂ ∂ ∂ = =ss is a monotonically increasing function of u and T sois a monotonically increasing function of u and T so ss→∞→∞, T, T→∞→∞ (this is not tr(this is not trivial,ivial,
related to the fact that second derivative of u with respect to s is positive) related to the fact that second derivative of u with respect to s is positive)
RT
RT
u
u
≈≈3
3
d) Show that d) Show that α α →→0 as P0 as P→→00 P PT
T
vv
vv
∂ ∂ ∂ ∂ = =1
1
α
α
S Svv
u
u
P
P
∂ ∂ ∂ ∂ − − = =( (
))
+ + − − = =R
R
ss
vv
vv
b
b
As
As
u
u
3
3
exp
exp
00 22 3 3 / / 4 4(
(
)
)
(
(
)
)
b
b
(
(
vv
vv
))
u
u
R
R
ss
vv
vv
b
b
As
As
vv
vv
Ab
Ab
P
P
00 44 / / 33 00 222
2
003
3
exp
exp
2
2
== −− −−
+ + − − − − − − = =0
0
2
2
22 → →
∂ ∂ ∂ ∂ = =
∂ ∂ ∂ ∂ p p P PT
T
u
u
bu
bu
P
P
T
T
vv
bu
bu
P
P
vv
vv
2
2
0 0 −− = = when whenP
P
→→0
0
Num
Num
erica
erica
l Prob
l Prob
lem –
lem –
Heli
Heli
um an
um an
d W
d W
ater
ater
R R == 8. 8.3143144;V4;V00==0.0.0303;U;U00==15000;15000; a ahh==0.0.00300346;bh46;bh==23.23.771010--66;;cchh==11..55;;HH**helium helium **LL a aww==0.0.54544;bw4;bw==30.30.551010--66;c;cww==33..11;;HH**wwataterer**LL P
Pllot3Dot3DAARR LoLoggAHAH Vh Vh-- b bhhLLHHUhUh++aahhêê Vh VhLLcchhEE++RRLoLogg@H@H V0 V0-- Vh Vh-- b bwwLL HHU0U0--UhUh++aawwêHêH V0 V0-- Vh VhLLLLcwcwDD,, Vh,0.Vh,0.0101V0,0.V0,0.9999V0V0<<,, Uh,0.Uh,0.0101U0,0.U0,0.9999U0U0<<,P,PllototRRangangee->-> Al AlllEE
Using the known form of the fundamental relation for the VdW fluid, we can plot Using the known form of the fundamental relation for the VdW fluid, we can plot entropy of the composite system as:
entropy of the composite system as:
0.01 0.01 0.02 0.02 5000 5000 10000 10000 20 2000 2 25500 0.01 0.01 0.02 0.02 Ü
Num
Num
erica
erica
l Prob
l Prob
lem –
lem –
Heli
Heli
um an
um an
d W
d W
ater
ater
C
CononttouourrPPllototAARR LoLoggAHAH Vh Vh-- b bhhLLHHUhUh++aahhêê Vh VhLLcchhEE++R LoRLogg@H@H V0 V0-- Vh Vh-- bw bwLL HHU0U0--UhUh++aaww êH êH V0 V0-- Vh VhLLLLccwwDD, V, Vh,0.h,0.0101V0,0.V0,0.9999V0V0<<, , UhUh,0.,0.0101UU0,0.0,0.9999U0 U0 ,P,PllototRRangeange->-> Al AlllEE
10000 10000 12000 12000 14000 14000
Same plot in the contour format: Same plot in the contour format:
0 0 00..00005 5 0.0.001 1 00..00115 5 00..002 2 00..00225 5 00..0033 0 0 2000 2000 4000 4000 6000 6000 8000 8000 Ü
Numerical Problem – Helium and Water
Now we can numerically solve for the volume and energy of helium that maximize the entropy of the composite (water + helium) system
m ve=FindMinim um A-R LogAH Vh- bhLHUh+ahê VhLchE-R Log@H V0- Vh- bwL HU0-Uh+awê H V0- VhLLcwD,8 Vh,0.01 V0,0.99 V0<,8Uh,0.001U0,0.999 U0<E
8-273.848,8VhØ0.0150799,UhØ4903.03<<
Vh1= Vhê.m ve@@2,1DD 0.0150799
Uh1=Uhê.m ve 2,2 4903.03
Substituting thus obtained values of the volume and energy into the VdW
equations of state, we find the values of temperature and pressure of helium in equilibrium T= Uh1 +ahê Vh1 chR 393.154 P=T R Vh1- bh - ahchR Uh1 Vh1 Vh1 ah Vh1 217109.
