Problem 1.8-3. Work and heat: J V V P W J V P V P U J W U Q

Problem 1.8-3. Work and heat  W  = − P A (V  B − V  A ) = −4 ⋅10 [ J ] 3

(A→B)

∆U  = 2.5(P BV  B − P AV  A ) = 10 [ J ]

Q = ∆U  − W  = 1.4 ⋅10 [ J ]

4

4

 B

∫

W  = − P (V )dV  = 7 ⋅10 [ J ]

(B→C)

. (C→A)

W  = 0

(A→B, parabola)

C 

C  −

 B  B  B

3

− . ⋅

3

−

Q = ∆U  = 2.5(P AV  A − PC V C  ) = −7.5 ⋅10 [ J ] 3

 B

∫

W  = − P (V )dV  = −2.67 ⋅10 [ J ] 3

 B

Q = ∆U  − W  = 12.67 ⋅10 [ J ] 3

− . ⋅

3

Consistent: a, c, e, g, i

Problem 1.10-1. Shapes of S vs U  Most difficult: difficult: b, c, f and i 1 / 3

b)

  R  

S  = 

  NU  

Inconsistent: Inconsisten t: b, d, f, h, j

S 2 / 3

    θ      V    2

U Inconsistent with the postulates because entropy does not scale linearly with the s ze o t e system 1 / 3

  R    λ  N λ U     2   θ      λ V   

2 / 3

λ  = λ 2 / 3

λ S  = 

1 / 2

c)

  R  S  =    θ  

   Rθ V   NU  + 2 v0  

2

1 / 2

    

Consistent with the postulates, zero of energy is arbitrary a rbitrary..

not true for arbitrary λ

S

U

Problem 1.10-1. Shapes of S vs U 

f)

  UV     S  =  NR ln 2   N   Rθ v0  

S

Inconsistent with postulate IV, S should be zero at ∂S/ ∂U→∞

U

2

i)

U  =

0

 R V 

exp

  NR 

Consistent with the postulates

U Then rotate it around the U/S bisector to obtain S vs U

First plot U vs S

S

S

U

Problem 1.10-3. Entropy maximum for a compos composite ite system  For a composite  composite system system (S=S1+S2)

( NV ) A = 27 ⋅10

S

( NV ) A = 8 ⋅10

−6

−6

U  = U  A + U B = 80 A+

A

S  const  ∂

= 3U  A + 2(U  − U  A )

1 / 3

1 / 3

[3U 

+ 2(U  − U  A )

Solving for UA

U  A =

0=

∂U  A

1 / 3  A

1 / 3

B

- need to maximize maximize with with respect respect to UA

]

0 = U  A− 2 / 3 − U 

1 + (2 / 3)

3 / 2

= 51.8[ J ]

2 3

−

(U  − U A ) 2 / 3

Problem 2.3-4. Restrictions due to the Postulates

n

m

r 

S  =  AU  V   N 

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:

λ S  =  A(λ U ) (λ V ) (λ  N ) = λ n + m + r  AU nV m N r  n

λ  = λ n+ m+ r 

m

r 

n + m + r  = 1

Problem 2.3-4. Restrictions due to the Postulates S n

m

r 

S  =  AU  V   N 

S must be increasing with U, so n>0 U

∂S 

00

Problem 2.6-3. Energy in Equilibrium (1)

U 

( 2)

+ U 

(1)

3

= 2.5 × 10 [ J ]

 N 

1

The equations of state of the two systems are:

(1)

( 2)

=2

 N 

3

 N 

=

2

T 

 R

(1)

=3

U 

(1)

1

(1)

( 2)

T 

U 

=

5 2

=? ( 2)

 R

 N 

( 2)

U 

Since the two systems are in thermal equilibrium:

(1)

T 

(1)

U 

( 2)

= T 

3  N (1) (1)

2 U  ( 2)

+ U 

3

= 2.5 ×10 [ J ]

=

5  N ( 2)

( 2)

U 

( 2)

2 U 

(1)

U 

+

15 6

(1)

U 

=

15 6

(1)

U 

3

= 2.5 × 10 [ J ]

