Critical constants Tc, Pc, Vc; law of corresponding states: Question bank

Definition Symbols you will need on this page

Before any trap uses them, here is every symbol, in plain words:

== $a$ == = the van der Waals attraction constant. It measures how strongly molecules pull on each other; a bigger $a$ means stronger Intermolecular Forces. It appears as the $a / V^{2}$ pressure-correction term in Van der Waals Equation of State.
== $b$ == = the van der Waals excluded-volume constant — roughly the volume the molecules themselves take up, the space unavailable to their motion.
== $R$ == = the universal gas constant ( $\approx 0.0821 L atm mol^{- 1} K^{- 1}$ ), the fixed proportionality that links pressure×volume to temperature in the Ideal Gas Equation.
== $P, V, T$ == = actual pressure, molar volume, temperature. Subscript $c$ (as in $P_{c}, V_{c}, T_{c}$ ) means "the value at the critical point."
Reduced variables rescale each property by its own critical value: $P_{r} = P / P_{c}$ , $V_{r} = V / V_{c}$ , $T_{r} = T / T_{c}$ . They are pure numbers (no units) telling you "what fraction of critical."
== $Z$ == = the compressibility factor $P V / R T$ ; $Z_{c}$ is its value at the critical point.

Recall The two equations referred to below (from the parent derivation)

Setting the first and second derivatives of $P$ to zero at the critical point gives: $\frac{R T}{( V - b ) ^{2}} = \frac{2 a}{V ^{3}} (equation 1: horizontal tangent)$ $\frac{2 R T}{( V - b ) ^{3}} = \frac{6 a}{V ^{4}} (equation 2: zero curvature)$ Dividing (2) by (1) cancels $R$ , $T$ and $a$ and yields $V_{c} = 3 b$ . These are the "equation (1)" and "equation (2)" the traps below point to.

True or false — justify

TF1. Above the critical temperature

T_{c}

, applying enough pressure will eventually liquefy the gas.

False. Above

T_{c}

the isotherm has no coexistence plateau, so compressing just yields a dense supercritical fluid with no meniscus — never a distinct liquid phase.

TF2. At the critical point the liquid and gas phases have the same density.

True. That equality is why the meniscus vanishes: the two phases become indistinguishable, so there is no density step (no surface) between them.

TF3. The van der Waals constant

b

equals the critical molar volume

V_{c}

False. The inflection conditions give

V_{c} = 3 b

; even at the critical point there is free space beyond the excluded volume

b

, so molecules are not packed to their hard-core limit.

TF4. The van der Waals equation predicts the same critical compressibility factor

Z_{c}

for every gas.

True.

Z_{c} = P_{c} V_{c} / R T_{c} = 3/8 = 0.375

contains no attraction constant

a

or excluded-volume constant

b

, so vdW forces one universal value — a clue that leads to corresponding states.

TF5. Because vdW predicts

Z_{c} = 0.375

, real gases have

Z_{c} \approx 0.375

too.

False. Real gases give

Z_{c} \approx 0.27

–

0.31

. The systematic gap shows vdW is only qualitatively correct near the critical point.

TF6. Two different gases at the same actual temperature and pressure are in corresponding states.

False. Corresponding states requires equal reduced values

T_{r} = T / T_{c}

and

P_{r} = P / P_{c}

, not equal absolute values; the same fraction of each gas's own critical conditions is what matters.

TF7. The reduced equation

(P_{r} + 3/ V_{r}^{2}) (3 V_{r} - 1) = 8 T_{r}

, written in the reduced variables

P_{r} = P / P_{c}

V_{r} = V / V_{c}

T_{r} = T / T_{c}

, still contains hidden gas-specific constants.

False. Every attraction constant

a

and excluded-volume constant

b

cancelled during reduction; the only "constants" left (3, 8, 1) are pure numbers, which is precisely why one chart serves all gases.

TF8. A gas with stronger intermolecular attractions tends to have a higher

T_{c}

True. Since

T_{c} = 8 a /27 R b

(with

a

the attraction constant,

b

the excluded volume,

R

the gas constant), a larger

a

pushes

T_{c}

up — stronger Intermolecular Forces survive faster molecular motion.

TF9. At the critical point the isotherm's slope

\partial P / \partial V

is positive.

False. It is exactly zero (horizontal tangent), together with zero curvature; a positive slope would violate mechanical stability.

TF10. The critical point is a single point on the

P

–

V

diagram, not a plateau.

True. As

T \to T_{c}

the coexistence plateau shrinks to one point of inflection

(P_{c}, V_{c})

; below

T_{c}

it is a flat segment, above

T_{c}

it never appears.

Spot the error

SE1. "To find the excluded-volume constant

b

from experiment I should use

V_{c} = 3 b

, since it's the simplest formula."

Error:

V_{c}

is the hardest critical constant to measure accurately. Prefer

b = R T_{c} / (8 P_{c})

, which uses the well-measured

T_{c}

and

P_{c}

(with

R

the gas constant) and cancels the attraction constant

a

SE2. "The critical point is defined by

\partial P / \partial V = 0

alone."

Error: A single horizontal tangent could be a local max or min. An inflection also needs

\partial^{2} P / \partial V^{2} = 0

; both conditions together pin the critical point.

