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Problem 2. Van der Waals equation of state
Solution
Part А. Non-ideal gas equation of state
A1. If 𝑉 = 𝑏 is substituted into the equation of state, then the gas pressure turns infinite. It is obvious that
this is the moment when all the molecules are tightly packed. Therefore, the parameter 𝑏 is approximately
equal to the volume of all molecules, i.e.
𝑏 = 𝑁𝐴 𝑑3
(A1.1)
A2. In the most general case the van der Waals equation of state can be rewritten as
𝑃𝑐 𝑉3 − (𝑅𝑇𝑐 + 𝑏𝑃𝑐 )𝑉2 + 𝑎𝑉 − 𝑎𝑏 = 0
(A2.1).
Since at the critical values of the gas parameters the straight line disappears, then, the solution of
(A2.1) must have one real triple root, i.e. it can be rewritten as follows
𝑃𝑐 (𝑉 − 𝑉𝑐 )3 = 0
(A2.2).
Comparing the coefficients of expression (A2.1) and (A2.2), the following set of equations is
obtained
3𝑃𝑐 𝑉𝑐 = 𝑅𝑇𝑐 + 𝑏𝑃𝑐
3𝑃𝑐 𝑉3𝑐 = 𝑎
(A2.3).
{
3
𝑃𝑐 𝑉𝑐 = 𝑎𝑏
Solution to the set (A2.3) is the following formulas for the van der Waals coefficients
27𝑅 2 𝑇𝑐2
𝑎=
(A2.4),
64𝑃𝑐
𝑅𝑇𝑐
𝑏 = 8𝑃
(A2.5).
𝑐
Alternative solution
The critical parameters are achieved in the presence of an inflection point in the isotherm, at which
the first and second derivatives are both zero. Therefore, they are defined by the following conditions
𝑑𝑃
( ) =0
(A2.6),
𝑑𝑉
𝑇
and
𝑑2 𝑃
(
) =0
(A2.7).
𝑑𝑉2 𝑇
Thus, the following set of equations is obtained
𝑅𝑇
2𝑎
− (𝑉 −𝑏𝑐 )2 + 3 = 0
𝑐
2𝑅𝑇𝑐
(𝑉𝑐 −𝑏)3
−
𝑉𝑐
6𝑎
𝑉4𝑐
=0
(A2.8),
𝑎
(𝑃 + ) (𝑉𝑐 − 𝑏) = 𝑅𝑇𝑐
{ 𝑐 𝑉2𝑐
which has the same solution (A2.4) and (A2.5).
A3. Numerical calculations for water produce the following result
𝑎𝑤 = 0.56
m6 ∙Pa
(A3.1).
mole2
m3
3.1 ∙ 10−5 mole
𝑏𝑤 =
A4. From equations (A1.4) and (A3.2) it is found that
(A3.2).
𝑏
𝑑𝑤 = √𝑁 = 3.7 ∙ 10−10 m ≈ 4 ∙ 10−10 m
3
𝐴
(A4.1).
Part B. Properties of gas and liquid
B1. Using the inequality 𝑉𝐺 ≫ 𝑏, the van der Waals equation of state can be written as
(𝑝0 +
𝑎
𝑉2𝐺
) 𝑉𝐺 = 𝑅𝑇
(B1.1),
which has the following solutions
𝑅𝑇
4𝑎𝑝0
0
𝑅2 𝑇2
𝑉𝐺 = 2𝑝 (1 ± √1 −
)
(B1.2).
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Smaller root in (B1.2) gives the volume in an unstable state on the rising branch of the van der Waals
isotherm. The volume of gas is given by the larger root, since at 𝑎 = 0 an expression for the volume of an
ideal gas should be obtained, i.e.
𝑅𝑇
4𝑎𝑝0
0
𝑅2 𝑇2
𝑉𝐺 = 2𝑝 (1 + √1 −
)
For given values of the parameters the value
𝑎𝑝0
(𝑅𝑇)2
(B1.3).
𝑎𝑝0
(𝑅𝑇)2
= 5.8 ∙ 10−3 . It can therefore be assumed that
≪ 1, then (B1.3) takes the form
𝑉𝐺 ≈
𝑎𝑝
𝑅𝑇
(1 − 2 02 )
𝑝0
𝑅 𝑇
=
𝑅𝑇
𝑎
− 𝑅𝑇
𝑝0
(B1.4).
B2. For an ideal gas
𝑅𝑇
𝑝0
𝑉𝐺0 =
(B2.1),
hence,
∆𝑉
( 𝑉 𝐺) =
𝐺0
𝑉𝐺0 −𝑉𝐺
𝑉𝐺0
1
= 2 ( 1 − √1 −
4𝑎𝑝0
2 2)
𝑅 𝑇
≈
𝑎𝑝0
𝑅2 𝑇2
= 0.58%.