(1)

U 

= 714[ J ]

Problem 2.7-3. The Indeterminate Problem of Thermodynamics a)

Show that

P

(1)

=P

( 2) (1, 2 )

= −P

Since the piston is adiabatic, δ Q (1,2) = 0 and thus dU 

(1, 2 )

dV 

(1, 2 )

Since the composite system is isolated dU  ≡ dU (1) + dU ( 2 ) = 0 (1)

− P dV 

(1)

−P

( 2)

dV 

( 2)

=0

Since the total volume of the composite system is constant dV  (1)

,

( 2)

(1)

= − dV 

( 2)

.

b) Show that thermodynamics does not tell us what the temperatures of the two systems are

The entropy maximum condition requires: (1)

(1)

dS  ≡ dS  (1)

dU 

+ dS 

(1)

+ P dV  (1)

T 

( 2)

=

(1)

+

dU 

(1)

P

+

(1) (1)

( 2)

dV 

(1)

T  T  ( 2) ( 2) ( 2) dU  + P dV  ( 2)

T 

+

dU 

( 2)

T 

+

( 2)

P

( 2)

T 

dV 

( 2)

=0

=0

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 maximum condition.

Problem 2.8-1. Semi-permeable partition

N1(1), N2(1)

N1(2), N2(2)

N1 can freely flow through the membrane while N2 cannot.

N1 (1)

Two different gases N1 and N2

( 2)

althou h one of the as com onents can freel penetrate the membrane

P is intensive parameter conjugate to volume, volume is constrained Equilibrium conditions: 

 µ 1(1) (1)

T 

(1)

T 

=

 µ 1( 2 ) ( 2)

Conservation conditions:  (1)

( 2)

 N 1 +  N 1

T 

( 2)

= T 

(1)

U 

( 2)

+ U 

(1)

( 2)

=  N 1 +  N 1

(

(1)

= U 

( 2)

+ U 

initial

)

initial

Problem 2.8-1. Semi-permeable partition

 U 3 / 2V    N 1   N 2        −  N 1 R ln  −  N 2 R ln  S  =  NA +  NR ln 5 / 2    N      N       N    1 T 

=

∂S  ∂U 

=

3 NR 2U 

U  =

3 NRT  2

Solve for T

 N  =  N 1 + N 2

(1)

T  (1)

U 

( 2)

+ U 

( 2)

= T 

= T 

(1)

= U 

T  = 272.73[ K ] 1

T 

=

∂S  ∂ N 1

= ...

Here, when differentiating do not forget that N=N1+N2

After some algebra:

3 / 2     U  V    =  A −  R +  R ln 3 / 2   T  2   N 1 ( N 1 +  N 2 )  

 µ 1

5

( 2)

+ U 

initial

Problem 2.8-1. Semi-permeable partition

(1)

( 2)

 N 1 +  N 1

Solve for N1

(

(1)

( 2)

=  N 1 +  N 1

)

initial

(1)

(1) 1

 µ 

=

µ 

T 

T 

To find P:

∂S  ∂V 

Similarly:

=

P

=

T 

 NR V 

P

(1)

= 680[kPa]

P

( 2)

= 567[kPa]

( 2)

 N 1 = N 1

( 2) 1

P

(1)

 N  ( =

(1) 1

(1)

)

+  N 2  RT 

V 

(1)

= 0.75

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;

@ D

T S_  := U0

@ D

P S_  := U0

S  V 

J2

S2  V 2

S

+

 Nu

N ExpB

S  Nu

F;

S B  Nu F;

Exp

@ @ D @ D<

< @D

<

Param etricPlot  T S ,P S , S,0,0. 8 ,Fram e -> True,TextStyle ->   FontFam ily -> "Tim es",FontW eight -> "Bold",FontSize -> 14 ,

@ D<

D

Fram eLabel->   Tem perature"" K ,Pressure" " Pa ,PlotRange -> Al l

14000 12000 10000             D

     a       P             @

     e      r     u      s      s      e      r       P

8000 6000 4000 2000 0 0

Ü

Graphics

 Ü

100

200 300 Temperature K

 @ D

400

500

Problem 3.3-1. Fundamental relation from the equations of state

T  =

Equations of state:

