In this problem we will investigate the van der was gas in the vicinity of the critical point. The van der Waals equation is (P+²) (V - Nb) = NkeT You have shown before that at the critical point Sa V₁ =3Nb, kaTc= 276' For the following it is convenient to re-write the van der Waals equation in the so-called reduced form, t = T/T, p= P/P, and v=V/V, so that the equation of state in reduced (dimensionless) variables becomes (check that!) St 3r-1 3 We will study the vicinity of the critical point where t and v are close to 1 and determine three critical exponents: (for the difference in volumes of gas and liquid), (for the isothermal compressibility), and & (for pressure). (a) Expand the reduced van der Waals equation in Taylor series in (-1) keeping terms up
In this problem we will investigate the van der was gas in the vicinity of the critical point. The van der Waals equation is (P+²) (V - Nb) = NkeT You have shown before that at the critical point Sa V₁ =3Nb, kaTc= 276' For the following it is convenient to re-write the van der Waals equation in the so-called reduced form, t = T/T, p= P/P, and v=V/V, so that the equation of state in reduced (dimensionless) variables becomes (check that!) St 3r-1 3 We will study the vicinity of the critical point where t and v are close to 1 and determine three critical exponents: (for the difference in volumes of gas and liquid), (for the isothermal compressibility), and & (for pressure). (a) Expand the reduced van der Waals equation in Taylor series in (-1) keeping terms up
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Transcribed Image Text:1. Critical behavior of the van der Waals gas
In this problem we will investigate the van der Waals gas in the vicinity of the critical point. The van
der Waals equation is
'+
(V - Nb) = NkBT.
You have shown before that at the critical point
θα
02
V 3Nb, kgTc =
Pe=
276'
2762
For the following it is convenient to re-write the van der Waals equation in the so-called reduced form,
t=T/T, p= P/P, and v=V/Ve, so that the equation of state in reduced (dimensionless) variables
becomes (check that!)
p=
St
3-1
3
2
We will study the vicinity of the critical point where t and are close to 1 and determine three critical
exponents: (for the difference in volumes of gas and liquid), (for the isothermal compressibility),
and 6 (for pressure).
(a)
(b)
(c)
(d)
(e)
Expand the reduced van der Waals equation in Taylor series in (-1) keeping terms up
to cubic. Argue that for t sufficiently close to 1 the term quadratic to (-1) becomes negligible
compared to others and may be dropped, so that near the critical point
3
p≈ (4t – 3) – 6(t − 1)(-1)-(-1).
Note: Do not hesitate to use Wolfram Alpha or Mathematica, etc. to do expansion or to take
derivatives. Just mention that in your solutions.
The resulting expression p(v) is antisymmetric about the point 1. Use this fact and
the Maxwell construction (equal-area construction in the case of van der Walls gas) to find an
approximation for the vapor pressure (an equilibrium pressure of liquid-vapor coexistence) as a
function of temperature. Then show that the difference in volumes of the gas and liquid phases
follows the scaling law (V₁– V₁) x (T-T.)³. Determine the critical exponent 3. Note, 3 is indeed
not 1/kgT.
Calculate the isothermal compressibility, K;
1 (av
, and show that it diverges as
xx T-Te for both limits: when T→ Te from below and for TT from above. Determine
the critical exponenty.
Show that at TT the pressure scales with volume as (P – Pc) x (V – V.)², where dis
another critical exponents. Determine 6.
Sketch graphs for (V,- Vi) vs T and x vs T near the critical point.
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