B3. Mechanical stability of a thermodynamic system is in power provided that
𝑑𝑃
( ) < 0.
𝑑𝑉
𝑇
(B2.2)
(B3.1)
The minimum volume, in which the matter can still exist in the gaseous state, corresponds to a point
in which
𝑑𝑃
𝑉𝐺𝑚𝑖𝑛 → (𝑑𝑉) = 0
(B3.2).
𝑇
Using the van der Waals equation of state (B3.2) is written as
𝑑𝑃
𝑅𝑇
2𝑎
( ) =−
2+ 3 = 0
𝑑𝑉
(𝑉−𝑏)
𝑇
𝑉
From (B3.2) and (B3.3), and with the help of 𝑉𝐺𝑚𝑖𝑛 ≫ 𝑏, it is found that
2𝑎
𝑉𝐺𝑚𝑖𝑛 = 𝑅𝑇
Thus,
𝑉𝐺
𝑉𝐺𝑚𝑖𝑛
2
=
𝑅2 𝑇 2
2𝑎𝑝0
= 86
(B3.3).
(B3.4).
(B3.5).
B4. Using the inequality 𝑃 ≪ 𝑎/𝑉 , the van der Waals equation of state is written as
𝑎
2 (𝑉𝐿 − 𝑏) = 𝑅𝑇,
𝑉𝐿
(B4.1)
whose solution is
𝑎
𝑉𝐿 = 2𝑅𝑇 (1 ± √1 −
4𝑏𝑅𝑇
)
𝑎
(B4.2).
In this case, the smaller root should be taken, since at 𝑇 → 0 the liquid volume 𝑉𝐿 = 𝑏 must be
obtained according to (B4.1), i.e.
𝑎
𝑉𝐿 = 2𝑅𝑇 (1 − √1 −
4𝑏𝑅𝑇
)
𝑎
≈ 𝑏 (1 +
𝑏𝑅𝑇
).
𝑎
(B4.3).
B5. Since (B4.3) gives the volume of the one mole of water its mass density is easily found as
𝜇
𝜇
𝜇
2 kg
𝜌𝐿 = 𝑉 =
(B5.1).
𝑏𝑅𝑇 ≈ 𝑏 = 5.8 ∙ 10 m3
𝐿
𝑏(1+ 𝑎 )
B6. In accordance with (B4.3) the volume thermal expansion coefficient is derived as
1 ∆𝑉
𝑏𝑅
𝑏𝑅
𝛼 = 𝑉 ∆𝑇𝐿 = 𝑎+𝑏𝑅𝑇 ≈ 𝑎 = 4.6 ∙ 10−4 К−1
𝐿
(B6.1).
B7. The heat, required to convert the liquid to gas, is used to overcome the intermolecular forces that create
negative pressure 𝑎/𝑉 2 , therefore,
𝑉 𝑎
1
1
𝐸 = 𝐿𝜇 ≈ ∫𝑉 𝐺 𝑉 2 𝑑𝑉 = 𝑎 (𝑉 − 𝑉 )
(B7.1),
𝐿
and using 𝑉𝐺 ≫ 𝑉𝐿 , (B7.1) yields
𝑎
𝐿 = 𝜇𝑉 =
𝐿
𝐿
𝑎
𝑏𝑅𝑇
𝜇𝑏(1+
)
𝑎
𝑎
𝐺
J
≈ 𝜇𝑏 = 1.0 ∙ 106 kg
(B7.2).
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B8. Consider some water of volume 𝑉. To make a monolayer of thickness 𝑑 out of it, the following work
must be done
𝐴 = 2𝜎𝑆
(B8.1).
Fabrication of the monomolecular layer may be interpreted as the evaporation of an equivalent
volume of water which requires the following amount of heat
𝑄 = 𝐿𝑚
(B8.2),
where the mass is given by
𝑚 = 𝜌𝑆𝑑
(B8.3).
Using (A4.1a), (B5.1) and (B7.2), one finally gets
𝑎
N
𝜎 = 2𝑏2 𝑑𝑤 = 0.12 ∙ 10−2 m
(B8.4).
Part С. Liquid-gas systems
C1. It follows from the Maxwell rule that
𝑉
∫𝑉 𝐺 𝑃(𝑉)𝑑𝑉 = 𝑝0 (𝑉𝐺 − 𝑉𝐿 )
𝐿
Evaluation of the integral leads to the following equation for determining 𝑝0
𝑉
𝑅𝑇
∫𝑉 𝐺 (𝑉−𝑏 −
𝐿
𝑎
𝑉 −𝑏
𝑉2
1
1
) 𝑑𝑉 = 𝑅𝑇 ln ( 𝑉𝐺−𝑏) + 𝑎 (𝑉 − 𝑉 ) = 𝑝0 (𝑉𝐺 − 𝑉𝐿 )
𝐿
𝐺
𝐿
(C1.1).
