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The ideal gas law works so well in most problems that it is easy to forget it is an approximation. JEE Advanced questions test exactly where it fails: the compressibility factor, the meaning of the van der Waals constants and the critical point. This post explains each with one worked example on carbon dioxide, and ends with a practice set.
Which exams include real gases?
The JEE (Advanced) 2026 brochure lists "Deviation from ideality, van der Waals equation" under States of Matter: Gases and Liquids. The chemistry units of the JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus do not include a states-of-matter unit; their physics units cover only the ideal ("perfect") gas through kinetic theory. Check the current year's documents, since syllabi change.
Why do real gases deviate from the ideal gas law?
An ideal gas has two built-in assumptions: its molecules take up no volume, and they do not attract or repel each other. OpenStax Chemistry 2e explains how each one fails.
At high pressure, molecules are crowded, so their own volume is a real fraction of the container. The gas is harder to compress than the ideal law predicts.
At moderate pressure and low temperature, attractions pull molecules slightly together, so they hit the walls less hard. Pressure is lower than the ideal value.
The compressibility factor measures the deviation: Z = PV/(nRT). Z = 1 for ideal behaviour, Z < 1 when attraction dominates and Z > 1 when molecular size dominates. Gases behave most ideally at low pressure and high temperature.
What do a and b mean in the van der Waals equation?
The 1910 Nobel Prize in Physics was awarded to Johannes Diderik van der Waals "for his work on the equation of state for gases and liquids". The equation is
(P + an²/V²)(V − nb) = nRT
b corrects the volume: V − nb is the free space molecules can move in. Units: L/mol.
a corrects the pressure for attraction. The term an²/V² is added because the measured pressure is lower than it would be without attraction. Units: L² atm/mol².
OpenStax's Table 9.3 gives, for CO₂, a = 3.59 L² atm/mol² and b = 0.0427 L/mol; for He, a = 0.0342 and b = 0.0237. A larger a means stronger attractions, which usually means easier liquefaction.
How different is the van der Waals answer for CO₂?
Example. Find the pressure of 1.00 mol of CO₂ in 0.500 L at 300 K, first with the ideal law, then with van der Waals. Take R = 0.08206 L atm/(mol K).
Ideal: P = nRT/V = (1.00 × 0.08206 × 300)/0.500 = 24.62/0.500 = 49.2 atm.
Van der Waals, rearranged: P = nRT/(V − nb) − an²/V².
Volume term: 24.62/(0.500 − 0.0427) = 24.62/0.4573 = 53.8 atm.
Attraction term: 3.59 × 1.00²/0.500² = 3.59/0.250 = 14.4 atm.
P = 53.8 − 14.4 = 39.5 atm.
Z = PV/(nRT) = (39.5 × 0.500)/24.62 = 0.80.
The b term alone would raise the pressure by about 4.6 atm, but the a term lowers it by 14.4 atm. At this density attraction wins, so Z < 1 and the ideal law overestimates by about a quarter.
How do critical constants follow from a and b?
Above the critical temperature a gas cannot be liquefied by pressure alone. Van der Waals constants predict the critical point:
T_c = 8a/(27Rb)
P_c = a/(27b²)
V_c = 3b (molar volume)
For CO₂: T_c = (8 × 3.59)/(27 × 0.08206 × 0.0427) = 28.72/0.0946 ≈ 304 K. The NIST Chemistry WebBook lists measured values near 304.2 K, so the estimate is close here. Also P_c ≈ 72.9 atm and V_c = 0.128 L/mol. Combining these gives Z_c = P_cV_c/(RT_c) = 3/8 for every van der Waals gas, a neat check on your algebra.
At low pressure, the equation gives Z ≈ 1 + (b − a/RT)P/RT. The a and b effects cancel at the Boyle temperature, T_B = a/(Rb). For helium this is a/(Rb) = 0.0342/(0.08206 × 0.0237) ≈ 18 K, so at room temperature b dominates and Z > 1 for helium, while for CO₂ attraction dominates at moderate pressures.
Common mistakes
Unit mismatch in R. With a in L² atm/mol², use R = 0.08206 L atm/(mol K), not 8.314.
Dropping n². The attraction term is an²/V², not a/V², unless n = 1.
Sign of the a term. The measured P is lower than ideal; in the rearranged form, subtract an²/V².
Treating Z < 1 as "more ideal". Any departure from 1 is non-ideal.
Expecting the formulas to be exact. Van der Waals is a model. It works well for T_c of CO₂ but less well for other quantities.
For the ideal-gas groundwork, see our posts on ideal gas problems and kinetic theory.
Practice routine
Repeat the example for He in the same conditions and explain why Z comes out above 1.
Compute T_c for N₂ from OpenStax's constants (a = 1.39, b = 0.0391).
Derive Z ≈ 1 + (b − a/RT)P/RT from the van der Waals equation for one mole, using the approximation V ≈ RT/P in the small terms.
If a derivation stalls, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revisit the ideal gas law and kinetic theory before trying again.
Start with question 1: it reuses every step of the worked example with new constants.