3 As 2

P=

v

 As

3

v2

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:

−

−

d  µ  =

d  µ  =

 As v

2

3

dv −

3 As 3 v

2

3 As v

dv −

2

ds

6 As 2 v

ds −

2 As 3 v

2

dv +

  As 3    d  µ  = − d    v  

 3 As 2  

  As 3  

  v  

  v  

3 As 2 v

2

ds

 µ  = −

 As 3 v

+ const 

Problem 3.3-1. Fundamental relation from the equations of state Molar form of the Euler equation of state in the energy representation is:

u = Ts − Pv + Substituting the equations of state into this equation:

u=

3 As 2 v

s−

 As 3 v

2

v−

 As 3

u=

v

+ const  =

 As 3 v

+ const 

 As 3 v

+ const 

Problem 3.3-1. Fundamental relation from the equations of state Now let us directly integrate the molar form of the equations of state:

du = Tds − Pdv

du =

3 As 2 v

ds −

  As 3    dv = d  2 v   v  

 As 3

du =

3 As 2 v

ds −

 As 3 v

2

u=

dv

 As v

3

+ const 

Problem 3.5-1. Fundamental relation from the equations of state P=

c) Equations of state:

u c + buv

T  =

v a + buv

u a + buv

Equations of state are functions of u and v, so convenient to work in the entropy representation in molar form:

=

  ∂s    ∂u  v 1

where

T 

=

 ∂s   ∂v u a u

=

1

P

T 

T 

P

+ bv

=

T 

c v

+ bu

Are these equations of state compatible?

  ∂ 2 s     ∂ 2 s     =    ∂u∂v   ∂v∂u 

 P   T     

∂

∂u

  1    T     

∂ =

∂v

b=b

Yes

Problem 3.5-1. Fundamental relation from the equations of state

ds =

ds =

a u

1 T 

du +

P T 

 a    c   + bv du +  + bu dv  u    v  

ds = 

dv

du + bv(du ) +

c v

dv + bu (dv ) = ad  ln (u ) + bv (du ) + cd ln (v ) + bu (dv )

vdu = d (uv ) - This is wrong because v is a variable, not a constant (same problem in 3.3-1)

ds = ad ln(u) + bv(du) + cd ln(v) + bu(dv) = ad ln(u) + cd ln(v) + b(u(dv) + v(du)) = ad ln(u) + cd ln(v) + bd (uv)

s = a ln (u ) + c ln (v ) + buv   + const 

Problem 3.5-6. VdW and ideal gas in equilibrium  In this problem, temperature, pressure and molar volumes of the van der Waals fluid and the ideal gas are the same in equilibrium

VdW:

P=

 RT 

−

v −b

a v

Ideal:

2

P=

 RT  v

v=

 RT  P

Substituting this molar volume into the equation of state for VdW:

 RT  P=  RT  P

−b

−

aP 2

( RT )2

After simple algebra:

Here P is the only unknown, so need to solve for P

 1  RT   7 −  = 3.5 ⋅10 [Pa ]  b a  

P =  RT 

Problem 3.6-1. Microwave Background Radiation  Fundamental relation Equation of state

S  =

4 3

b1 / 4U 3 / 4V 1 / 4

Substituting U into expression for S:

Isentropic process: S initial=Sfinal=S

4 3

bV iT i 3 =

4

4

U  = bVT 

4 bV  f  T  f 3 = 2 bV iT  f 3 3 3

S  =

4 3

bVT 3

2V i = V  f  T i 3 = 2T  f 3

T  f  =

1 3

2

T i

Note that energy of the background radiation is not conserved, it constantly decreases! Where does it go?

