Post on 06-Nov-2021
Equations of State
Chapter 2
Advanced Thermodynamics (M2794.007900)
Min Soo Kim
Seoul National University
2/15
2.2 Equation of State of an Ideal Gas
ππ =π
ππ π (2.1)
m: mass of gas
M: molecular weight
R: universal constant
(8.314 Γ 103J
kilomoleβK)
ππ = ππ π (2.2)
β’ Since π β‘π
πis the number of kilomoles of the gas, the equation of state of
an ideal gas is
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2.2 Equation of State of an Ideal Gas
β’ In Equation (2.2) we note that the extensive variable V divided by n, the number
of kilomoles of the gas, is the specific volume π£.
β’ Thus the equation of state can be written ππ£ = π π.
β’ The projections of the surface π π, π£, π = 0 on the π β π£ plane, π β π plane, and
the π£ β π plane are shown in Figure 2.1
Figure 2.1 Diagrams for an ideal gas. (a) the isotherms are equilateral hyperbolae; (b) the isochores
are straight lines; (c) the isobars are also straight lines.
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2.3 Van Der Waalsβ Equation for a Real Gas
π +π
π£2π£ β π = π π (2.3)
a and b: characteristic constants
β’ The term π
π£2arises from the intermolecular forces due to the overlap of electron
clouds.
β’ The constant b takes into account the finite volume occupied by the molecules.
β’ Multiplication of Equation (2.3) by π£2 yields the equation
ππ£3 β ππ + π π π£2 + ππ£ β ππ = 0 (2.4)
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β’ Equation (2.4) is a cubic equation in π£ with three roots, only one of which needs to
be real.
β’ In Figure 2.2 some isotherms calculated from the van der Waals equation have
been drawn.
Figure 2.2 Isotherms for
a Van Der Waalsβ gas.
2.3 Van Der Waalsβ Equation for a Real Gas
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2.4 π· β π β π» Surfaces for Real Substances
β’ Figure 2.3 is a schematic diagram of the π β π£ β π surface for a substance
that contracts on freezing
Figure 2.3 π β π£ β π surface for a substance that contracts on freezing
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2.4 π· β π β π» Surfaces for Real Substances
β’ Notice the regions(solid, liquid, gas or vapor) in which the substance can
exist in a single phase only.
β’ Elsewhere two phases can exist simultaneously in equilibrium, and along
the so-called triple line, all three phases can coexist.
Figure 2.4 π β π diagrams for (a) a substance that contracts on freezing; and
(b) a substances that expands on freezing
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2.5 Expansivity and Compressibility
β’ Suppose that the equation of state of a given substance is written in the form
π£ = π£ π, π (2.5)
β’ Taking the differential, we obtain
ππ£ = (ππ£
ππ)πππ + (
ππ£
ππ)πππ (2.6)
π½ β‘1
π£(ππ£
ππ)π (2.7)
β’ The expansivity, or coefficient of volume expansion, is given by
β’ This is the fractional change of volume resulting from a change in
temperature, at constant pressure.
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2.5 Expansivity and Compressibility
β’ Similarly, the isothermal compressibility is defined as
π β‘ β1
π£(ππ£
ππ)π (2.8)
β’ This is the fractional change in volume as the pressure changes, with
the temperature held constant.
β’ The negative sign is used since the volume always decreases with
increasing pressure (at constant temperature)
β’ For an ideal gas, π£ = π π/π
π½ =1
π£
π
π=
1
π(2.9)
π = β1
π£β
π π
π2=
1
π(2.10)
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2.7 Specific heat
β’ Specific heat: the amount of heat required to change a unit mass of a substance
by one degree in temperature
πΆπ, πΆπ β The properties we can measure
π’ = π’ π, π£ (2.11)
ππ’ =ππ’
ππ)π£ππ +
ππ’
ππ£)πππ£ (2.12)
πΏπ = βπ + πΏπ= βπ + πππ£
πΆπ£
If there is no volume change
(constant v)
= βπ (2.13)
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2.7 Specific heat
π1βπ2 π1βπ2
V constant V constant
Q = π2π
More easy to measure
heat and internal
energy in difference
πΆπ£ =ππ’
ππ)π£ =
ππ
ππ)π£ πΆπ£ can be measured
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2.7 Specific heat
β β‘ π’ + ππ£
= β(π, π) (2.14)
πβ =πβ
ππ)πππ +
πβ
ππ)πππ (2.15)
πΆπ
πΏπ)πΌπππ = πππ = ππ’ + πππ£
= πβ β π£ππ (2.16)
If there is no pressure change
(constant P)
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2.7 Specific heat
π1βπ2
Constant pressure
Amount of heat added = ββ
πΆπ β‘πβ
ππ)π =
ππ
ππ)π πΆπ can be measured
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2.8 Maxwell equation
ππ’ = πππ β πππ£ (2.17)
π’ = π’(π , π£) (2.18)
ππ’ =ππ’
ππ )π£ππ +
ππ’
ππ£)π ππ£ (2.19)
T P
π
ππ£(ππ’
ππ )β =
π
ππ (ππ’
πβ)π
ππ
ππ£)π = β
ππ
ππ )π£ (2.20)
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2.8 Maxwell equation
π½ =π(π, π£)
π(π, π )= 1
P
Wnet
v
T
s
Wnet
Jacobian
ππ₯ππ¦ = πππππ
π₯ = ππππ π
y= ππ πππ
π(π, π )
π(π£, π )=π(π, π£)
π(π£, π )= β
π π, π£
π π , π£= β
ππ
ππ )π£