INGENIA

THM-05

Antoine saturation pressure

log10 Psat = A − B/(T+C).

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Phase changeAntoine 1888

Governing equation

log⁡10Psat=A−BT+C\log_{10}P_{\mathrm{sat}}=A-\dfrac{B}{T+C}

T∈[0,180]T \in \left[0, 180\right]

where

A
A (—)
B
B (—)
C
C (—)
T
Temperature (°C)
P_{\mathrm{sat}}
Saturation pressure (kPa)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

log⁡10Psat=A−BT+C\log_{10}P_{\mathrm{sat}}=A-\dfrac{B}{T+C}

T∈[0,180]T \in \left[0, 180\right]

T=0T = 0Psat=0.606P_{\mathrm{sat}} = 0.606

②

Domain of the variable — blue

Tmin⁡=0,Tmax⁡=180T_{\min} = 0,\quad T_{\max} = 180

T∈[0,180]T \in \left[0, 180\right]

T=0T = 0T=0T = 0

③

Response curve — green

log⁡10Psat=A−BT+C\log_{10}P_{\mathrm{sat}}=A-\dfrac{B}{T+C}

T∈[0,180]T \in \left[0, 180\right]

T=0T = 0Psat=0.606P_{\mathrm{sat}} = 0.606

④

Rate of change — amber

dPsatdT\frac{\mathrm{d}P_{\mathrm{sat}}}{\mathrm{d}T}

T∈[0,180]T \in \left[0, 180\right]

T=1.132T = 1.132Psat′=0.048P_{\mathrm{sat}}^{\prime} = 0.048

Lecture brief

Historical brief

Carnot (1824), Clausius entropy, the first law and later van der Waals and Gibbs potentials turned heat into a state science. The sheets compute work, efficiency and vapour pressure. This sheet (THM-05 — Antoine saturation pressure) is the form associated with Antoine 1888. Working symbols: AA, BB, CC, TT →\rightarrow PsatP_{\mathrm{sat}}. A three-parameter empirical integral of Clausius–Clapeyron, standard for vapour-pressure tables.

Purpose

Purpose: compute PsatP_{\mathrm{sat}} from AA, BB, CC, TT in Thermodynamics via log⁡10Psat=A−BT+C\log_{10}P_{\mathrm{sat}}=A-\dfrac{B}{T+C} log10 Psat = A − B/(T+C). Use it when a real thermodynamics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given A=8.071 —A = 8.071\,\mathrm{—}, B=1730.630 —B = 1730.630\,\mathrm{—}, C=233.426 —C = 233.426\,\mathrm{—}, T=60.000 ∘CT = 60.000\,\mathrm{^{\circ}C}, the governing relation log⁡10Psat=A−BT+C\log_{10}P_{\mathrm{sat}}=A-\dfrac{B}{T+C} yields Psat=19.870 kPaP_{\mathrm{sat}} = 19.870\,\mathrm{kPa}. T in °C, P in mmHg, water-like constants. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Outputs

  • Saturation pressure P_{\mathrm{sat}}19.870 kPa
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Narration of this film

T in °C, P in mmHg, water-like constants.

A three-parameter empirical integral of Clausius–Clapeyron, standard for vapour-pressure tables.

Reading speed

Watch on YouTube