Problem 3.7-2. Rubber Band  Equation of state of the rubber band is:

U  = cL0T 

Since T is constant, U is also constant:

dT  = 0

From the conservation of energy:

dU  = W  + Q = 0

dU  = 0

Q = − W  Calculating work from the second equation of state:

δ W  = bT 

 L − L0  L1 − L0

We obtain:

δ Q = −bT 

dL

 L − L0  L1 − L0

dL

Γ  = bT 

 L − L0  L1 − L0

Problem 3.8-1. Paramagnet  The fundamental relation of a simple paramagnetic system is:

  S   I 2   U  =  NRT 0 exp + 2 2    NR  N   I 0   Equations of state:

T  =

  S  I 2    = T  ex  + ∂S   I , N    NR  N  I 0  

 ∂U  

2   S  2 IRT 0  I     ∂U    Be =  exp + 2 2  = 2    ∂ I    S , N   NI 0   NR  N   I 0  

    S  S   I 2   2 I 2    ∂U    µ  =  − 2 2  exp + 2 2  =  RT 0 1 −     NR  N   I   NR  N   I   ∂ N   S , I  0   0      

Problem 3.8-1. Paramagnet  The Euler equation:

U  = TS  +  BeV  +  N 

Substituting the equations of state into the Euler equation:

  S  U  =  RT 0 exp

+

 I 2    S 



0

+

2 I 2

+1−

0

2   S   I    U  =  RT 0 exp + 2 2   NR  N   I  0    

This coincides with the postulated fundamental relation

S 

−

2 I 2  



0

Problem 3.9-6. Model of a solid insulator 

Fundamental relation:

   

2

u =  As 4 / 3 exp b(v − v0 ) +

s    3 R 

a) Show that at s→0, T→0

T  =

s   4 1 / 3 s 4 / 3     2  =  A exp b(v − v0 ) +  s + 3 R  3 3 R   ∂s  

∂u

Clearly T→0 as s→0 independent on the value of v

Problem 3.9-6. Model of a solid insulator  b) Show that cv ~ T3 at T→0

Let us calculate T/u:

T 

  4

=

+

1  



s     2 u =  As exp b(v − v0 ) +  3 R    s   4 1 / 3 s 4 / 3     2  T  =  A exp b(v − v0 ) +  s + 3 R  3 3 R     1     4 T  = u  +  4 / 3

Since s→0 at T→0, we can simplify the expression for T at low temperature:

u~s

On the other hand, from the fundamental relation at low T, we obtain: Substituting into the expression for T:

Since

T  ~

u s

T  ~ s 4

u ~ Ts ~ T 

1 / 3

T  ~

u

4 / 3

3

s ~ T 

4   ∂u   ∂T  cv =  ~ T 3  ~  ∂T   v ∂T 

This is valid for real insulators – heat capacity capacity vanishes as T 3 at low temperatures

s

Problem 3.9-6. Model of a solid insulator  c) Show that cv ~ R at T→∞ tr ivial, s is a monotonically increasing function of u and T so s→∞, T→∞ (this is not trivial, related to the fact that second derivative of u with respect to s is positive)

1   u   4 T  = u  + ≈  3s 3 R  3 R

u ≈ 3 RT 

  ∂u   cv =    ∂T   v

≈

∂ (3 RT ) ∂T 

≈ 3 R

d) Show that α →0 as P→0

s    ∂u    2 4 / 3 ( ) u  As exp b v v = − +    α  =  0  3 R   ∂v  S    v  ∂T   P P s     2 4 / 3 v = v0 − P = −2 Ab(v − v0 ) As exp b(v − v0 ) +  = −2b(v − v0 )u 2bu 3 R    P   ∂u     ∂v   when P→0   =  →0 2   ∂T   P 2bu  ∂T   p

1   ∂v  

P = −

Numerica Num ericall Problem Problem – Heli Helium um and and Water  Water  Using the known form of the fundamental relation for the VdW fluid, we can plot entropy of the composite system as: R  =   8.3144;V 3144;V0 0 = 0.03;U 03;U0 0 = 15000; ah = 0.00346;bh 00346;bh = 23.7 10-6;ch = 1.5;H* helium  aw

