Deviation From Ideal Gas Behaviour

Q: What is the compressibility factor Z of a gas?

The compressibility factor is Z = pV/nRT, equal to the actual volume of a gas divided by the volume it would have if ideal at the same pressure and temperature. For an ideal gas Z = 1. Z below 1 means attractions dominate and the gas is easier to compress; Z above 1 means molecular size dominates.

Q: Why do real gases deviate from ideal gas behaviour?

The ideal gas model assumes molecules have no volume and do not attract each other. Real molecules attract, which lowers the pressure on the walls, and they occupy space, which reduces the free volume. Both effects become important at high pressure and low temperature, when molecules are close together and slow.

Q: What is the physical significance of van der Waals constants a and b?

The constant a measures the strength of attraction between gas molecules and appears in the pressure correction a n squared over V squared. The constant b is the excluded volume per mole, about four times the actual volume of the molecules, and appears in the volume correction V − nb. Larger a means easier liquefaction.

Q: What is Boyle temperature?

Boyle temperature is the temperature at which a real gas obeys Boyle's law over a wide pressure range, because attraction and molecular size effects cancel. For a van der Waals gas it equals a/Rb. Above it Z is greater than 1 at all pressures; below it Z first dips below 1.

Q: What is critical temperature and why must a gas be cooled below it to liquefy?

Critical temperature is the highest temperature at which a gas can be liquefied. Above it, molecules move too fast for attractions to hold them together, so any pressure only compresses the gas. Below it, pressure brings molecules close enough to condense. For carbon dioxide the critical temperature is 30.98 °C.

Q: Why do hydrogen and helium show only positive deviation at room temperature?

Hydrogen and helium have very weak intermolecular attractions, so their van der Waals constant a is tiny. Their Boyle temperatures, about 110 K and 23 K, are far below room temperature, so the molecular size term always wins and Z stays above 1 as pressure rises.

Q: Are real gases and the van der Waals equation in the JEE Advanced 2026 syllabus?

Yes. The JEE Advanced syllabus lists deviation from ideality and the van der Waals equation under States of Matter: Gases and Liquids. The chapter has been removed from NCERT Class 11 and from the JEE Main and NEET syllabi, so these ideas are examined in JEE Advanced.

Q: What types of real gas questions are asked in JEE Advanced?

JEE Advanced asks for the sign of deviation from Z vs p graphs, Z from given data, van der Waals pressure calculations, critical constants and Boyle temperature from a and b, ordering gases by ease of liquefaction, and limiting forms of the van der Waals equation when a or b is neglected.

ChemistryStates Of MatterFor JEE aspirants

Real gases show deviation from ideal gas behaviour because their molecules attract one another and occupy space. The size of the deviation is measured by the compressibility factor $Z = \frac{p V}{n R T}$ , which is 1 for an ideal gas. The van der Waals equation corrects pressure (constant $a$ ) and volume (constant $b$ ), and it explains the Boyle temperature, the critical constants and why gases liquefy only below $T_{c}$ . Deviation from ideal gas behaviour is a regular JEE Advanced topic.

On this page1Real vs ideal2Compressibility factor Z3Why gases deviate4van der Waals equation5Z from vdW6Boyle temperature7Liquefaction8Critical constants9Continuity of state10Examples

Key Formulas - Quick Reference

★ Must learn Compressibility factor: $Z = \frac{p V}{n R T} = \frac{V _{real}}{V _{ideal}}$ ; ideal gas $Z = 1$ ; $Z < 1$ attraction dominates, $Z > 1$ repulsion dominates.
★ Must learn van der Waals equation: $(p + \frac{a n ^{2}}{V ^{2}}) (V - nb) = n R T$ ; for 1 mol $(p + \frac{a}{V _{m}^{2}}) (V_{m} - b) = R T$ .
Excluded volume: $b = 4 N_{A} \cdot \frac{4}{3} π r^{3}$ (four times the actual volume of 1 mol of molecules).
★ Must learn Limits: low $p$ ( $b$ negligible) $Z = 1 - \frac{a}{V _{m} R T}$ ; high $p$ ( $a$ negligible) $Z = 1 + \frac{p b}{R T}$ .
★ Must learn Boyle temperature: $T_{B} = \frac{a}{R b}$ ; at $T_{B}$ a real gas behaves ideally over a wide pressure range.
★ Must learn Critical constants: $V_{c} = 3 b$ , $p_{c} = \frac{a}{27 b ^{2}}$ , $T_{c} = \frac{8 a}{27 R b}$ , and $Z_{c} = \frac{p _{c} V _{c}}{R T _{c}} = \frac{3}{8}$ .
From $T_{c}$ and $p_{c}$ : $b = \frac{R T _{c}}{8 p _{c}}$ , $a = \frac{27 R ^{2} T _{c}^{2}}{64 p _{c}}$ ; also $T_{B} = \frac{27}{8} T_{c}$ .
Liquefaction needs $T < T_{c}$ ; larger $a$ (higher $T_{c}$ ) means easier liquefaction.

