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Chemistry · JEE Advanced only

Deviation from Ideality and Van der Waals Equation (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 4 to 5 hours. Spend about 1.5 hours understanding the concepts and derivations, 1 hour memorizing formulas and critical constants relationships, and 2 hours solving numerical problems involving Z, Van der Waals equation, and comparing real vs ideal behavior..

Last updated: 26 October 2026

Direct answer

Real gases do not behave like ideal gases, especially at high pressures and low temperatures.

Real gases do not behave like ideal gases, especially at high pressures and low temperatures. The Van der Waals equation corrects the ideal gas equation by accounting for the actual volume of gas molecules and the attractive forces between them.

Official syllabus

What the JEE Advanced syllabus says

“Deviation from ideality, van der Waals equation”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Equation of state of a perfect gas.

JEE Main only covers the ideal gas equation, which assumes no intermolecular forces and zero molecular volume. For Advanced, you need to understand why real gases deviate and how the Van der Waals equation fixes these assumptions.

Key ideas

What you must understand

  • Real gases deviate from ideal behavior because molecules have finite size and attract each other
  • At high pressure, molecular volume matters because molecules are squeezed close together
  • At low temperature, intermolecular attractions become significant because molecules move slowly
  • The 'a' constant corrects for intermolecular attractions (pressure correction)
  • The 'b' constant corrects for molecular volume (volume correction, also called co-volume or excluded volume)
  • At very high temperatures and very low pressures, real gases behave almost ideally
Formulas

Formulas and key facts

  • Van der Waals equation: (P + an²/V²)(V − nb) = nRT
  • For 1 mole: (P + a/Vm²)(Vm − b) = RT where Vm is molar volume
  • Compressibility factor: Z = PV/nRT (Z = 1 for ideal gas)
  • For real gases: Z > 1 when repulsion dominates, Z < 1 when attraction dominates
  • At Boyle temperature, Z = 1 over a range of pressures
  • Boyle temperature: TB = a/Rb
  • Critical constants: Tc = 8a/27Rb, Pc = a/27b², Vc = 3b
  • b = 4 × (volume of one mole of molecules) = 4 × NA × (4/3)πr³
Avoid

Common mistakes

  • Confusing the signs: the pressure correction adds to P (because actual pressure is reduced by attractions), while volume correction subtracts from V (because available volume is less)
  • Forgetting that 'a' has units of atm·L²/mol² and 'b' has units of L/mol, leading to unit errors in calculations
  • Assuming Z is always less than 1 for real gases, but Z can be greater than 1 at very high pressures where repulsion dominates
  • Mixing up when to use n² (for total moles) versus using molar volume form of the equation
Worked example

A JEE Advanced-level example

Question. One mole of a Van der Waals gas at 27°C has a molar volume of 0.05 L. If a = 0.1 atm·L²/mol² and b = 0.02 L/mol, calculate the pressure of the gas. Also find the compressibility factor Z. (R = 0.082 L·atm/mol·K)

Show solution
  1. Given: n = 1 mol, T = 27°C = 300 K, Vm = 0.05 L, a = 0.1 atm·L²/mol², b = 0.02 L/mol
  2. Using Van der Waals equation for 1 mole: (P + a/Vm²)(Vm − b) = RT
  3. Calculate a/Vm² = 0.1/(0.05)² = 0.1/0.0025 = 40 atm
  4. Calculate Vm − b = 0.05 − 0.02 = 0.03 L
  5. Calculate RT = 0.082 × 300 = 24.6 L·atm
  6. So (P + 40)(0.03) = 24.6, which gives P + 40 = 820, therefore P = 780 atm. For Z: Z = PVm/RT = (780 × 0.05)/24.6 = 39/24.6 = 1.585

Answer: Pressure P = 780 atm, Compressibility factor Z = 1.59 (approximately)

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Deviation from Ideality and Van der Waals Equation in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is Equation of state of a perfect gas.

How long does Deviation from Ideality and Van der Waals Equation take to prepare for JEE Advanced?

About 4 to 5 hours. Spend about 1.5 hours understanding the concepts and derivations, 1 hour memorizing formulas and critical constants relationships, and 2 hours solving numerical problems involving Z, Van der Waals equation, and comparing real vs ideal behavior., if you already know the related JEE Main topic.

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