Real gases — deviations from ideality, compressibility factor Z: Question bank

Quick reminders before you start:

$Z = \frac{P V _{m}}{R T}$ — real world $P V_{m}$ divided by ideal-world $R T$ . Match $\Rightarrow Z = 1$ .
$Z = \frac{V _{m}}{V _{m} - b} - \frac{a}{R T V _{m}}$ — a size term (raises $Z$ ) fighting an attraction term (lowers $Z$ ).
$b$ = space molecules physically occupy; $a$ = strength of their mutual pull; $T_{B} = \frac{a}{R b}$ .

Figure — Real gases — deviations from ideality, compressibility factor Z

Intuition What the compass figure shows (read this even if you can't see the image)

Picture a horizontal see-saw with a yellow pivot at its centre.

Horizontal axis = which correction term wins the tug of war. Slide left and the pink attraction term $- a / (R T V_{m})$ dominates; slide right and the blue size term $V_{m} / (V_{m} - b)$ dominates.
Left arm (pink) is labelled " $Z < 1$ (low P)" — attraction pulls the score below 1.
Right arm (blue) is labelled " $Z > 1$ (high P)" — finite size heaves the score above 1.
Yellow pivot at the centre is labelled " $Z = 1$ " — the balance point, reached either by a truly ideal gas or by a real gas at an accidental crossing (including the Boyle-temperature low- $P$ crossing).
Vertical axis carries no number; it only holds the text labels. Height above the bar = "name of the winning term"; depth below the bar = "what the term evaluates to and the resulting $Z$ verdict." So reading top-to-bottom on either side gives you effect → algebra → $Z$ . There are no separate quadrants to decode: left half ⇒ $Z < 1$ , centre ⇒ $Z = 1$ , right half ⇒ $Z > 1$ , full stop.

True or false — justify

Every real gas has

Z < 1

at low enough pressure.

False. Only if attraction wins the low-

P

slope, i.e.

a / R T > b

. Below its Boyle temperature this holds, but for

H_{2}

He

at room

T

we have

b > a / R T

, so

Z > 1

even at low pressure.

Z = 1

means the gas is behaving as an ideal gas.

Mostly true, with a catch.

Z = 1

means

P V_{m} = R T

at that point, but a real gas can hit

Z = 1

accidentally where the size and attraction corrections happen to cancel — it is not ideal everywhere, just at that crossing.

A gas at its Boyle temperature is ideal at all pressures.

False. At

T_{B}

only the initial (low- $P$ ) slope of

Z

P

is zero. At higher

P

the size term

V_{m} / (V_{m} - b)

still climbs, so

Z

eventually rises above 1.

If two gases have the same

Z

at the same

P, T

, they deviate from ideality identically.

False. The same net

Z

can come from very different balances of

a

and

b

; one gas might have strong attraction and large size that nearly offset, another weak versions of both.

Raising temperature (at fixed

P

) always pushes

Z

toward 1.

True. At fixed

P

the low-density vdW form is

Z \approx 1 + \frac{b P}{R T} - \frac{a}{R T V _{m}}

; both correction terms carry a

1/ T

, so heating shrinks each one and

∣ Z - 1∣

decreases monotonically —

Z

moves toward 1. It approaches 1 without necessarily reaching it, but the direction is always inward.

The van der Waals

a

has units of pressure.

False.

a

carries

atm\cdotL^{2} mol^{- 2}

; only the combination

a / V_{m}^{2}

has pressure units.

a

alone is a coupling strength, not a pressure.

Z = 1

crossing on its way up, a gas has zero intermolecular attraction.

False. Attraction is still present; the crossing means the attraction term and size term are momentarily equal in magnitude, not that either vanished.

Spot the error

"At high pressure molecules are closer, so attraction dominates and

Z

falls below 1."

The error is ignoring finite volume. At high

P

the

V_{m} / (V_{m} - b)

term explodes as

V_{m} \to b

, so the size effect wins and

Z

rises above 1 for every real gas.

He

shows

Z > 1

, therefore helium has the strongest intermolecular forces."

Backwards.

Z > 1

from small pressure means

b > a / R T

, which requires a tiny

a

— helium's attraction is among the weakest of all gases.

"Since

Z < 1

means 'more compressible', a

Z < 1

gas will keep compressing forever more easily."

No — as you squeeze further,

V_{m}

drops toward

b

and the size term takes over. The easy-compression regime is temporary;

Z

turns around and climbs.

"The correction to pressure is

a / V_{m}

because each molecule feels one pull."

It is

a / V_{m}^{2}

. There are two density factors — one for the molecule hitting the wall, one for the neighbours pulling it back — a two-body interaction, hence the square.

"We subtract

b

from pressure and add

a / V_{m}^{2}

to volume in the vdW equation."

Swapped.

b

(excluded volume) is subtracted from

V_{m}

;

a / V_{m}^{2}

(internal pressure) is added to

P

. Each correction fixes its own kind of lie.

b

equals the actual volume of one mole of molecules."

It is roughly four times the true molecular volume per mole, because two hard spheres exclude a sphere of radius equal to their contact distance, not just their own bodies.

"Because

P V = n R T

was derived for ideal gases,

Z = P V / n R T

can never exceed 1."