1. Do Real Gases Obey the Ideal Gas Equation?

If $p V = n R T$ held exactly, a plot of $p V$ (or of $Z = p V / n R T$ ) against pressure at constant temperature would be a flat line. Measured data are not flat:

Figure 1: Z vs

p

at 320 K (computed from the van der Waals equation). All gases start at

Z = 1

H_{2}

and He rise at once;

N_{2}

dips slightly (minimum 0.98),

C H_{4}

to 0.83 and

C O_{2}

to 0.38 near 100 bar. At high pressure every gas has

Z > 1

$H_{2}$ and He (at room temperature): $Z$ rises steadily above 1 from the start. This is positive deviation.
$N_{2}$ , $C H_{4}$ , $C O_{2}$ , CO: $Z$ first falls below 1 (negative deviation), reaches a minimum that is deeper for more easily liquefied gases, then rises and crosses 1.
At very low pressure every gas has $Z \to 1$ ; at very high pressure every gas has $Z > 1$ .

The same deviation shows up on a $p$ vs $V$ plot: at high pressure the measured volume of a real gas is larger than the ideal value, while at low pressure the two curves merge.

2. The Compressibility Factor, Z

Compressibility factor:

Z = \frac{p V}{n R T}

. Since

V_{ideal} = \frac{n R T}{p}

Z

is also the ratio of the actual volume of the gas to the volume it would have if it were ideal at the same

p

and

T

Z = \frac{V _{real}}{V _{ideal}}

Figure 2:

Z = \frac{V _{real}}{V _{ideal}}

at the same

p

and

T

. Attractions let the gas shrink below the ideal volume (

Z < 1

); the molecules' own volume keeps it above (

Z > 1

Value	Meaning	Compressibility	Typical case
$Z = 1$	ideal behaviour	as predicted	any gas as $p \to 0$ ; any gas at $T_{B}$ over a range
$Z < 1$	attraction dominates	easier to compress than ideal	$C O_{2}$ , $N H_{3}$ , $C H_{4}$ at moderate $p$
$Z > 1$	repulsion (finite size) dominates	harder to compress than ideal	$H_{2}$ , He at room $T$ ; every gas at very high $p$

Key idea

Z

tells you which force wins: below 1, attraction; above 1, the molecules' own size.

3. Why Real Gases Deviate

Two postulates of the kinetic theory are only approximations, and both fail when molecules are crowded together (high pressure) or moving slowly (low temperature):

No attraction between molecules. If this were true, gases would never liquefy. They do, so attractions exist.
Negligible volume of molecules. If this were true, liquids would be easy to compress. They are not, so molecules have a definite size and repel strongly at contact.

3.1 Pressure correction (attraction)

A molecule about to strike the wall has neighbours only on the inside, so it is pulled back and hits the wall with less force. The measured pressure is therefore lower than the ideal pressure. The pull depends on the number of attracting molecules and the number being attracted, so the correction goes as $(n / V)^{2}$ :

p_{ideal} = p_{real} + \frac{a n ^{2}}{V ^{2}}

3.2 Volume correction (finite size)

Repulsive forces are short range and act only when molecules nearly touch. They make the molecules behave like small, hard spheres. The space that moving molecules can actually use is the container volume minus the volume taken up by the molecules, $V - nb$ .