(C1.2).
Using the inequalities 𝑏, 𝑉𝐿 ≪ 𝑉𝐺 and formulas (B2.1) and (B4.3), this equation is simplified to
𝑅𝑇 𝑎
𝑎
𝑅𝑇 ln (
) − = 𝑅𝑇
(C1.3).
𝑝 𝑏2 𝑅𝑇
𝑏
0
(C1.3) is simply rearranged as
ln 𝑝0 = ln
𝑎
𝑏2
𝑎
− 𝑏𝑅𝑇 − 1
(C1.4).
Consequently, expressions for 𝐴 and 𝐵 are obtained as follows
𝑎
𝐴 = ln (𝑏2 ) − 1
(C1.5),
𝑎
𝐵 = − 𝑏𝑅
C2. Collecting formulas (C1.4), (C1.5) and (C1.6), the vapor pressure is found as
𝑎
𝑝0 = 2
= 6.2 ∙ 105 Pa
𝑎
𝑏 exp(𝑏𝑅𝑇+1)
(C1.6).
(C2.1).
C3. At equilibrium, the pressure in the liquid and gas should be equal at all depths. The pressure 𝑝 in the
fluid at the depth ℎ is related to the pressure of saturated vapor above the flat surface by
𝑝 = 𝑝0 + 𝜌𝐿 𝑔ℎ
(C3.1).
The surface tension creates additional pressure defined by the Laplace formula as
2𝜎
∆𝑝𝐿 = 𝑟
(C3.2).
The same pressure 𝑝 in the fluid at the depth ℎ depends on the vapor pressure 𝑝ℎ over the curved
liquid surface and its radius of curvature as
2𝜎
𝑝 = 𝑝ℎ + 𝑟
(C3.3).
Furthermore, the vapor pressure at different heights are related by
𝑝ℎ = 𝑝0 + 𝜌𝑆 𝑔ℎ
(C3.4).
Solving (C3.1)-( C3.4), it is found that
2𝜎
ℎ = (𝜌 −𝜌 )𝑔𝑟
(C3.5).
𝐿
𝑆
Hence, the pressure difference sought is obtained as
2𝜎 𝜌𝑆
2𝜎 𝜌
∆𝑝𝑇 = 𝑝ℎ − 𝑝0 = 𝜌𝑆 𝑔ℎ = 𝑟 𝜌 −𝜌
≈ 𝑟 𝜌𝑆 .
𝐿
𝑆
𝐿
(C3.6).
Note that the vapor pressure over the convex surface of the liquid is larger than the pressure above
the flat surface.
C4. Let 𝑃𝑒 be vapor pressure at a temperature 𝑇𝑒, and 𝑃𝑒 − ∆𝑃𝑒 be vapor pressure at a temperature 𝑇𝑒 −
∆𝑇𝑒. In accordance with B4, when the ambient temperature falls by an amount of ∆𝑇𝑒 the saturated vapor
pressure changes by an amount
𝑎
∆𝑃𝑒 = 𝑃𝑒 𝑏𝑅𝑇 2 ∆𝑇𝑒
(C4.1).
𝑒
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In accordance with the Thomson formula obtained in part C3, the pressure of saturated vapor above
the droplet increases by the amount of ∆𝑝𝑇 . While a droplet is small in size, the vapor above its surface
remains unsaturated. When a droplet has grown up to a certain minimum size, the vapor above its surface
turns saturated.
Since the pressure remains unchanged, the following condition must hold
𝑃𝑒 − ∆𝑃𝑒 + ∆𝑝𝑇 = 𝑃𝑒
(C4.2).
Assuming the vapor is almost ideal gas, its density can be found as
𝜇𝑃
𝜌𝑆 = 𝑅𝑇𝑒 ≪ 𝜌𝐿
(C4.3).
𝑒
From equations (C4.1)-( C4.3), (B5.1) and (C3.6) one finds
2𝜎 𝜇𝑃𝑒
𝑎∆𝑇
= 𝑃𝑒 2𝑒
𝑟 𝑅𝑇
𝑏𝑅𝑇𝑒
𝑒
(C4.4).
Thus, it is finally obtained that
𝑟=
2𝜎𝑏 2 𝑇𝑒
𝑎∆𝑇𝑒
= 1.5 ∙ 10−8 m
(C4.5).