Van der Waals Equation: Critical Constants and Boyle Temperature
The ideal gas law fails at high pressure and low temperature, and the van der Waals equation is the first correction every physical chemistry course teaches. Numerical questions in IIT-JAM, GATE and CSIR-NET ask for the pressure of a real gas, the critical constants from a and b, or the Boyle temperature. This page shows how each is computed, with the arithmetic checked.
The equation
For n moles: (P + n2a ÷ V2)(V − nb) = nRT
Pressure form (1 mole): P = RT ÷ (Vm − b) − a ÷ Vm2
The constant b (L mol−1) corrects for the finite volume of molecules, so the free volume is Vm − b. The constant a (L2 atm mol−2, or Pa m6 mol−2) corrects for attraction between molecules, which lowers the pressure on the wall. Use R = 0.082057 L atm mol−1 K−1 when a is in L2 atm mol−2. The values of a and b differ slightly between data tables, so use the values given in the question.
Critical constants
At the critical point the P–V isotherm has a point of inflection, so (∂P/∂V)T = 0 and (∂2P/∂V2)T = 0. Solving these two conditions together with the equation gives:
Zc = PcVc ÷ RTc = 3/8 = 0.375
Boyle temperature: TB = a ÷ (Rb) (so TB = 27Tc ÷ 8)
The Boyle temperature is the temperature at which the gas behaves ideally at low pressure, because the attraction and size corrections cancel. Real gases usually have a smaller Zc than 0.375, so the model is useful but approximate.
Worked example 1: critical constants of CO2
Question. Take a = 3.59 L2 atm mol−2 and b = 0.0427 L mol−1 for CO2 (typical textbook values). Find Vc, Pc, Tc and TB.
Vc. 3 × 0.0427 = 0.1281 L mol−1.
Pc. b2 = 0.04272 = 0.001823. 27b2 = 0.04923. Pc = 3.59 ÷ 0.04923 = 72.9 atm.
Tc. 8a = 28.72. 27Rb = 27 × 0.082057 × 0.0427 = 0.09460. Tc = 28.72 ÷ 0.09460 = 303.6 K.
TB. a ÷ (Rb) = 3.59 ÷ (0.082057 × 0.0427) = 3.59 ÷ 0.003504 = 1024.6 K. Check: 27 × 303.6 ÷ 8 = 1024.7 K, which agrees within rounding.
Worked example 2: pressure of a real gas
Question. 1.00 mol of CO2 (same a and b) occupies 0.500 L at 300 K. Compare the van der Waals pressure with the ideal pressure.
Ideal. P = RT ÷ V = 0.082057 × 300 ÷ 0.500 = 49.23 atm.
van der Waals. V − b = 0.500 − 0.0427 = 0.4573 L. RT ÷ (V − b) = 24.617 ÷ 0.4573 = 53.83 atm. a ÷ V2 = 3.59 ÷ 0.250 = 14.36 atm. P = 53.83 − 14.36 = 39.47 atm.
Compressibility factor. Z = PV ÷ RT = 39.47 × 0.500 ÷ 24.617 = 0.802. Z is below 1, so attraction dominates and the gas is more compressible than an ideal gas at this condition.
Worked example 3: finding a and b from critical data
Question. A gas has Tc = 304 K and Pc = 73.0 atm. Find a and b.
Rearrange. From Tc = 8a ÷ (27Rb) and Pc = a ÷ (27b2): dividing gives Tc ÷ Pc = 8b ÷ R, so b = RTc ÷ (8Pc). Then a = 27b2Pc.
b. b = 0.082057 × 304 ÷ (8 × 73.0) = 24.945 ÷ 584 = 0.04271 L mol−1.
a. a = 27 × (0.04271)2 × 73.0 = 27 × 0.0018241 × 73.0 = 3.595 L2 atm mol−2.
These agree with the values used in example 1, which is a useful cross-check.
When does the ideal gas law work?
At large volume the corrections shrink. For the same gas, 1.00 mol in 10.0 L at 300 K: ideal P = 0.082057 × 300 ÷ 10.0 = 2.462 atm; van der Waals P = 24.617 ÷ 9.9573 − 3.59 ÷ 100 = 2.4722 − 0.0359 = 2.436 atm. The ideal value is only about 1.0 % higher. So the ideal law is a good approximation at low pressure, but not at 0.5 L.
Common mistakes
- Using the wrong R. If a is in L2 atm mol−2, R = 0.082057 L atm mol−1 K−1. With a in Pa m6 mol−2, use R = 8.314 J mol−1 K−1 and volumes in m3.
- Mixing up the signs. The attraction term lowers pressure (−a/V2) and the size term reduces volume (V − b). Putting them the other way round is a common slip.
- Forgetting n2. For n moles the attraction term is n2a ÷ V2, not na ÷ V2.
- Taking Vc = b. Vc is 3b.
- Using the equation for liquids and for the two-phase region. Below Tc the equation shows loops that are not physical; real isotherms have a flat tie line.
Exam summary
| Quantity | Formula | Unit (a in L2 atm mol−2) |
|---|---|---|
| Vc | 3b | L mol−1 |
| Pc | a ÷ 27b2 | atm |
| Tc | 8a ÷ 27Rb | K |
| TB | a ÷ Rb | K |
| Zc | 3/8 | no unit |
Check the current official syllabus of your exam for the depth expected, for example whether reduced equations of state are asked.
The ideal gas tool solves PV = nRT. Use it to find the ideal-gas value first, then compare it with your van der Waals answer to see how large the deviation is.
Open the Ideal Gas Law (PV = nRT) Calculator →Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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