Figure 3: Two faulty postulates, two corrections. Attraction lowers the measured pressure, so

p_{ideal} = p + \frac{a n ^{2}}{V ^{2}}

. Finite size removes space, so the volume available for motion is

V - nb

The constant $b$ is not simply the volume of the molecules. Two molecules of radius $r$ cannot bring their centres closer than $2 r$ , so each pair shuts out a sphere of radius $2 r$ :

Figure 4: The centre of a second molecule cannot come closer than

2 r

, so each pair excludes

\frac{4}{3} π (2 r)^{3} = 8 v

. Shared between two molecules, that is

4 v

each:

b = 4 N_{A} \cdot \frac{4}{3} π r^{3}

Quick Recall: tap to check

A gas has

Z = 0.8

at some

p

and

T

. Is it easier or harder to compress than an ideal gas?

Easier: its volume is only 80% of the ideal value, so attractions dominate.

Why is the pressure correction proportional to

n^{2} / V^{2}

The inward pull on the molecules near the wall is proportional to both the number of molecules striking the wall and the number pulling them back, each

\propto n / V

b

equal to the volume of 1 mol of molecules?

No, it is about four times that volume:

b = 4 N_{A} \cdot \frac{4}{3} π r^{3}

4. The van der Waals Equation

Putting both corrections into $p V = n R T$ gives the van der Waals equation (1873):

(p + \frac{a n ^{2}}{V ^{2}}) (V - nb) = n R T

Figure 5: The van der Waals equation keeps

n R T

and corrects the two ideal-gas quantities. Units follow from each bracket:

\frac{a n ^{2}}{V ^{2}}

is a pressure, so

a

is in

bar L^{2} mol^{- 2}

;

nb

is a volume, so

b

is in L mol

^{- 1}

$a$ and $b$ are the van der Waals constants, characteristic of each gas. In this model they do not depend on temperature or pressure.

Gas	He	$H_{2}$	$N_{2}$	$O_{2}$	$C H_{4}$	$C O_{2}$	$N H_{3}$	$H_{2} O$
$a$ / bar L $^{2}$ mol $^{- 2}$	0.035	0.248	1.37	1.38	2.28	3.64	4.23	5.54
$b$ / L mol $^{- 1}$	0.024	0.027	0.039	0.032	0.043	0.043	0.037	0.030

Constant a (attraction)

Measures the strength of intermolecular attraction. Larger for polar, H-bonded or large, easily polarised molecules ( $H_{2} O$ , $N H_{3}$ , $C O_{2}$ ). Larger $a$ means easier liquefaction and a higher $T_{c}$ .

Constant b (bulk)

Measures molecular size: four times the actual volume of 1 mol of molecules. Larger for bigger molecules. It makes the gas harder to compress than an ideal gas.

Exam Trick a for attraction, b for bulk. Rank gases for ease of liquefaction by

a

(or

T_{c}

), never by

b

. And watch the units:

a

carries

L^{2} mol^{- 2}

because

\frac{a n ^{2}}{V ^{2}}

must be a pressure.

5. Explaining the Z Curves with the van der Waals Equation

For 1 mol, solve the van der Waals equation for $p$ and multiply by $\frac{V _{m}}{R T}$ :

Z = \frac{p V _{m}}{R T} = \frac{V _{m}}{V _{m} - b} - \frac{a}{R T V _{m}}

Low pressure ( $V_{m}$ large, $b ≪ V_{m}$ ): $Z \approx 1 - \frac{a}{V _{m} R T}$ , so $Z < 1$ . Attraction wins; the dip is deeper for larger $a$ and lower $T$ .
High pressure ( $V_{m}$ small, the $b$ term wins): $Z \approx 1 + \frac{p b}{R T}$ , a straight line rising with $p$ . Every gas ends with $Z > 1$ .
$H_{2}$ and He: $a$ is so small that the attraction term never wins at room temperature, so $Z = 1 + \frac{p b}{R T} > 1$ from the start.
Very low pressure or very high temperature: both corrections become negligible and $Z \to 1$ .

Key idea

Negative deviation = the $a$ term winning; positive deviation = the $b$ term winning. The pressure and temperature decide which.

6. Boyle Temperature

Boyle temperature ( $T_{B}$ ): the temperature at which a real gas obeys the ideal gas law (Boyle's law) over an appreciable range of pressure. For a van der Waals gas,

T_{B} = \frac{a}{R b}

Figure 6: One gas, four temperatures (

N_{2}

, van der Waals). Below

T_{B}

the dip deepens as

T

falls (minimum 0.53 at 150 K). At

T_{B} = \frac{a}{R b} = 426

K the curve leaves

Z = 1

with zero slope. Above

T_{B}

Z > 1

at all pressures, like

H_{2}

and He at room temperature.

