3. Gibbs Phase Rule
• Had already briefly discussed it without giving the
name
• F = C – P + 2, where C is the number of
components or chemical constituents, P the
p
number of phases, and F the number of degrees of
freedom
• F is the number of intensive variables which may
i h b fi i i bl hi h
be freely chosen, and must be so chosen before
the system is in a determined state
• For C = 1, F = 3 – P, F= 2 if P = 1, 1 if P = 2, and
0 if P = 3
• Does not make any reference to ‘amounts’
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4. T-v liquid-vapor equilibrium - Quality
As one moves from left to right in the saturation region at a given T and
P, the relative amount of vapor grows at the expense of the liquid.
, p g p q
Quality x = mg/m, fraction of total mass present as vapor, varies from 0 to
1 along tie line
At any point on the line, v = (1-x)vf + xvg = vf + xvfg where vf and vg are
g
the specific volumes of the liquid and vapor phases respectively, and vfg
= vg - vf
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5. State of liquid-vapor system
• T and P not independent in two phase equilibrium
regions f pure substance
i for b t
• Only one degree of freedom, and so only one
needs t be specified, d
d to b ifi d does it h
have t b one of
to be f
these?
• Quality x as an ‘intensive’ property to ‘fully fix’
x, intensive fully fix
state of two phase system along with T or p
• State postulate – two independent intensive
variables fix state of pure substance
• Example
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6. Example: Water saturation region
Temp (C) Sat P (kPa) vf (m3/kg) vf g(m3/kg) vg (m3/kg)
95 84.6 0.001040 1.98186
100 101.3 0.001044 1.67185 1.67290
105 120.8 0.001047 1.41936
Three rigid containers of volume 1.0 m3, each contain water at 100ºC,
with respective quality values 0.500, 0.001263, and 0.00010. Find the
mass in each. The contents of each container are heated till they just
become homogeneous. Identify the final state in each case, also
finding the final pressure.
Container Quality v(m3/kg) Remark Final State
1 0.500 0.83697 > vc Sat. vapor
2 0.001263 0.003155 = vc Critical
3 0.000100 0.001211 < vc Sat. liquid
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7. Example (contd.) – final state
Temp (C) Sat P (kPa) vf (m3/kg)
230 2794.9
2794 9 0.001209
0 001209
Container 3 235 3060.1 0.001219
231 2848 0.001211
Temp (C) P (kPa) v (m3/kg)
Container 2
374.1
374 1 22089 0.003155
0 003155
Temp (C) Sat P (kPa) vg(m3/kg)
120 198.5 0.89186
Container 1 125 232.1 0.77059
122.3
122 3 214 0.83697
0 83697
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8. Example (contd.)
• The contents of container 1, now in the saturated vapor
state at 122. C are now heated at constant volume to 150 C.
Find the final pressure
The final state is in the superheated steam region
Temp (C) P (kPa) vg(m3/kg)
150 200 0.93964
150 300 0.63388
150 233.6 0.83697
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9. The Ideal Gas
• PV = nȒT or Pṽ = ȒT
• Ȓ is the universal gas constant 8.3145 kJ kmole-1 K-1
1 1
• n = m/M
• PV = mRT, where R is the gas constant per unit mass of a
g
given g
gas, 0.2968 kJ/kg K for N2, 0.1889 kJ/kg K for CO2 etc
g g
• Obeyed as a limiting law by gases at low density where
inter-molecular forces are negligible
• What constitutes low density, and by how much does a
real gas d i t f
l deviate from id l b h i
ideal behaviour at given T and P?
t i d
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10. Compressibility Factor Z
• Z = Pv/RT
• Z = 1 for ideal gas. Note below that Z → 1 as P → 0
g
• What do Z > 1 and Z < 1 signify?
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