SKH Thermodynamics Labs

Lab 06 · Vapor–liquid–liquid equilibrium

When can two liquids and a vapor coexist?

A 1.00 mol binary feed with z₁ = 0.50 at fixed temperature. The liquid follows symmetric Margules with A = 3, the vapor is ideal, and P₁sat = 80 kPa, P₂sat = 40 kPa.

Find the three-phase pressure, read the three phase compositions, and explain why their amounts are not fixed.

2027 prototype
Review the student workflow

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Core: predict, calculate, read the diagrams and explain the result. Steps 1–5.

Graduate: the core steps plus the missing amount specification, a pressure sweep and the heterogeneous azeotrope in Step 4.

Graduate route. Complete Steps 1–3 to verify the three-phase state. In Step 4, fix the amounts with an extra specification, sweep the pressure across the three-phase pressure, and test when the vapor lies inside the liquid gap. Go to the graduate investigation.

Choose a conditionAll named cases use A = 3, P₁sat = 80 kPa and P₂sat = 40 kPa at the same fixed temperature.

Picture three coexisting phases

Compositions are fixed; amounts are not
A closed vessel with a vapor above two liquid layers The vessel holds three phases at one temperature and pressure: a vapor with composition y1 at the top, and two liquid layers, alpha (component-1-poor) and beta (component-1-rich). Vapor V · y₁Liquid β · x₁ᵝLiquid α · x₁ᵅ

At one T, a binary with three phases has F = 2 − 3 + 2 = 1 degree of freedom, so T fixes P and all three compositions. Which liquid is on top depends on density, which the model does not calculate.

Pressure–composition diagram of the starting system with its three-phase line Above the three-phase pressure, a two-liquid region spans the middle compositions. Below it, a liquid plus vapor region lies on each side and vapor fills the bottom. At the three-phase pressure, liquid alpha, liquid beta and vapor coexist on one horizontal line. Lα + LβLα + VLβ + VVaporLαLβ 00.514080P₃ Mole fraction of component 1 · x₁ or y₁Pressure / kPa y₁x₁ᵅx₁ᵝ

Calculated for the starting system. Above P₃ = 113.2 kPa two liquids coexist; below it, each liquid coexists with vapor. At P₃ the three points x₁ᵅ, y₁ and x₁ᵝ lie on one horizontal line.

The feed composition z₁ chooses where on the three-phase line you are. Two material balances cannot fix three amounts, so the amounts form a family of solutions.

1. Set the system, pressure and feed

Basis · 1.00 mol feed · fixed T
kPa
kPa

Hypothetical pure-component vapor pressures at the same fixed temperature, 10–200 kPa.

mol/mol

Optional. Your choice does not change the calculation.

Add a calculated result to the comparison before changing a setting.

Model assumptions

The liquid Gibbs energy uses the selected activity model; the vapor is ideal. Both are placed on one Gibbs-energy reference using the pure-component saturation pressures, so the stable phases are those on the lowest common tangent of the liquid and vapor curves. Poynting corrections and temperature dependence are neglected. The system is hypothetical and does not represent a named chemical pair.

Open the current Lab 06 for its original controls and examples.

2. Identify the phases, compositions and amounts

Calculate the state to see its phases, compositions, amounts and checks.

4. Check your understanding

Write first, then compare

Use Steps 1–3 to answer each question in your own words. Then open the explanation and compare. Your answers are saved with your study.

A. Why are the three compositions the same for every feed at P₃?

Calculate the starting case (z₁ = 0.50) and “Same pressure · z₁ = 0.20”. Compare x₁ᵅ, x₁ᵝ and y₁.

Compare with the explanation

x₁ᵅ = 0.0707, x₁ᵝ = 0.9293 and y₁ = 0.6667 are identical for both feeds. Only the amounts change.

For two components and three phases, F = 2 − 3 + 2 = 1. Fixing T uses that one degree of freedom, so P₃ and all three compositions are fixed. This extends Labs 01 and 04: compositions come from the equilibrium conditions, amounts come from the material balance.

B. Why are the three amounts not unique?

At the starting case, move the slider under the phase table and watch the three amounts.

Compare with the explanation

There are three unknown amounts but only two balances: total moles, fα + fβ + fV = 1, and component 1, fαx₁ᵅ + fβx₁ᵝ + fVy₁ = z₁. For z₁ = 0.50 any vapor amount from 0 to 0.72 mol satisfies both.

The phase rule fixes intensive variables only. One more extensive specification, such as the total enthalpy or volume, is needed to choose the amounts. Lab 07 supplies it with an energy balance.

C. Why do only two phases remain above or below P₃?

Calculate “Higher pressure · 140 kPa” and “Lower pressure · 100 kPa”, and compare the Gibbs curves in Step 3 with the starting case.

Compare with the explanation

The ideal-vapor Gibbs energy contains ln(P/Pisat), so raising P lifts the vapor curve while the liquid curve hardly changes. At 140 kPa the vapor curve lies above the liquid common tangent: two liquids, no vapor. At 100 kPa it drops below that tangent, and a new tangent between one liquid and the vapor gives liquid + vapor (V ≈ 0.79 mol).

Only at P₃ does one straight line touch both liquid minima and the vapor curve at once.

4. Investigate what fixes a three-phase state

Master's core work

Calculate a state in Step 1 before running an investigation.

A. Supply the missing specification

  1. Calculate the starting case at P₃ and read the allowed range of the vapor amount.
  2. Specify a vapor amount and solve the two balances for fα and fβ by hand. Check against the lab.
  3. Explain which physical specification (heat duty, vessel volume) would replace your chosen vapor amount in a real flash.
mol

B. Sweep the pressure across P₃

  1. Keep the feed and run the sweep from 90 to 140 kPa. P₃ is inserted automatically.
  2. Track the phase identity, the compositions and the vapor amount. Identify where each changes continuously and where it jumps.
  3. Explain why the vapor amount at P₃ is a range rather than a number.

C. When does the vapor lie inside the liquid gap?

  1. Run the comparison. It keeps the liquid model and sets P₂sat = 10 kPa while P₁sat/P₂sat ranges from 1 to 20.
  2. Find where y₁ crosses x₁ᵝ. For symmetric Margules, show that y₁ = r/(1 + r) at P₃, with r = P₁sat/P₂sat.
  3. Explain why a vapor inside the gap makes a decanter useful in heterogeneous azeotropic distillation.

D. Defend a limited conclusion

The record check only confirms that the sections and calculations are present. Scientific reasoning still needs instructor review.

5. Compare and save

  1. Calculate and save the four named cases.
  2. Compare the compositions at P₃ for z₁ = 0.50 and 0.20, and the phases at 140 and 100 kPa.
  3. Answer questions A–C in Step 4 using the saved comparison.
  4. Download the complete study, reopen it, and reproduce a saved case.
  1. Calculate and save the four named cases.
  2. Complete investigations A–C and write the graduate claims.
  3. Download the complete study, reopen it, and reproduce a saved case and its investigation record.

Your calculated cases will appear here after you add them.

Starting case loaded. Press Calculate equilibrium.

Your inputs, comparisons and explanations are kept in this open page. Download one study file to keep the complete record and reopen it later. Results are recalculated on restore.