ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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A positive-slope segment in the reduced van der Waals model

Physical scope

An exact derivative calculation in the dimensionless reduced van der Waals model below its model critical temperature; no real-fluid measurement or universal equation of state is asserted.

Example

Adopt the reduced van der Waals equation as a phenomenological model:

p (v, t) = \frac{8 t}{3 v - 1} - \frac{3}{v^{2}}, v > \frac{1}{3}, t > 0,

where $p = P / P_{c}$ , $v = V / V_{c}$ , and $t = T / T_{c}$ use positive model scales $P_{c}, V_{c}, T_{c}$ . Define a model critical point by the simultaneous isothermal conditions $\partial_{v} p = \partial_{v}^{2} p = 0$ in this domain. The equation is a model premise; its extension to liquid-like densities is not a universal physical law.

Facts & Assumptions

Given: The reduced model and domain in the Example, with $P_{c}, V_{c}, T_{c} > 0$ .

[A1]

The isothermal local mechanical-stability criterion for this model is not needed for the calculation below. The claimed result is only the sign of the model derivative.

[F1]

Phase branches and coexistence domains: A phase branch and its chart are model-specified; a branch equation is not asserted to be universal.

Verification

1.1givenalgebra

At fixed $t$ , differentiating the rational equation gives $\frac{\partial p}{\partial v} = - \frac{24 t}{(3 v - 1)^{2}} + \frac{6}{v^{3}} .$ At $t = 0.9$ and $v = 1$ , this derivative is $- 24 (0.9) / 4 + 6 = 0.6 > 0$ .

2.1givenstep 1.1algebra

The second derivative is $\partial_{v}^{2} p = 144 t / (3 v - 1)^{3} - 18 / v^{4}$ . At a simultaneous zero, the first derivative gives $t = (3 v - 1)^{2} / (4 v^{3})$ and the second gives $t = (3 v - 1)^{3} / (8 v^{4})$ . Equality and $v > 1 / 3$ imply $2 v = 3 v - 1$ , hence $v = 1$ , $t = 1$ , and $p = 1$ . Conversely these values make both derivatives zero. Thus $T_{c}$ is the model critical temperature in this stated sense, and $t = 0.9 < 1$ is below it.

3.1A1F1step 1.1algebra∎

The derivative is continuous near $(t, v) = (0.9, 1)$ because $3 v - 1 \neq 0$ there. Hence it remains positive on some open neighborhood within $v > 1 / 3$ . Since $P = P_{c} p$ and $V = V_{c} v$ with positive scales, $(\partial P / \partial V)_{T} = (P_{c} / V_{c}) (\partial p / \partial v)_{t} > 0$ there as well. This verifies a positive-slope branch of this phenomenological model only.

Depends on

Phase branches and coexistence domains

A positive-slope segment in the reduced van der Waals model

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources