IIT-JAM Gaseous State — Kinetic Theory, Real Gases and Critical Constants
At school level the gaseous state means PV = nRT. At IIT-JAM level it means everything that happens when that equation stops working: the three molecular speeds, the shape of the Maxwell distribution, the van der Waals correction terms, the compressibility factor, the critical constants and the Boyle temperature. The unit is compact, entirely formula-driven, and it rewards a student who can carry a calculation through without unit errors. Here is the whole thing, with the arithmetic done in full.
The ideal gas baseline
Dalton: Ptotal = ΣPi, and Pi = xi Ptotal
Graham: rate of effusion ∝ 1/√M (at the same T and P)
Choose the value of R that matches your pressure unit and you will avoid most numerical errors in this chapter. If pressure is in bar and volume in litres, use 0.083145. If you are computing an energy or a molecular speed, use 8.314 and put the molar mass in kilograms per mole. That last point is where most speed calculations go wrong.
Kinetic theory and the three speeds
urms = √(3RT/M) · uavg = √(8RT/πM) · ump = √(2RT/M)
Fixed ratio: ump : uavg : urms = 1 : 1.128 : 1.225
The order is easy to remember from the numerators — 2, 8/π ≈ 2.546, 3 — so the most probable speed is always the smallest and the root-mean-square speed always the largest. Note that the average kinetic energy depends only on temperature, not on the identity of the gas, while the speeds depend on molar mass. That distinction is a favourite one-mark question.
Example 1 — the three speeds of nitrogen at 300 K.
M(N₂) = 2 × 14.007 = 28.014 g/mol = 0.028014 kg/mol.
3RT = 3 × 8.314 × 300 = 7482.6 J/mol.
3RT/M = 7482.6 ÷ 0.028014 = 267 102 m² s−2.
urms = √267 102 = 516.8 m/s.
Now use the fixed ratios instead of recomputing from scratch:
ump = urms × √(2/3) = 516.8 × 0.81650 = 422.0 m/s
uavg = urms × √(8/3π) = 516.8 × 0.92132 = 476.1 m/s
Check the ratio: 422.0 : 476.1 : 516.8, divided by 422.0, gives 1 : 1.128 : 1.225 ✓. Also note the size of the answer — a few hundred metres per second is the right order of magnitude for a light gas at room temperature, so an answer of 5 m/s or 50 000 m/s means a unit error, almost always the grams-versus-kilograms one.
The Maxwell–Boltzmann distribution, described in words
The distribution of molecular speeds is a curve that starts at zero, rises to a peak at ump, and falls away with a long tail at high speed. Its behaviour under changes of temperature and molar mass is examined every year:
| Change | Peak position | Peak height | Curve width | Consequence |
|---|---|---|---|---|
| Raise the temperature (same gas) | Moves right (ump ∝ √T) | Falls | Broadens | Far more molecules in the high-speed tail — this is why reaction rates rise so steeply with T |
| Heavier gas (same temperature) | Moves left (ump ∝ 1/√M) | Rises | Narrows | Heavier molecules move more slowly and their speeds are less spread out |
Two properties are true of every such curve: the total area under it is fixed at 1, because every molecule has some speed; and it is not symmetric about the peak, which is exactly why uavg and urms both lie to the right of ump.
Real gases — the van der Waals equation
a corrects for intermolecular attraction (pressure term) · b corrects for molecular volume (volume term)
Compressibility factor Z = PV ÷ nRT (Z = 1 exactly for an ideal gas)
Reading Z is a standard question. Z < 1 means the gas is more compressible than ideal, because attractive forces dominate — typical at moderate pressure. Z > 1 means it is less compressible than ideal, because the finite size of the molecules dominates — typical at high pressure, where the b term takes over. Hydrogen and helium have very weak attractions, so their Z exceeds 1 across a wide range at ordinary temperatures.
Example 2 — ideal versus real pressure. Find the pressure exerted by 1.00 mol of CO₂ confined to 1.00 L at 300 K, first as an ideal gas and then with the van der Waals equation. Take a = 3.64 bar L² mol−2 and b = 0.0427 L mol−1 (constants of this kind are always supplied in the question).
Ideal: P = nRT/V = (1.00 × 0.083145 × 300) ÷ 1.00 = 24.94 bar.
van der Waals: rearrange to P = nRT/(V − nb) − an²/V².
V − nb = 1.000 − (1.00 × 0.0427) = 0.9573 L.
First term = 24.9435 ÷ 0.9573 = 26.0561 bar.
Second term = a n²/V² = 3.64 × (1.00)² ÷ (1.00)² = 3.64 bar.
P = 26.0561 − 3.64 = 22.42 bar.
Compressibility factor: Z = PV/nRT = (22.4161 × 1.000) ÷ 24.9435 = 0.899.
Interpretation, which is what the marks are for: Z < 1, the real pressure is lower than the ideal prediction, so attractive forces are dominating at these conditions. Molecules approaching the wall are pulled back by their neighbours and strike it less hard. The volume correction b is present but is the smaller effect here.