Z

is defined by that ratio but measured with real

P, V

; nothing forbids

P V > n R T

. Whenever the real product beats the ideal prediction,

Z > 1

Why questions

Why is the attraction term in

Z

divided by

T

(as

a / R T V_{m}

) while the size term has no explicit

T

Attraction competes with thermal kinetic energy: hotter molecules fly past each other, so the effect of

a

weakens as

T

rises — an explicit

1/ T

. The size term

b

has no explicit

T

because a molecule's body never shrinks; it only feels

T

indirectly, through

V_{m}

changing.

In the low-

P

expansion, why are we allowed to keep only the

1/ V_{m}

term and throw away

1/ V_{m}^{2}

and higher?

Because at low pressure

V_{m}

is large, so

b / V_{m}

is a small number; writing

\frac{V _{m}}{V _{m} - b} = \frac{1}{1 - b / V _{m}} = 1 + \frac{b}{V _{m}} + (\frac{b}{V _{m}})^{2} + \dots

, each further term is a smaller power of that small quantity. Truncating at first order (the virial expansion in powers of

1/ V_{m}

) captures the initial slope; the dropped terms only matter once

P

grows.

Why does the naive "high

P

= more deviation down" intuition fail specifically at very high

P

Because attraction has a ceiling: once molecules are packed, they are already as close as their hard cores allow, so the pull can't intensify further, but the volume they occupy becomes a large fraction of the container and dominates.

Why does

Z < 1

happen at low pressure rather than high pressure?

At low

P

molecules are far apart, so their finite size is negligible (

b ≪ V_{m}

) and the only surviving correction is the long-range attraction, which lowers measured pressure and drags

Z

below 1.

Why is

T_{B} = a / (R b)

the exact temperature where the low-

P

slope vanishes?

The low-

P

expansion gives

Z \approx 1 + \frac{1}{V _{m}} (b - \frac{a}{R T})

; the slope is zero when

b = a / R T_{B}

, and solving that bracket for

T

gives

T_{B} = a / (R b)

Why can a large, polarizable molecule like

CO_{2}

dip further below

Z = 1

than

H_{2}

Polarizability means strong induced-dipole attraction, so its

a

is large. A big

a / R T

overwhelms

b

at ordinary temperatures, deepening the

Z < 1

dip.

Why do we divide measured

P V_{m}

R T

rather than compare

P V_{m}

to some absolute number?

R T

is the ideal prediction for

P V_{m}

, so the ratio is a fair "real vs ideal" score that stays dimensionless and comparable across all gases and conditions.

Above

T_{B}

, why is the initial slope of

Z

P

positive even though

Z \to 1

P \to 0

Two different statements. As

P \to 0

every real gas has

Z \to 1

— that is the fixed starting point (infinite dilution, no corrections). Above

T_{B}

the attraction term is too weak to cancel

b

, so leaving that point the curve slopes upward:

Z

climbs above 1 immediately. Heating doesn't move the

P \to 0

anchor at 1; it fixes the sign of the slope out of it.

Edge cases

What is

Z

for a truly ideal gas at 500 atm and 50 K?

Exactly ==

1

==, by construction — an ideal gas has no

a

and no

b

, so

P V_{m} = R T

holds at every

P

and

T

, however extreme.

V_{m} \to b

(extreme compression), what happens to

Z

The term

V_{m} / (V_{m} - b) \to \infty

, so

Z \to + \infty

; the gas becomes essentially incompressible because molecules are touching — you cannot force them any closer.

V_{m} \to \infty

(near-zero pressure), what does

Z

approach and why?

Z \to 1

: both correction terms carry a

1/ V_{m}

that vanishes, so the gas recovers ideal behaviour when molecules are infinitely dilute.

If a gas had

a = 0

but

b \neq = 0

(hard spheres, no pull), what does

Z

look like?

Z = V_{m} / (V_{m} - b) > 1

always — with no attraction the only deviation is the size term, so

Z

exceeds 1 at every pressure and never dips below.

If a gas had

b = 0

but

a \neq = 0

(point particles that attract), what does

Z

look like?

Z = 1 - \frac{a}{R T V _{m}} < 1

always — pure attraction lowers

Z

below 1 at all finite densities, deepening as

V_{m}

shrinks, with no rising branch.

Does a Boyle temperature exist for an ideal gas?

No.

T_{B} = a / (R b)

requires nonzero

a

and

b

; an ideal gas has

a = b = 0

, so the formula is

0/0

and there is simply nothing to correct.

At its critical temperature and pressure, is a real gas's

Z

equal to 1?

No — for the van der Waals model specifically the critical compressibility is

Z_{c} = 3/8 \approx 0.375

. Real gases scatter around

Z_{c} \approx 0.27

–

0.31

, so

3/8

is a model prediction, not a universal number — do not generalize it to every gas. See Critical Constants and Liquefaction.

Recall One-line self-test

Cover everything above. Can you state, for each of $Z < 1$ , $Z = 1$ , $Z > 1$ : which term wins, at what pressure regime, and one gas/condition that shows it? If yes, you've mastered the traps.

Related build-up: Ideal Gas Law · Intermolecular Forces · Kinetic Theory of Gases · Boyle Temperature · Critical Constants and Liquefaction.