blundell and blundell clausius-clapeyron eq.bingweb.binghamton.edu/~suzuki/thermostatfiles/16.10 pt...
TRANSCRIPT
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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