Below $T_{B}$ : $Z$ first decreases with pressure, passes through a minimum, then rises above 1.
At $T_{B}$ : the attraction and size effects cancel at low and moderate pressure, and $Z \approx 1$ .
Above $T_{B}$ : $Z > 1$ at all pressures; attractions are too feeble to matter.

$T_{B}$ depends on the nature of the gas. $H_{2}$ (about 110 K) and He (about 23 K) are far above their Boyle temperatures at room temperature, which is why they show only positive deviation. Real gases therefore behave most ideally at low pressure and high temperature.

JEE Advanced Where $T_{B}$ comes from. At low pressure put

V_{m} \approx \frac{R T}{p}

into the van der Waals expression for

Z

and expand:

Z \approx 1 + (b - \frac{a}{R T}) \frac{p}{R T}

. The initial slope

\frac{d Z}{d p}

is negative when

\frac{a}{R T} > b

and zero when

T = \frac{a}{R b} = T_{B}

. Combining with

T_{c} = \frac{8 a}{27 R b}

gives

T_{B} = \frac{27}{8} T_{c} \approx 3.4 T_{c}

. The model also predicts

Z_{c} = \frac{3}{8} = 0.375

for every gas; real gases give 0.23 to 0.31 (0.29 for

N_{2}

), a reminder that the van der Waals equation is only approximate near the critical point.

7. Liquefaction of Gases and Critical Constants

Thomas Andrews obtained the first complete $p$ - $V$ - $T$ data for a substance in both gas and liquid states by compressing $C O_{2}$ at several fixed temperatures. Every real gas behaves the same way:

Figure 7: Isotherms of

C O_{2}

(van der Waals model, flat parts from the equal-area rule). On compressing at 21.5 °C, liquid appears at B, pressure stays fixed while gas condenses, and at C all is liquid; after C the steep line shows how incompressible the liquid is. The flat part shrinks as

T

rises and vanishes at the critical point E; above

T_{c}

no pressure produces a liquid.

At 50 °C the isotherm looks like that of an ideal gas: no pressure, however high, produces a liquid.
At 21.5 °C the gas is compressed until point B, where liquid first appears. Further compression condenses more gas at constant pressure until point C, where all is liquid.
Beyond C the line is almost vertical: a small volume change needs a huge pressure because liquids are nearly incompressible.
The flat (two-phase) part gets shorter as temperature rises and shrinks to a single point, the critical point, at $T_{c} = 30.98$ °C for $C O_{2}$ .

Critical temperature ( $T_{c}$ ): the highest temperature at which a gas can be liquefied; above it no pressure produces a liquid. Critical pressure ( $p_{c}$ ): the pressure needed to liquefy the gas at

T_{c}

. Critical volume ( $V_{c}$ ): the molar volume at

T_{c}

and

p_{c}

A gas below its critical temperature can be liquefied by pressure alone and is called a vapour; above $T_{c}$ it is a gas. $C O_{2}$ below 30.98 °C is carbon dioxide vapour.

Substance	$T_{c}$ / K	$p_{c}$ / bar	$V_{c}$ / dm $^{3}$ mol $^{- 1}$
$H_{2}$	33.2	12.97	0.0650
He	5.3	2.29	0.0577
$N_{2}$	126.0	33.9	0.0900
$O_{2}$	154.3	50.4	0.0744
$C O_{2}$	304.10	73.9	0.0956
$H_{2} O$	647.1	220.6	0.0450
$N H_{3}$	405.5	113.0	0.0723

Figure 8: Critical temperature measures intermolecular attraction.

N H_{3}

H_{2} O

and (just)

C O_{2}

have

T_{c}

above 298 K, so squeezing them at room temperature gives a liquid. He,

H_{2}

N_{2}

O_{2}

and

C H_{4}

(once called permanent gases) must first be cooled below

T_{c}

So called permanent gases ( $H_{2}$ , He, $N_{2}$ , $O_{2}$ ), which show continuous positive deviation at room temperature, need cooling as well as compression. Cooling slows the molecules; compression brings them close; together they let the attractions hold the molecules in a liquid.