Numerical Problem – Helium and Water
T0=300 300 eq1=Uh+ahê Vh==chR T0 Uh+ 0.00346 Vh == 3741.48 eq2=Uw+aw H V0- VhL==cw R T0 0.544Now, to find the equilibrium state of the system we use two temperature
equations of state and the pressure equation of state for the two VdW fluids:
Uw+ 0.03-Vh == . eq3= R Vh- bh - ahchR UhVh2+ah Vh == R V0- Vh- bw - aw cw R Uw H V0- VhL2+awH V0- VhL 8.3144 -0.0000237+Vh - 0.0431517 0.00346Vh+UhVh2== -14.0214 0.544H0.03-VhL+UwH0.03-VhL2+ 8.3144 0.0299695-Vh
NSolve@8eq1,eq2,eq3<,8 Vh,Uh,Uw<D
99UhØ3668.54-599.249 I,Uw Ø7714.26+0.00343921 I,VhØ6.9252µ10-7-5.68961µ10-6I=,
9UhØ3668.54+599.249 I,Uw Ø7714.26-0.00343921 I,VhØ6.9252µ10-7+5.68961µ10-6I=,8UhØ3741.25,U wØ7695.87,VhØ0.0151061<,
8UhØ3741.36,Uw Ø4753.18,VhØ0.0298174<,8UhØ3741.36,UwØ -7106.08,VhØ0.0299633<=
Physically meaningful solutions have real-valued volumes and energies. There are three such solutions.
Numerical Problem – Helium and Water
Three physically meaningful solutions:
8
UhÆ3741.25095377012062`,UwÆ7695.86692894271351`,VhÆ0.0151061206931622615`<
,8
UhÆ3741.36396037953744`,UwÆ4753.17737520313787`,VhÆ0.029817401540839814`<
,8
UhÆ3741.36452555181753`,UwÆ -7106.08071201483948`,VhÆ0.0299633385449733715`<
1 2 3They correspond to three volume ratios of helium to water:
1.01425 163.295 817.298
Some of you have stated that the first solution (Vh/Vw = 1.01425) is the one that should be experimentally observed as it maximizes the total entropy of helium plus water. This is incorrect since the system is not adiabatically isolated from the environment in part (b) of this problem. Therefore, the total entropy of helium, water AND the environment must be maximized by the correct solution. At this point, you do not have enough information to decide which of the three solutions is correct – we will learn this later in the course. It will turn out that the right solution is Vh/Vw =817.298. This solution minimizes energy of the system. This solution is consistent with our intuition that water vapor should condense into liquid at room temperature and pressure close to atmospheric, and thus occupy small volume. Note that the internal energy of water for solution #3 is negative, which means that the absolute value of the (negative) potential energy exceeds the kinetic energy and thus the system is bound (remember mechanics and astronomy), or “condensed” in the language of thermodynamics.
Problem 4.2
3 Free Expansion of Ideal Gas
dV
V
NR
dV
T
P
dS
dN
T
dV
T
P
dU
T
dS
1
In general:
Since U of ideal gas is independent on volume (dU=0), and N = const in the process:
In a series of infinitesimal free expansions, entropy changes by:
i f V VV
V
NR
dV
V
NR
S
f iln
Problem 4.2
4 Temperature and entropy change in free expansion
s
Av
T
2
The equations of state:
0ln
2
s
s
Av
P
Given the initial temperature
T 0, molar volume
v0, and final molar volume
v f,
find the final temperature and increase in molar entropy
s.
