Problem and solution

Blundell and Blundell

Concept of Thermal Physics

Chapter 28 Phase transitions

Masatsugu Sei Suzuki, SUNY at Binghamton

(Date: November 10, 2019)

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28-1

((Solution))

410101.1 s kg/m3, 410065.1 l kg/m3,

41090827.01 s

sV m3/kg, 410938971.0

1 l

lV m3/kg,

5.24fL kJ/kg (latent heat of fusion for lead)

600273327 MT K

7

4

3

1033.1100307.0

105.24

600

11

sl

f

sl

sl

VV

L

TVV

SS

dT

dP Pa/K =131.3 atm/K

where

1 atm = 1.01315 x 105 Pa.

When 99P atm,

75.03.131

99

3.131

dPT C

which means that the melting temperature increases as

75.600MT K = 327.75 C

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28-2

((Solution))

L: latent heat of vaporization

lg VVV

Thus we have the Clausius-Clapeyron equation,

VT

L

dT

dP

Two approximations:

(i) lg VV

glg VVVN

(ii)

nRTPVg (ideal gas law)

Using these two approximations, we have

22RT

PL

nRT

LP

TV

L

VT

L

dT

dP v

g

where vL is the latent heat of vaporization per mol. Suppose that vL is independent of T.

2T

dT

R

L

P

dP v

or

RT

LP vln +constant

or

)exp()( 0RT

LPTP v

or

RT

LPP v 0lnln

At P = 1 atm = 1013 mbar, T = 373 K. At P = 615 mbar, what is the temperature T?

373

1ln)1013ln( 0

R

LP v

)273(

1ln)615ln( 0

xR

LP v

Then we have

)273

1

373

1()

615

1013ln(

xR

Lv

with

R = 8.3144598 J/(mol K)

vL 2260 kJ/kg = (2260 x 18) J/mol = 40.680 kJ/mol

So we have the boiling temperature as

T = 359.329 K = 86.3 C.

((Another approach))

3

3

1 1( )

2257 10 /

(373K)(1.672m /kg)

3620 Pa/K

0.0357atm/K

v

steam water

LdP

dTT

J kg

When the pressure is 615 mbar, the corresponding pressure is

607.025.1013

615P atm

The phase of change is

393.01607.0 P atm

The change of the boiling temperature is

8911100' BT C.

since

11K0.0357atm/

393.0

K0.0357atm/

atmPT

((Note))

The specific volumes of water and steam at 100 C and 1 atm are

310043.11 water

m3/kg

672.11

steam

m3/kg

Latent heat of vaporization:

2257vL kJ/ kg

One atmosphere

1 atm=1.01325 x 105 Pa

Triple point (TP) of water:

REFERENCE

A. Rex, Finn’s Thermal Physics, CRC Press (2017)

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28-3

((Solution))

(i) Clapeyron-Clausius equation for fusion

KT 273

31000.1 lV m3/kg, 31009.1 sV m3/kg,

7

3104.1

1009.0273

11

f

sl

f

sl

slL

VV

L

TVV

SS

dT

dP Pa/K

4104.109.0273 fL Pa = 343.98 x 103 J/kg

We consider the equation of thermal diffusion: dx

dTA

dt

dQ

(ii) Thermal diffusion equation

x

TA

dt

dxAL

dt

dQiceicef

or

6

23510364.0

10109174.0104398.3

5.03.2

x

T

Ldt

dx

icef

ice

m/s

or

14.3dt

dx cm/day

where

x = x0 = 1.0 cm at t = 0

K 5.0T

31009.1

1ice = 0.9174 x 103 kg/m3

Lf = 3.4398 x 105 J/kg

ice= 2.3 W/(m K).

(iii) Thermal equilibrium

water= 0.56 W/(m K).

In thermal equilibrium (heat flux-1 is equal to heat flux-2)

xxwaterice

1

25.0

51.0x m

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28.5

((Solution))

L: latent heat of vaporization/mole

lg vvv

Thus we have the Clausius-Clapeyron equation,

vldP

dT T v

Two approximations:

(i) lg vv

g

g

glgN

Vvvvv

(ii)

TkNPV Bgg (ideal gas law)

or

TkPv Bg

Using these two approximations, we have

2 2 2

v v v A v v

g B A B B

l l l P N l P L PdP

dT T v Tv k T N k T R T

where vL is the heat of vaporization per mol,

v A vL N l .