7.1 Critical constants from a and b

The critical isotherm has a horizontal point of inflection at the critical point ( $\frac{d p}{d V} = 0$ and $\frac{d ^{2} p}{d V ^{2}} = 0$ ). Applying these to the van der Waals equation gives:

V_{c} = 3 b, p_{c} = \frac{a}{27 b ^{2}}, T_{c} = \frac{8 a}{27 R b}

Exam Trick Higher $T_{c}$ = stronger attraction = liquefies first on cooling. Cooling a mixture from a high temperature, the gas with the higher

T_{c}

condenses first;

N H_{3}

(405.5 K) before

C O_{2}

(304.1 K) before

N_{2}

(126 K).

Key idea

Liquefaction needs $T < T_{c}$ , then enough pressure. Above

T_{c}

, pressure only compresses the gas.

Quick Recall: tap to check

C O_{2}

at 40 °C is compressed to 200 bar. Does it liquefy?

No: 40 °C is above

T_{c}

(31 °C), so it stays a dense supercritical fluid.

Find

T_{B}

of a van der Waals gas whose

T_{c} = 200

T_{B} = \frac{27}{8} T_{c} = 675

Which has the larger

a

N H_{3}

N_{2}

? Why?

N H_{3}

: it is polar and hydrogen-bonded, so its molecules attract strongly (higher

T_{c}

too).

8. Continuity of State

A gas can be turned into a liquid without ever seeing two phases. Take the route around the critical point:

Figure 9: Continuity of state. Heat the gas at constant volume (A to F), compress it above

T_{c}

(F to G), then cool at constant volume (G to D). It becomes liquid after crossing the critical isotherm at H, yet no boiling or condensing surface ever appears. Gas and liquid are two ends of one fluid state.

Along A to F to G to D the substance is always a single phase, yet it starts as a gas and ends as a liquid. Gas and liquid are therefore not fundamentally different: a liquid is a very dense gas. Both are called fluids, and they can be told apart only below $T_{c}$ , inside the two-phase dome, where a visible surface separates them. At the critical temperature that surface disappears and the densities of liquid and vapour become equal.

9. Solving Real-Gas Problems

Figure 10: Two temperatures decide almost everything:

T_{B} = \frac{a}{R b}

for the sign of the deviation, and

T_{c} = \frac{8 a}{27 R b}

for liquefaction. Since

T_{B} = 3.375 T_{c}

, a gas below its

T_{c}

is always below its

T_{B}

too.

9.1 The whole concept at a glance

Figure 11: The whole concept on one page. Every branch comes from two facts: molecules attract (a) and molecules take up space (b).

10. Solved Examples

Solved Example 1

The critical temperatures of ammonia and carbon dioxide are 405.5 K and 304.10 K. Which gas liquefies first when both are cooled from 500 K towards their critical temperatures?

Solution:

Ammonia. Cooling from 500 K reaches 405.5 K before 304.1 K, so $N H_{3}$ can be liquefied first. $C O_{2}$ needs more cooling. The higher $T_{c}$ reflects the stronger intermolecular attraction (hydrogen bonding) in $N H_{3}$ .

Solved Example 2

At 300 K and 100 bar, the molar volume of methane is 0.200 L mol

^{- 1}

. Find

Z

and say which force dominates.

Solution:

Z = \frac{p V _{m}}{R T} = \frac{100 \times 0.200}{0.08314 \times 300} = \frac{20.0}{24.94} = 0.80

$Z < 1$ : attraction dominates, so the real volume is 20% smaller than the ideal volume.

Solved Example 3

Calculate the pressure of 1.00 mol of

C O_{2}

in a 1.00 L vessel at 300 K using (a) the ideal gas equation and (b) the van der Waals equation (

a = 3.64

bar L

^{2}

mol

^{- 2}

b = 0.0427

L mol

^{- 1}

Solution:

(a) $p = \frac{R T}{V} = 0.08314 \times 300 = 24.94$ bar.

(b) p = \frac{R T}{V - b} - \frac{a}{V ^{2}} = \frac{24.94}{0.9573} - 3.64 = 26.05 - 3.64 = 22.41 bar

The real pressure is about 10% lower ( $Z = 0.90$ ): the attraction term outweighs the size term here.

