SKH Thermodynamics Labs

Lab 09 · Vapor nonideality and the second virial coefficient

When is an ideal vapor good enough?

The Lab 01 mixture at 350 K and 95 kPa, z₁ = 0.50, with an ideal liquid. The vapor is described twice: as an ideal gas and with second virial coefficients B₁₁ = −120, B₁₂ = −90 and B₂₂ = −180 cm³ mol⁻¹.

Compare the two vapors, find where the virial correction enters, and decide how large BP/RT must be before it matters.

2027 prototype
Review the student workflow

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Core: predict, calculate, compare the two vapors and explain the result. Steps 1–5.

Graduate: the core steps plus a term-by-term decomposition, the cross coefficient B₁₂ and a scale study in Step 4.

Graduate route. Complete Steps 1–3 to compare the two vapors. In Step 4, decompose the correction to each K-value, test the cross coefficient and scale B toward the truncation limit. Go to the graduate investigation.

Choose a caseAll cases use the Lab 01 mixture at 350 K with an ideal liquid.

Where the vapor correction enters

Three small factors, partly cancelling
The gamma–phi equilibrium equation with its correction factors labeled y_i phi_i P equals x_i gamma_i phi_i-sat P_i-sat Pi_i. Phi_i corrects the vapor mixture, phi_i-sat corrects the pure saturated vapor in the liquid reference, gamma_i is the liquid activity coefficient and Pi_i is the Poynting factor. With phi_i, phi_i-sat and Pi_i equal to one this is the modified Raoult's law of Labs 01 and 02. γᵢ · liquid (Labs 01–02)Πᵢ · Poynting yᵢφᵢP=xᵢγᵢφᵢˢᵃᵗPᵢˢᵃᵗΠᵢ vapor mixturepure saturated vapor ln φᵢ = (2Σⱼ yⱼBᵢⱼ − Bmix) P/RTln φᵢˢᵃᵗ = Bᵢᵢ Pᵢˢᵃᵗ/RT · ln Πᵢ = vᵢᴸ(P − Pᵢˢᵃᵗ)/RT

With φᵢ = φᵢˢᵃᵗ = Πᵢ = 1 this is the modified Raoult's law used in Labs 01, 02 and 08.

Correction terms in ln K for component 1 in the starting case ln phi1-sat is minus 0.0062. Minus ln phi1 is plus 0.0033. The net correction to ln K1 is minus 0.0028, a change of about 0.3 percent in K1. ln φ₁ˢᵃᵗ− ln φ₁net ln K₁ +0.0033−0.0062−0.0028 (−0.3% in K₁) 0

Calculated for the starting case. The correction to the pure saturated vapor and the correction to the vapor mixture have opposite signs here and largely cancel.

Both vapors use the same liquid model and the same saturation pressures. Any difference in the results comes only from the vapor description.

1. Set the conditions and coefficients

Basis · 1.00 mol feed
K
kPa
mol/mol
Second virial coefficients · cm³ mol⁻¹

Constant over 300–390 K. Synthetic teaching values, not real-compound data. B₁₂ is supplied, not computed from a mixing rule.

Optional. Your choice does not change the calculation.

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

Model and validity

The pressure-form second-virial vapor: Z = 1 + BmixP/RT with Bmix = y₁²B₁₁ + 2y₁y₂B₁₂ + y₂²B₂₂. The liquid reference includes φᵢˢᵃᵗ consistently, so B = 0 recovers the ideal vapor exactly. The lab refuses states with max|Bᵢⱼ| · max(P, Pᵢˢᵃᵗ)/RT > 0.1, where neglected higher virial terms may matter; passing this guard is not an accuracy guarantee. Saturation pressures are the Lab 01 synthetic correlations (150 and 60 kPa at 350 K).

Open the current Lab 09 for temperature-dependent B, your own Antoine constants and the Gibbs-curve view.

2. Compare the two vapors

Calculate to compare the ideal and virial vapors side by side.

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. What does Z = 0.996 mean?

Calculate the starting case and read Z. Compute Bmix at the vapor composition and check Z yourself.

Compare with the explanation

At y₁ = 0.6148, Bmix = 0.378(−120) + 2(0.6148)(0.3852)(−90) + 0.148(−180) = −114.7 cm³ mol⁻¹. Then Z = 1 + (−114.7 × 10⁻⁶)(95 000)/(8.314 × 350) = 0.9963.

A negative B means net attraction between molecules: the vapor occupies about 0.4% less volume than an ideal gas at the same T and P.

B. Why does the result change so little?

Compare the bubble pressure and β of the two vapors, and read the correction terms in the right-hand figure.

Compare with the explanation

The bubble pressure falls from 105.00 to 104.85 kPa (0.14%) and β from 0.494 to 0.488. From yᵢφᵢP = xᵢγᵢφᵢˢᵃᵗPᵢˢᵃᵗ, K₁ changes by the factor φ₁ˢᵃᵗ/φ₁ = exp(−0.0062 + 0.0033) = 0.997.

Both the vapor mixture and the pure saturated vapor in the reference are nonideal in the same direction, so their corrections largely cancel. At low pressure, |B|P/RT is below 1%, and the ideal vapor is a good approximation.

C. When does the vapor correction matter?

Load “Ten times larger |B|” and compare Z, β and the bubble pressure with the ideal vapor.

Compare with the explanation

With B ten times larger, Z = 0.963, β falls from 0.494 to 0.438 and the bubble pressure is 1.5% below the ideal-vapor value. The size indicator |B|P/RT reaches 0.093, close to the 0.1 limit of the second-virial truncation.

Large |B| (strongly attracting or associating vapors such as carboxylic acids) and higher pressure make the vapor correction important. Beyond the limit, an equation of state is needed.

4. Investigate the vapor correction

Master's core work

Calculate a state in Step 1 before running an investigation.

A. Decompose the correction to each K-value

  1. Tabulate ln φᵢ, ln φᵢˢᵃᵗ and ln Πᵢ for both components at the calculated state.
  2. Show that ln Kᵢ(virial) − ln Kᵢ(ideal) = ln φᵢˢᵃᵗ + ln Πᵢ − ln φᵢ at the same liquid composition.
  3. Explain why the corrections have opposite signs for the two components here.

B. How much does the cross coefficient matter?

  1. Keep B₁₁ and B₂₂ and compare the supplied B₁₂ with the geometric mean −(B₁₁B₂₂)1/2 and with ±30%.
  2. Compare β, y₁ and the bubble pressure.
  3. Explain why a combining rule for B₁₂ is an assumption that needs mixture data.

C. Scale B toward the truncation limit

  1. Multiply all three coefficients by 1, 3, 5, 8, 10 and 15.
  2. Compare Z, β and the bubble-pressure error of the ideal vapor with the size indicator.
  3. Explain why the lab refuses the largest case, and what model you would use instead.

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. Confirm that all B = 0 reproduces the ideal vapor exactly.
  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 Compare the two vapors.

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.