SE3. "Since

Z = 1

for an ideal gas, an ideal gas has

Z_{c} = 1

at its critical point."

Error: An ideal gas has no attractions and never liquefies, so it has no critical point at all —

Z_{c}

is undefined for it.

SE4. "

P_{c} = a /27 b^{2}

, so raising the attraction constant

a

always lowers the critical pressure."

Error: Raising

a

increases

P_{c}

(it sits in the numerator). Confusing which variable is on top is the trap; more attraction demands more pressure to reach criticality.

SE5. "Corresponding states means all gases have identical

P

–

V

curves."

Error: They have identical reduced

P_{r}

–

V_{r}

curves. In real (absolute) variables each gas is scaled by its own

P_{c}, V_{c}, T_{c}

, so the raw curves differ.

SE6. "Dividing the second derivative equation by the first eliminates

R

and

a

, so it's an approximation."

Error: It is exact algebra, not an approximation — it cleanly yields

V_{c} = 3 b

because the cancellation leaves pure geometry with no lost information.

SE7. "Because real

Z_{c} < 0.375

, the vdW equation overestimates the critical volume."

Error (subtle): A lower

Z_{c}

means the real product

P_{c} V_{c} / R T_{c}

is smaller than vdW predicts; you cannot single out

V_{c}

as the culprit — all three constants and the equation's form contribute.

Why questions

WHY1. Why do we set both the first and second derivatives to zero at the critical point?

The plateau collapsing to a point means the tangent is horizontal (first derivative zero) and the curve stops curving (second derivative zero) — an inflection with a flat tangent needs both.

WHY2. Why does dividing equation (2) by equation (1) — the two derivative equations shown in the collapsible box above — leave a purely geometric relation?

Both equations share the factors

R

(the gas constant),

T

, and

a

(the attraction constant); dividing cancels them, so only volume terms survive, giving

2/ (V - b) = 3/ V

and hence

V_{c} = 3 b

WHY3. Why is a universal

Z_{c}

from vdW considered a clue rather than just a curiosity?

If one dimensionless combination is the same for all gases, it hints that rescaling every gas by its own critical constants erases individuality — exactly the law of corresponding states.

WHY4. Why can we predict an unknown gas's behaviour from a single generalized

Z

-vs-

P_{r}

chart?

Because in reduced variables the equation of state has no gas-specific constants, so every gas plotted against

T_{r}

and

P_{r}

falls (approximately) on one master surface.

WHY5. Why does increasing the excluded-volume constant

b

(bigger molecules) lower the critical temperature?

T_{c} = 8 a /27 R b

b

is in the denominator; larger molecular size raises the excluded volume, weakening the effective pull-per-space and letting the gas resist liquefaction until lower temperatures.

WHY6. Why does the meniscus physically disappear rather than just becoming faint?

The meniscus is the density boundary between liquid and gas; when their densities converge to equality at

T_{c}

, there is literally no boundary left to see.

WHY7. Why is

T_{c}

called a "hard ceiling" for liquefaction by pressure?

Above

T_{c}

molecular kinetic energy overwhelms attractions for any density, so no pressure can produce two coexisting phases — cooling below

T_{c}

first is mandatory.

Edge cases

EC1. What happens to the reduced equation of state at the critical point itself, where the reduced variables are

T_{r} = P_{r} = V_{r} = 1

It must be satisfied:

(1 + 3) (3 - 1) = 8 = 8 \times 1

✓ — the reduction is built to make the critical point land at all-ones.

EC2. If a hypothetical gas had attraction constant

a = 0

(no attractions), what happens to

T_{c}

and

P_{c}

Both become zero:

T_{c} = 8 a /27 R b = 0

and

P_{c} = a /27 b^{2} = 0

. With no attractions there is no critical point and no liquefaction, matching the Ideal Gas Equation limit.

EC3. If the excluded volume

b = 0

(point molecules) but

a \neq = 0

, what happens to

T_{c}

T_{c} = 8 a /27 R b \to \infty

: with no excluded volume the attractions can always win at high enough density, so the "critical ceiling" diverges — an unphysical degenerate case.

EC4. In the reduced equation, what is the smallest physically allowed reduced volume

V_{r} = V / V_{c}

V_{r} > 1/3

, because the factor

(3 V_{r} - 1)

must stay positive — it is the reduced form of

V > b

, the volume can never fall to the excluded volume.

EC5. Two gases share the same

T_{r}

but sit at different

P_{r}

. Are they in corresponding states?

No. Corresponding states requires matching reduced temperature and reduced pressure simultaneously; only then do they share the same

V_{r}

and same

Z

EC6. At exactly

T = T_{c}

, does compressing the gas ever show a flat plateau?

No flat segment — only a single inflection point at

V_{c}

. The plateau exists strictly below

T_{c}

; at

T_{c}

it has already shrunk to that lone point.

Recall Quick self-test

One vanishing derivative is enough to define the critical point. ::: False — you need both $\partial P / \partial V = 0$ and $\partial^{2} P / \partial V^{2} = 0$ . Equal absolute $T$ and $P$ put two gases in corresponding states. ::: False — equal reduced $T_{r}$ and $P_{r}$ do. Real gases obey $Z_{c} = 3/8$ . ::: False — real $Z_{c} \approx 0.27$ – $0.31$ ; only vdW predicts $0.375$ .