Solved Example 4

For

N_{2}

a = 1.370

bar L

^{2}

mol

^{- 2}

and

b = 0.0387

L mol

^{- 1}

. Calculate its critical constants and Boyle temperature from the van der Waals equation.

Solution:

$T_{c} = \frac{8 a}{27 R b} = \frac{8 \times 1.370}{27 \times 0.08314 \times 0.0387} = 126.2$ K; $p_{c} = \frac{a}{27 b ^{2}} = \frac{1.370}{27 \times ( 0.0387 ) ^{2}} = 33.9$ bar.

$V_{c} = 3 b = 0.116$ L mol $^{- 1}$ ; $T_{B} = \frac{a}{R b} = \frac{1.370}{0.08314 \times 0.0387} = 426$ K.

The measured $T_{c}$ (126.0 K) and $p_{c}$ (33.9 bar) match well, but the measured $V_{c}$ is 0.090 L and the measured $T_{B}$ is about 327 K, so the model is rougher for these two.

Solved Example 5

Estimate the radius of an

N_{2}

molecule from

b = 0.0387

L mol

^{- 1}

Solution:

$b = 3.87 \times 1 0^{- 5}$ m $^{3}$ mol $^{- 1} = 4 N_{A} \cdot \frac{4}{3} π r^{3}$ , so the volume of one molecule is $\frac{3.87 \times 1 0 ^{- 5}}{4 \times 6.022 \times 1 0 ^{23}} = 1.61 \times 1 0^{- 29}$ m $^{3}$ .

$r = (\frac{3 \times 1.61 \times 1 0 ^{- 29}}{4 π})^{1/3} = 1.57 \times 1 0^{- 10}$ m $\approx 157$ pm.

Solved Example 6

Which gas has

Z > 1

at all pressures at

0^{\circ} C

?
(A)

C O_{2}

(B)

C H_{4}

(C)

H_{2}

(D)

N H_{3}

Solution:

Answer: (C). The Boyle temperature of $H_{2}$ (about 110 K) is far below 273 K, so its attraction term never wins and it shows only positive deviation. The other three are below their Boyle temperatures and dip below $Z = 1$ first.

Solved Example 7

For a van der Waals gas whose constant

a

can be neglected, a plot of

Z

against

p

at constant

T

is:
(A) a horizontal line at

Z = 1

(B) a straight line of positive slope

\frac{b}{R T}

Solution:

Answer: (B). With $a = 0$ , $p (V_{m} - b) = R T$ , so $p V_{m} = R T + p b$ and $Z = 1 + \frac{b}{R T} p$ : a line starting at 1 with slope $\frac{b}{R T}$ .

Practice Questions

Critical temperatures of $C O_{2}$ and $C H_{4}$ are $31.1^{\circ} C$ and $- 81.9^{\circ} C$ . Which has stronger intermolecular forces and why? (Ex. 5.22)Answer: $C O_{2}$ : a higher $T_{c}$ means stronger attractions, so it can be liquefied at a higher temperature.
Explain the physical significance of the van der Waals parameters. (Ex. 5.23)Answer: $a$ measures intermolecular attraction (pressure correction); $b$ is the excluded volume, about four times the actual volume of 1 mol of molecules (volume correction).
1.00 mol of a gas occupies 20.0 L at 273 K and 1 atm. Find $Z$ and name the dominant force.Answer: $Z = \frac{1 \times 20.0}{0.0821 \times 273} = 0.892$ : attraction dominates.
Give the SI units of $a$ and $b$ .Answer: $a$ : Pa m $^{6}$ mol $^{- 2}$ (= N m $^{4}$ mol $^{- 2}$ ); $b$ : m $^{3}$ mol $^{- 1}$ .
Estimate $a$ and $b$ for $C O_{2}$ from $T_{c} = 304.1$ K and $p_{c} = 73.9$ bar.Answer: $b = \frac{R T _{c}}{8 p _{c}} = 0.0428$ L mol $^{- 1}$ ; $a = \frac{27 R ^{2} T _{c}^{2}}{64 p _{c}} = 3.65$ bar L $^{2}$ mol $^{- 2}$ .
Boyle temperature of $C H_{4}$ ( $a = 2.283$ , $b = 0.0428$ )? Will its $Z$ be above or below 1 at 300 K and 50 bar?Answer: $T_{B} = \frac{a}{R b} = 642$ K; 300 K is below $T_{B}$ , so $Z < 1$ (about 0.90).
Under what conditions does the van der Waals equation reduce to $p V = n R T$ ?Answer: Low pressure and high temperature: $V ≫ nb$ and $\frac{a n ^{2}}{V ^{2}} ≪ p$ .