=

0.544;bw 544;bw

=

30.5 10

6;cw

-

=

L

*

3.1;H* w ater*L

Plot3DAR Log LogAH Vh- bh  bhL HUh + ahê VhLchE + R Log Log@H V0 - Vh - bw  bw L HU0- Uh+ aw êH V0 - VhLLcw D, Vh,0.01V0,0. 01 V0,0.99V0 99 V0< , Uh,0.01U0,0. 01U0,0.99U0 99U0< ,Pl ,PlotRange ange -> AllE

250

200 10000

0.01

5000

0.02

Ü

Surf aceG eGraph phics

Ü

Numerica Num ericall Problem Problem – Heli Helium um and and Water  Water  Same plot in the contour format:

Cont ontour ourPlotAR Log LogAH Vh- bh  bhL HUh + ah ê VhLchE + R Log Log@H V0 - Vh - bw  bw L HU0- Uh+ aw êH V0 - VhLLcw D, Vh,0. Vh,0.01V0,0. 01 V0,0.99V0 99 V0 AllE

14000

12000

10000

8000

6000

4000

2000

0 0

Ü

0.005

C ont ontour ourG raphi aphics

Ü

0.01

0.015

0.02

0.025

0.03

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 = Fi ndMinim um  A-R LogAH Vh - bhLHUh + ahê VhLchE - R Log@H V0-  Vh- bw L HU0- Uh+ aw ê H V0 - VhLLcw D,8 Vh,0.01 V0,0.99 V0 0

~ 27 T  < 32

Therefore, vdW model fails even qualitatively for

T  <

27 8a 32 27bR

T  <

a

4bR

T  <

a

4bR

Problem 9.7-1 Phase diagram of a solution 

Upper boundary

T  = T 0 − (T 0 − T 1 ) x A

Lower boundary

T  = T 0 − (T 0 − T 1 ) x A (2 − x A )

Initially:

gas

2

 x A0 =

 N  A  N  A + N  B

=

T

C

1

D

2 S

The temperature where boiling first occurs: 0

xA

1   1  3 T  B = T 0 − (T 0 − T 1 )  2 −  = T 0 − (T 0 − T 1 ) 2   2  4 The molar fraction of A in the vapor is given by:

T 0 − (T 0 − T 1 ) x A = T  B = T 0 − 2

3 4

(T 0 − T 1 )

2

 x A =

3 4

 x A = 0.866

1

Problem 9.7-3 Binary liquid composition  T

Mole fraction of A in the binary liquid is  xl A Mole faction of A in the gas phase is  xg A g

T*

g  A

 x =

G

L

l

 N  A

l  A

 x =

g

Where:

 N  A l

l

 N  - mole number of liquid A+B g  N  - mole number of gas A+B

xA  x Al

 x A0

l  A l  B

-

 N  - mole number of liquid B l

g

 N  + N  =  N - total mole number

 x Ag

Mole fractions of liquid and gas A are related via the phase diagram on the left: l  A

g  A

g  A

 x =

T 0 − Dx = T 0 − Cx Therefore:

 N  Ag g

 N 

g  A

= x =

 D

l  A

 D C 

 D N  Al

 x = C  C 

l

l

x A (1)

Problem 9.7-3 Binary liquid composition  0

From the initial condition of all material being liquid with mole fraction of A:  x A and taking into account that the total mole fraction of A does not change during boiling:

0  A

 x =

 N  Al + N  Ag

From (1):

(2)

 N  g  A

=

l  A l

C  N 

l

 N − N 

Substituting in (2): l

l  A

 N  + 0  A

 x =

 D  N  A l

C  N   N 

( N − N  ) l

   D   1     N  Al  N l    D ( N − N l )  l 1 +  =  x A f 1 +  − 1  = l l  N   N    C   N      C    f    

Where we have introduced the liquid fraction of material:

 f  =

 N l  N 

Problem 9.7-3 Binary liquid composition 

   D   1    0 l  x A =  x A f 1 +  − 1   C   f         

l  A

 x =

 x A0

   D     f  + (1 −  f ) C     

Since the fraction of material remaining in liquid is  f=1/2 and  D = 3C 

 x A =

 x

0

 1 3C   1    + 1 −    2 C    2  

=

 x

0

2

There is also a one-line geometric solution to this problem. What is it?