Solution:
Since equations of state involve v and s as independent extensive variables, it is
convenient to find the fundamental relation in the energy representation:
Pdv
Tdu
du
dv
s
s
Av
ds
s
Av
du
0 2ln
2
integrating
C
s
s
Av
u
0 2ln
Problem 4.2
4 Temperature and entropy change in free expansion
In the free expansion process, no work is done on
the system and no heat is transferred to the system
if it is adiabatically insulated form the environment.
Therefore, the energy of the system does not
change in the free expansion process:
i f
u
u
C
s
s
Av
u
0 2ln
0 2 0 2 0ln
ln
s
s
Av
s
s
Av
i f fNote that s
0is not the initial entropy, it is just a constant (this may be a source
of some confusion). Otherwise we would get
s f =s0in the irreversible free
Problem 4.2
4 Temperature and entropy change in free expansion
0 2 0 2 0ln
ln
s
s
Av
s
s
Av
i f f 2 2 0 0 0ln
ln
f v f v is
s
s
s
2 2 0 0 0 f v f v is
s
s
s
0 0 2 2 0s
s
s
s
v f v i f
2 2 0 0 0 f v v i fs
s
s
s
f f fs
Av
T
2
2 2 0 0 0 2 f v v i f fs
s
s
Av
T
2 2 0 2 0 0 0 0 2 f v v f fAv
s
T
s
Av
T
0 2 0 0 0 2 0 0 2 2 0T
Av
s
T
Av
s
s
f v v
0 2 0 2T
Av
T
Av
s
s
s
f f i f
Problem 4.4
1 Heat Exchange between Two Identical Objects
1 1 2
2 20
1 20 10
U
C
T
dT
C
T
dT
f f T T T TConservation of energy demands:
where
T fis the final temperature of each body. Therefore:
2 2 20 20 2 10 102
2
2
2
fT
fB
AT
T
B
AT
T
B
AT
Solving for
T f2
307
.
1
[
]
2
1
2 20 10 2 20 10K
B
A
T
T
T
T
B
A
T
f
Problem 4.4
1 Heat Exchange between Two Identical Objects
f f f f f f T T T T T T T T T T T TdT
T
BT
A
dT
T
BT
A
T
dT
C
T
dT
C
T
Q
T
Q
S
20 10 20 10 20 10 2 2 2 1 1 1 2 2 2 1 1 1 2 2 1 1 1
The change of entropy of the two systems in this process can be found from the
expression below, where we have taken into account that no work is done in the
process
20 20 10 10ln
ln
B
T
T
T
T
A
T
T
B
T
T
A
S
f f f
f
K
J
T
T
T
B
T
T
T
A
S
ln
f2
f 10 201
.
6
20 10 2Integrating, we obtain:
Problem 4.5
6 Maximum work theorem
P
V
A
B
C
adiabat
isobar
isochore
1. The system delivers a non-zero work during the adiabatic process A – B: W AB > 0 .
2. Since the process is adiabatic, no heat flows between
the system and the reversible heat source (RHS): Q AB = 0.
3. From the conservation of energy and the above two statements, the internal energy of the system decreases in this process U AB < 0.
4. Since energy is a monotonically increasing function of temperature, temperature of the system decreases in the process A – B: TB < T A.
5. Since temperature is a function of the state only, it also decreases in the process A – C – B.
6. Since W ACB < W AB and U AB = U ACB, we conclude from the conservation of energy that heat flows from the system to the reversible heat sink during the A-C-B process Q ACB>0.
7. Entropy of the composite system (system + reversible heat source) increases every time there is
heat transfer between the system and the RHS and Tsystem TRHS(irreversible process).
8. For example, at point B’ on the isobar close to point B, Tsystem < T A TRHSbecause positive heat has been transferred to the RHS during the ACB’ process Q ACB>0.