Critical constants from a and b
At the critical point the isotherm has a point of inflection, so both (∂P/∂V)T and (∂²P/∂V²)T vanish. Solving those two conditions with the van der Waals equation gives three results you should be able to quote instantly:
Critical compressibility Zc = PcVc/RTc = 3/8 = 0.375, for every van der Waals gas
Boyle temperature TB = a ÷ Rb = 27Tc/8
Example 3 — critical constants and Boyle temperature for the gas in Example 2.
Vc = 3b = 3 × 0.0427 = 0.1281 L/mol.
Pc = a ÷ 27b² : b² = (0.0427)² = 0.00182329, and 27 × 0.00182329 = 0.04922883.
Pc = 3.64 ÷ 0.04922883 = 73.94 bar.
Tc = 8a ÷ 27Rb : 27 × 0.083145 = 2.244915, and 2.244915 × 0.0427 = 0.09585787.
Tc = (8 × 3.64) ÷ 0.09585787 = 29.12 ÷ 0.09585787 = 303.8 K.
Verify with Zc: PcVc = 73.94 × 0.1281 = 9.4717, and
RTc = 0.083145 × 303.8 = 25.2578.
Zc = 9.4717 ÷ 25.2578 = 0.3750 = 3/8 ✓ — exactly the universal value,
which confirms all three constants at once. Use this check in the exam; it costs ten seconds.
TB = a ÷ Rb = 3.64 ÷ (0.083145 × 0.0427) = 3.64 ÷ 0.00355029 =
1025 K.
Cross-check: TB = 27Tc/8 = 27 × 303.8 ÷ 8 = 1025 K ✓.
The Boyle temperature is the temperature at which the attractive and repulsive corrections cancel over a range of low pressures, so the gas behaves almost ideally. Above TB a gas cannot be liquefied by pressure alone — and above Tc it cannot be liquefied at all, which is why Tc for a gas that is liquid-friendly at room temperature is around 300 K while a permanent gas has a very low Tc.
Mean free path — how far a molecule gets between collisions
Example 4. Estimate the mean free path of N₂ at 300 K and 1.00 × 105 Pa, taking the collision diameter as d = 3.7 × 10−10 m (given in the question).
Numerator: kBT = 1.380649 × 10−23 × 300 = 4.1419 × 10−21 J.
d² = (3.7 × 10−10)² = 1.369 × 10−19 m².
√2 × π = 4.44288.
Denominator = 4.44288 × 1.369 × 10−19 × 1.00 × 105 = 6.0823 ×
10−14.
λ = 4.1419 × 10−21 ÷ 6.0823 × 10−14 = 6.81 × 10−8 m ≈
68 nm.
That is roughly 180 times the molecular diameter itself — which is precisely why the ideal gas model works so well at ordinary pressure. Halve the pressure and λ doubles; the inverse proportionality is the examinable part.
Common mistakes that cost marks
- Using M in g/mol inside a speed formula. With R = 8.314 J K−1 mol−1 the molar mass must be in kg/mol, or your answer is out by a factor of about 31.6 (that is √1000).
- Mixing R values. Pick 0.083145 for bar-litre work and 8.314 for joules, and never use one in the other's formula.
- Putting the corrections on the wrong side. The pressure correction is added (P + an²/V², because real pressure is lower than ideal) and the volume correction is subtracted (V − nb, because free space is smaller).
- Forgetting n² and n. The full form is (P + an²/V²)(V − nb) = nRT; the familiar one-mole version hides both.
- Claiming Z < 1 always. It is true at moderate pressures for most gases, but Z rises above 1 at high pressure, and stays above 1 across a wide range for H₂ and He at ordinary temperatures.
- Confusing Tc with TB. They differ by the fixed factor 27/8, and their meanings are completely different.
- Treating the Maxwell curve as symmetric and therefore setting uavg = ump.
How this unit is examined
| Question shape | What it tests | Route to the answer |
|---|---|---|
| Compute urms, uavg or ump | Unit discipline | M in kg/mol, then apply the 1 : 1.128 : 1.225 ratio |
| Compare speeds of two gases | u ∝ 1/√M at fixed T | Take the ratio; no need to compute either speed |
| Real vs ideal pressure, and Z | van der Waals arithmetic | P = nRT/(V − nb) − an²/V², then Z = PV/nRT |
| Critical constants from a and b | Quoting three results | Vc = 3b, Pc = a/27b², Tc = 8a/27Rb; verify with Zc = 3/8 |
| Boyle temperature | TB = a/Rb | Cross-check with 27Tc/8 |
| Effect of T or M on the distribution | Conceptual reasoning | Peak position ∝ √(T/M); area is always 1 |
| Mean free path or collision frequency | Handling powers of ten | λ ∝ T/P; check the answer is tens of nanometres at 1 bar |
For the exact syllabus scope, question count and marking scheme, work from the current official IIT-JAM notification rather than from any summary — including this one.
Verify the ideal-gas half of every problem instantly. Almost every real-gas question starts by computing the ideal answer and then comparing. The free Ideal Gas Law calculator solves PV = nRT for whichever variable you leave blank and keeps the units consistent, so you can confirm the 24.94 bar in Example 2 before you spend time on the van der Waals correction.
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