Suppose that vL is independent of T.

2T

dT

k

L

P

dP

B

v

or

Tk

LP

B

vln +constant

or

)exp()( 0Tk

LPTP

B

v

or

)exp()( 0RT

LPTP v .

or

RT

LPP v 0lnln

where vL is the latent heat per mol.

The least-squares fit of the data to a straight line yields

6.170.5302 R

Lv

where R = 8.3144596 J/(K mol). Then we have

6

6.5

7

7.5

8

8.5

9

9.5

0.003 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037

Clausius-Clapeyron

lnP

1/T

y = m1 + m2 * M0

ErrorValue

0.05929825.845m1

17.584-5302m2

NA0.00027826Chisq

NA0.99998R

3144596.85302vL =4.4083 x 104 J/mol

For water 1kg = 1000/18 mol = 55.5555 mol

64 10449.2104083.418

1000vL J/kg

We consider the latent heat around the tri-critical point (Tc, Pc, Vc) at which the three phases

(sold=ice, gas, and water) meet together.

)( lgcv SSTL , J/kg 10776.2)( 6 sgcs SSTL ; latent heat for sublimation

Then we have the latent heat fL at the tri-critical point,

56 1027.310)449.2776.2()( fslcvs LSSTLL J/kg

We note that

181.667(J/kg)18

1000(J/mol) ff LL (J/mol)

The buoyant force is balanced with the gravitational force,

gVVgV sl )( 211

where the total volume of ice is 21 VVVs , and the volume of ice submerged in the water is 1V .

So we have

4

5

4

111

1

2 V

V

s

l

leading to

31000.1 l kg/m3, 31080.0 s kg/m3,

31000.1 lV m3/kg, 31025.1 sV m3/kg,

6

3

5

1079121.41025.0

1027.3

273

11

sl

f

sl

sl

VV

L

TVV

SS

dT

dP Pa/K

When the temperature drops from 0 C to -2 C, the pressure P increases to

61058242.9 P Pa = 94.58 atm

where

1 atm = 1.01315 x 105 Pa.

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28.6

((Solution))

31000.1 l kg/m3, 310917.0 s kg/m3,

31000.11 l

lV m3/kg, 310091.1 sV m3/kg,

334fL kJ/kg (latent heat of fusion for water)

7

3

5

1034.110091.0

1034.3

273

11

sl

f

sl

sl

VV

L

TVV

SS

dT

dP Pa/K = -132.2 atm/K.

where

1 atm = 1.01315 x 105 Pa.

For a very heavy skater (100 kg), only making the contact with ice over an area 10 cm x 1 mm =

10-4 m2.

410

8.9100

P = 9.8 x 106 Pa = 96.7 atm

Then

Fig. TP (triple point): T = 0 C (273.16 K) and P = 0.00637 atm. The stating point : T = -5 C

and P = 1 atm.

We now suppose that the temperature is -5 C. At 1 atmosphere, the ice is in the state on the P-T

plane. There is no water present. When the ice skater puts pressure on the ice, the state moves

TP

Log P atm

T C 05

1 atm

106.7 atm

661 atm

Ice

Water

dP dT 132.2 atm K

along the constant temperature line ( ). In theory, as soon as the phase boundary is

reached at the state , some ice melts so that the edge of the skate sinks in fractionally, with the

load now spread over a large area stabilizing the pressure. The state thus remains fixed at , with

the liberated water acting as a lubricant.

The question must be asked: Is 106.7 atm or so sufficient to achieve this goal? From

dT

dP -132.2 atm/K.

So a pressure increase of (132.2 atm/K) x 5 K = 661 atm would be required to go to from to . Thus this explanation for the success of the skater appears inadequate.

((Note))

Thermal conductivity:

3.2ice W/(m K)

56.0water W/(m K)

The gas constan:

3145.8R J/(K mol)

Latent heat of fusion:

334fL kJ/kg

Latent heat of vaporization:

2260vL kJ/ kg

The density

31000.1 water kg/m3.

310917.0 ice kg/m3.

3100951.011 waterice

m3/kg

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