Common Mistakes to Avoid

Watch out

Reading $Z < 1$ as repulsion. $Z < 1$ means attraction wins (the gas is more compressible than ideal); $Z > 1$ means size (repulsion) wins.
Taking $b$ as the actual volume of the molecules. It is about four times that volume.
Ranking ease of liquefaction by $b$ . Use $a$ or $T_{c}$ : larger $a$ , higher $T_{c}$ , easier liquefaction.
Trying to liquefy a gas above $T_{c}$ by raising the pressure. Above $T_{c}$ no pressure gives a liquid.
Writing the pressure correction as $\frac{a}{V}$ or $\frac{an}{V ^{2}}$ . It is $\frac{a n ^{2}}{V ^{2}}$ (for 1 mol, $\frac{a}{V _{m}^{2}}$ ).
Saying $H_{2}$ and He can never show $Z < 1$ . They do, below their (very low) Boyle temperatures.
Confusing critical temperature with boiling point. Boiling point depends on external pressure; $T_{c}$ is the top of the liquid-vapour curve.
Using $T_{B} = \frac{8 a}{27 R b}$ (that is $T_{c}$ ). $T_{B} = \frac{a}{R b}$ , which is $\frac{27}{8}$ times larger.

Frequently Asked Questions

What is the compressibility factor Z of a gas?

The compressibility factor is Z = pV/nRT, equal to the actual volume of a gas divided by the volume it would have if ideal at the same pressure and temperature. For an ideal gas Z = 1. Z below 1 means attractions dominate and the gas is easier to compress; Z above 1 means molecular size dominates.

Why do real gases deviate from ideal gas behaviour?

The ideal gas model assumes molecules have no volume and do not attract each other. Real molecules attract, which lowers the pressure on the walls, and they occupy space, which reduces the free volume. Both effects become important at high pressure and low temperature, when molecules are close together and slow.

What is the physical significance of van der Waals constants a and b?

The constant a measures the strength of attraction between gas molecules and appears in the pressure correction a n squared over V squared. The constant b is the excluded volume per mole, about four times the actual volume of the molecules, and appears in the volume correction V − nb. Larger a means easier liquefaction.

What is Boyle temperature?

Boyle temperature is the temperature at which a real gas obeys Boyle's law over a wide pressure range, because attraction and molecular size effects cancel. For a van der Waals gas it equals a/Rb. Above it Z is greater than 1 at all pressures; below it Z first dips below 1.

What is critical temperature and why must a gas be cooled below it to liquefy?

Critical temperature is the highest temperature at which a gas can be liquefied. Above it, molecules move too fast for attractions to hold them together, so any pressure only compresses the gas. Below it, pressure brings molecules close enough to condense. For carbon dioxide the critical temperature is 30.98 °C.

Why do hydrogen and helium show only positive deviation at room temperature?

Hydrogen and helium have very weak intermolecular attractions, so their van der Waals constant a is tiny. Their Boyle temperatures, about 110 K and 23 K, are far below room temperature, so the molecular size term always wins and Z stays above 1 as pressure rises.

Are real gases and the van der Waals equation in the JEE Advanced 2026 syllabus?

Yes. The JEE Advanced syllabus lists deviation from ideality and the van der Waals equation under States of Matter: Gases and Liquids. The chapter has been removed from NCERT Class 11 and from the JEE Main and NEET syllabi, so these ideas are examined in JEE Advanced.

What types of real gas questions are asked in JEE Advanced?

JEE Advanced asks for the sign of deviation from Z vs p graphs, Z from given data, van der Waals pressure calculations, critical constants and Boyle temperature from a and b, ordering gases by ease of liquefaction, and limiting forms of the van der Waals equation when a or b is neglected.

Previous year questions on Deviation From Ideal Gas Behaviour

2 questions from past papers, each with a step-by-step solution.

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