SKH Thermodynamics Labs

Lab 11 · Gas adsorption equilibrium

Where does the gas go?

A rigid 1.0 L vessel holds 20 g of a synthetic adsorbent and 0.030 mol of a single gas at 300 K. Without the adsorbent the pressure would be 74.8 kPa. The adsorbent follows a Langmuir isotherm with qs = 2.0 mol kg⁻¹, b = 0.10 kPa⁻¹ at 300 K and an isosteric heat of 25 kJ mol⁻¹.

Find the equilibrium pressure and how much gas sits on the solid. Then predict what heating the closed vessel does.

2027 prototype
Review the student workflow

View

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

Graduate: the core steps plus a fit-range study, the isosteric heat from three isotherms and a temperature-swing study in Step 4.

Graduate route. Complete Steps 1–3 for the starting case. In Step 4, fit isotherm models to measured-style data over two ranges, recover the isosteric heat, and test how it controls the gas released on heating. Go to the graduate investigation.

Choose a caseAll cases use the same vessel and gas inventory.

Two equations meet at equilibrium

Isotherm and material balance
A closed vessel with an adsorbent bed The gas inventory n splits between the free gas, PV over RT, and the adsorbed phase, m times q of P. At 300 K, 20 percent is in the gas and 80 percent on the solid. At 340 K, 36 percent is in the gas. gas · PV/RT adsorbed · m q(P) n = PV/RT + m q(P) 300 K340 K 20%80% adsorbed36%64% adsorbed

Teal is the free gas and amber the adsorbed amount, for the starting case and after heating to 340 K.

Isotherms and the vessel material-balance line Uptake q against pressure P. The 300 K Langmuir isotherm rises and flattens toward 2 mol per kg. The 340 K isotherm lies lower. The material-balance line falls from q = 1.5 at P = 0. It crosses the 300 K isotherm at 15.0 kPa and q = 1.20, and the 340 K isotherm at 30.3 kPa and q = 0.96. A linear isotherm with the same initial slope crosses the balance line at 6.8 kPa. crossings: 300 K at 15.0 kPa, q = 1.20340 K at 30.3 kPa, q = 0.96 Langmuir · 300 K340 Klinear, q = 0.2P · 6.8 kPamaterial balance 01020304001.02.0 Pressure P / kPa q / mol kg⁻¹

Model-generated for the starting case. Heating barely moves the balance line but lowers the isotherm, so the crossing moves to higher pressure.

The isotherm says how much the solid holds at a given pressure. The material balance says how the fixed inventory must be shared. Equilibrium is the single pressure that satisfies both.

1. Set the vessel and the isotherm

Single gas · ideal gas phase
K
L
g
mol
kJ mol⁻¹

b(T) = b(300 K)·exp[(Q/R)(1/T − 1/300)]; the same factor applies to K. Synthetic teaching values.

Optional. Your choice does not change the calculation.

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

Model and validity

Absolute uptake of one gas, ideal gas phase, adsorbed-phase volume neglected, constant isosteric heat and qs independent of temperature. Pressures are limited to 200 kPa. The equilibrium is unique because both the gas and the adsorbed amounts increase with pressure.

Open the current Lab 11 to fit Freundlich and BET models to your own isotherm data.

2. Read the result

Calculate to see the pressure and how the gas is shared.

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. Check the material balance

For the starting case, take the calculated pressure and verify n = PV/RT + m q(P).

Compare with the explanation

At P = 14.99 kPa, the gas holds (14 990 Pa)(0.001 m³)/(8.314 × 300) = 0.0060 mol. The uptake is q = 2.0 × 1.499/2.499 = 1.20 mol kg⁻¹, so the solid holds 0.020 kg × 1.20 = 0.0240 mol. The sum is 0.0300 mol.

80% of the gas is on the solid, and the pressure is 15.0 kPa instead of 74.8 kPa. The coverage θ = 0.60 is far from the low-pressure limit.

B. Why does a linear isotherm give the wrong pressure?

Load “Linear isotherm, same initial slope” (K = qsb = 0.2) and compare the pressure.

Compare with the explanation

The linear isotherm gives 6.8 kPa and q = 1.36 mol kg⁻¹, against 15.0 kPa and 1.20 mol kg⁻¹ for Langmuir. Both agree only as P → 0.

At θ = 0.6 the sites are filling, so each extra kPa adds less uptake. A linear isotherm ignores this saturation, overestimates what the solid can hold, and so underestimates the pressure by more than half.

C. What does heating the closed vessel do?

Load “Heated to 340 K”. Compare the pressure with the starting case and with an empty vessel heated by the same amount.

Compare with the explanation

The pressure rises from 15.0 to 30.3 kPa, a factor of 2.0. An empty vessel would rise only by 340/300 = 1.13, from 74.8 to 84.8 kPa. The solid releases 0.0047 mol, and the gas share rises from 20% to 36%.

Adsorption is exothermic, so b falls with temperature: b(340 K) = 0.1 × exp[(25 000/8.314)(1/340 − 1/300)] = 0.031 kPa⁻¹. This is the basis of temperature-swing regeneration.

4. Investigate the isotherm evidence

Master's core work

Calculate a state in Step 1 before running an investigation.

A. Does the fitted range matter?

  1. Fit Henry, Langmuir and Freundlich models to the 300 K isotherm, first for q ≤ 0.8 mol kg⁻¹ and then for all seven points.
  2. Use each fit to predict the vessel pressure and the uptake at 60 kPa, outside the data.
  3. Explain why a small RMSE inside the range does not guarantee a reliable vessel prediction.

B. Recover the isosteric heat

  1. Compute Qst = −R d ln P/d(1/T) at fixed loading from isotherms at 300, 320 and 340 K.
  2. Repeat with only two temperatures, and with the 340 K isotherm recorded as 342 K.
  3. Explain which change you can detect from the data and which you cannot.

C. How much gas does heating release?

  1. Heat the closed vessel from 300 K to 400 K with Q = 20, 25 and 30 kJ mol⁻¹.
  2. Compare the pressure and the gas released from the solid.
  3. State how uncertainty in Q translates into uncertainty in the released gas.

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. Double the adsorbent mass and predict the direction of the pressure change before calculating.
  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 the 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.