9. Since Tsystem TRHSduring the B’ - B process, the B’ – B process is irreversible and thus the whole A-C-B process is irreversible.
Problem 4.5
11 Heat Flow between Two Different Objects
2
20 2 2 10 2 2 2 1 1 2 2 1 12
2
20 10 20 10T
T
b
T
T
a
dT
bT
dT
aT
dT
T
C
dT
T
C
U
f f T T T T T T T T f f f f
The change of energy of the two bodies only depends on the initial and final
states and is given by:
Maximum work can be obtained from the two bodies in a reversible process
where the total change of entropy of the two bodies is zero:
10
20
2 2 2 1 1 1 2 2 1 1 2 12
0
20 10 20 10T
T
b
T
T
a
dT
T
C
dT
T
C
T
Q
T
Q
S
S
S
f f T T T T T T T T f f f f
In the above expression we replaced
Qwith
dUsince the process was carried out
at constant volumes of both bodies, so no work was done by the bodies
d i r e c t l y.
The bodies act as hot and cold bodies of a typical heat engine.
Solving for
T f:
b
a
bT
aT
T
f2
2
20 10
2 20 102
b
T
T
a
ab
U
Substituting the expression for
T fThe maximum work is negative of
U:
0
2
2 20 10 max
T
T
b
a
ab
W
Problem 4.6
2 Efficiency of the heat pump
When calculating efficiencies (and many other quantities) in thermodynamics,
one needs to use the absolute temperature scale (Kelvin scale).
Indeed, there is nothing special about zero Celsius of Fahrenheit, but if you use
Celsius or Fahrenheit in the equation above, your efficiency always turns t o
zero if the hot system is at zero degrees independent on the temperature of the
cold body. This is unphysical because zero of Celsius or Fahrenheit
temperature scale is to a large degree arbitrary.
The Kelvin scale is the absolute temperature scale (zero Kelvin is the lowest
temperature possible) and should always be used in thermodynamics. Convert
temperature to Kelvins before using it in a thermodynamic equation.
c h h
T
T
T
W
Q
Efficiency of a reversible heat pump:
K
T
h
294
.
26
K
T
c
283
.
15
5
.
26
Problem 5.3
−1 Fundamental Relation for Ideal Gas in Different Representations
+ = −5 / 2 0 0 2 / 3 0 0ln
N
N
V
V
U
U
NR
Ns
S
Fundamental relation in the entropy representation for monoatomic ideal gas:
Solving for U, we get internal energy representation of the fundamental relation:
−
=R
s
NR
S
N
N
V
V
U
U
3
2
3
2
exp
0 3 / 5 0 3 / 2 0 0For calculation of the Helmholtz potential, calculate temperature:
−
= ∂ ∂ =R
s
NR
S
N
N
V
V
NR
U
S
U
T
3
2
3
2
exp
3
2
5 / 3 0 0 3 / 2 0 0Problem 5.3
−1 Fundamental Relation for Ideal Gas in Different Representations
Therefore:U
NRT
2
3
= and
+ = −5 / 2 0 0 2 / 3 0 02
3
ln
N
N
V
V
U
NRT
NR
Ns
S
TS
U
F
= − Substituting these expressions into the Legendre transform for S:
+ − = −5 / 2 0 0 2 / 3 0 02
3
ln
2
3
N
N
V
V
U
NRT
R
s
NT
NRT
F
Problem 5.3
−1 Fundamental Relation for Ideal Gas in Different Representations
For calculation of the enthalpy, calculate pressure:
−
= ∂ ∂ − =R
s
NR
S
N
N
V
V
U
V
U
P
3
2
3
2
exp
3
2
5 / 3 0 0 3 / 5 3 / 2 0 03
2
−
=R
s
NR
S
N
N
V
P
U
V
5
2
5
2
exp
3
2
0 0 5 / 2 0 5 / 3 0Express volume as a function of pressure:
The appropriate Legendre transformation is:
PV
PV
PV
PV
U
H
2
5
2
3
= + = + =Problem 5.3
−1 Fundamental Relation for Ideal Gas in Different Representations
(
)
−
= =R
s
NR
S
N
N
PV
U
PV
H
5
2
5
2
exp
3
2
2
5
2
5
0 0 5 / 2 0 5 / 3 0For the Gibbs potential calculation:
U
NRT
2
3
= 3 / 5 3 / 2
−
= ∂ =R
NR
N
V
NR
S
T
3
3
exp
3
0 0 0 0
−
= ∂ ∂ − =R
s
NR
S
N
N
V
V
U
V
U
P
3
2
3
2
exp
3
2
5 / 3 0 0 3 / 5 3 / 2 0 0TS
NRT
PV
TS
NRT
PV
TS
U
G
= − + = − + = −2
5
2
3
NRT
PV
=Problem 5.3
−1 Fundamental Relation for Ideal Gas in Different Representations
+ − = −5 / 2 0 0 2 / 3 0 0ln
2
5
N
N
V
V
U
U
NR
Ns
T
NRT
G
+ − = −5 / 2 0 0 2 / 3 0 02
3
ln
2
5
N
N
PV
NRT
U
NRT
R
s
TN
NRT
G
Problem 5.3
−
5 Calculation of Gibbs potential and
α
,
κ
T, c
P(
)
1 / 3 3 / 1 0 2NVU
v
R
S
=θ
We are given the fundamental relation:
Solve it for U to obtain the fundamental relation in the energy representation:
3 2 0
S
NV
R
v
U
=θ
To obtain the Gibbs potential via a Legendre transformation:
(
T
P
)
U
TS
PV
G
,
= − +We need to express U, S and V as functions of T and P. There fore, we need three equations to eliminate U, S and V :
3 2 0
S
NV
R
v
U
=θ
2 2 03
S
NV
R
v
S
U
T
=θ
∂ ∂ = 3 2 2 0S
NV
R
v
S
U
P
=θ
∂ ∂ − =Problem 5.3
−5 Calculation of Gibbs potential and
α
,
κ
T, c
PSolving these three equations for U, S and V , we obtain:
P
v
NT
R
U
θ
0 3 227
=P
v
NT
R
S
θ
0 2 29
= 2 0 3 227
v
P
NT
R
V
θ
=Substituting them into the Legendre transformation for G and simplifying:
(
)
P
v
PV
TS
U
P
T
G
θ
027
,
= − + = −Writing the differential of G:
dG
= −SdT
+VdP
+dN
we note that:
P
v
NT
R
T
G
S
N P 0θ
2 2 ,9
=
∂ ∂ − = 2 0 3 2 ,27
2
P
v
NT
R
P
G
V
N Tθ
=
∂ ∂ =Problem 5.3
−5 Calculation of Gibbs potential and
α
,
κ
T, c
PRecalling the definitions of α , κ T , c P:
T
P
v
NT
R
P
v
NT
R
P
v
NT
R
V
P
v
NT
R
T
V
T
V
V
N P N P3
9
2
27
2
1
9
2
1
27
2
1
1
0 2 2 2 0 3 2 0 2 2 , 2 0 3 2 , = = =
∂ ∂ =
∂ ∂ =θ
θ
θ
θ
α
P
P
v
NT
R
P
v
NT
R
P
v
NT
R
V
P
v
NT
R
P
V
P
V
V
N P N P T2
27
4
27
2
1
27
4
1
27
2
1
1
3 0 3 2 2 0 3 2 3 0 3 2 , 2 0 3 2 , = = =
∂ ∂ − =
∂ ∂ − =θ
θ
θ
θ
κ
P
v
T
R
P
v
NT
R
T
N
T
T
S
N
T
c
N P N P Pθ
θ
0 2 2 , 0 2 2 ,9
2
9
=
∂ ∂ =
∂ ∂ =Problem 5.3
−
6 Fundamental Relation in the Enthalpy Representation
The second good fundamental relation:
2 / 1 2 0 2 2 / 1
