Critical constants Tc, Pc, Vc; law of corresponding states: Worked examples

Before we start, one reminder of the toolkit we will re-use everywhere (everything already earned in the parent note):

The scenario matrix

Every exam question on this topic is one (or a blend) of the cells below. The worked examples that follow are tagged with the cell they hit, and together they cover every cell.

Cell	Case class	What is given → asked	Trap to watch
A	Forward: constants → $a, b$	$T_{c}, P_{c}$ known → find $a, b$	pick the pair that dodges $V_{c}$
B	Backward: $a, b$ → constants	$a, b$ known → find $T_{c}, P_{c}, V_{c}$	plug straight in
C	Cross-check via $Z_{c}$	three constants given → is data self-consistent?	real $Z_{c} \neq = 0.375$
D	Reduced equation of state	$T_{r}, V_{r}$ → find $P_{r}$ (or vice-versa)	no $a, b$ should survive
E	Corresponding states matching	two gases, one $T_{r}$ → actual $T$ 's / same $Z$	equal $T_{r}$ AND $P_{r}$ needed
F	Limiting / degenerate	$a \to 0$ , $b \to 0$ , $V_{r} \to \infty$	reduces to ideal gas
G	Above $T_{c}$ (no liquid)	$T > T_{c}$ → can it liquefy?	hard ceiling; supercritical
H	Word / real-world	design/lab framing	translate words → symbols
I	Exam twist	ratio or "which gas liquefies easiest?"	compare via $a / b$ , $T_{c}$

Example 1 — Cell A (forward, dodge $V_{c}$ )

Worked example Ammonia: find

$a$ and $b$ For NH₃, $T_{c} = 405.5 K$ , $P_{c} = 111.3 atm$ . Find van der Waals $a, b$ . Use $R = 0.0821 L atm mol^{- 1} K^{- 1}$ .

Forecast: $b$ is a small "excluded volume" (~0.03–0.05 L/mol), $a$ a few atm·L²/mol². NH₃ has strong hydrogen bonding, so expect $a$ bigger than for CO₂ ( $a \approx 3.6$ ).

Step 1 — Get $b$ from $b = \frac{R T _{c}}{8 P _{c}}$ . Why this step? This inversion came from dividing $T_{c} / P_{c}$ , which cancels $a$ — so we never touch the poorly-measured $V_{c}$ . $b = \frac{0.0821 \times 405.5}{8 \times 111.3} = \frac{33.29}{890.4} = 0.0374 L mol^{- 1}$

Step 2 — Get $a$ from $P_{c} = \frac{a}{27 b ^{2}} \Rightarrow a = 27 P_{c} b^{2}$ . Why this step? Now $b$ is known, and this equation isolates $a$ directly. $a = 27 \times 111.3 \times (0.0374)^{2} = 27 \times 111.3 \times 0.001399 = 4.20 atm L^{2} mol^{- 2}$

Verify: Units: $27 \cdot [atm] \cdot [L mol^{- 1}]^{2} = atm L^{2} mol^{- 2}$ ✓. And $a = 4.20 > 3.6$ (CO₂), matching NH₃'s stronger intermolecular attraction. ✓

Example 2 — Cell B (backward: $a, b$ → all three constants)

Example 3 — Cell C (self-consistency check via $Z_{c}$ )

Worked example Is this reported data internally consistent?

A datasheet lists a gas with $P_{c} = 40.0 atm$ , $V_{c} = 0.120 L mol^{- 1}$ , $T_{c} = 200 K$ . Does it obey the van der Waals prediction?

Forecast: If the numbers are vdW-consistent, $Z_{c}$ must equal $3/8 = 0.375$ . Real data usually gives $\approx 0.27$ – $0.30$ , so I expect below 0.375.

Step 1 — Compute $Z_{c} = \frac{P _{c} V _{c}}{R T _{c}}$ . Why this step? $Z_{c}$ is the single dimensionless number that tests vdW against reality — it strips out gas-specific size. $Z_{c} = \frac{40.0 \times 0.120}{0.0821 \times 200} = \frac{4.80}{16.42} = 0.292$

Step 2 — Compare to $3/8$ . Why this step? The compressibility factor $Z_{c} = 0.375$ is the vdW ideal; the shortfall measures how imperfect vdW is for this real gas.

Verify: $0.292$ sits squarely in the real-gas band $0.27$ – $0.31$ . The data is physically realistic but not vdW-exact — exactly what the parent note warned. ✓

Example 4 — Cell D (reduced equation, no $a, b$ )

Example 5 — Cell E (corresponding states matching) — with figure

Worked example Two gases at the same reduced state

Argon has $T_{c} = 151 K$ ; nitrogen has $T_{c} = 126 K$ . At what actual temperature is each in the corresponding state $T_{r} = 1.30$ ? If both are also at $P_{r} = 2.0$ , what can you say about their compressibility factors $Z$ ?

Forecast: Ar (higher $T_{c}$ ) will need the higher actual temperature. Both share the same reduced coordinates, so their $Z$ should be equal.

Step 1 — Actual temperature $T = T_{r} T_{c}$ . Why this step? Reduced temperature is defined as $T_{r} = T / T_{c}$ , so we just multiply back up. $T_{Ar} = 1.30 \times 151 = 196.3 K, T_{N_{2}} = 1.30 \times 126 = 163.8 K$

Step 2 — Compare compressibility factors. Why this step? The Law of Corresponding States says all gases at equal $(T_{r}, P_{r})$ occupy equal $V_{r}$ — and since $Z = \frac{P V}{R T}$ can be rewritten $Z = Z_{c} \frac{P _{r} V _{r}}{T _{r}}$ , equal reduced coordinates force equal $Z$ . $Same (T_{r}, P_{r}) \Rightarrow same V_{r} \Rightarrow same Z$

Verify: Different actual temperatures (196.3 K vs 163.8 K) but identical reduced state — that is the whole point of corresponding states. Look at the figure: two different real isotherms collapse onto one curve when axes are reduced. ✓

Figure — Critical constants Tc, Pc, Vc; law of corresponding states

Example 6 — Cell F (limiting / degenerate: recover the ideal gas)

Worked example What happens as

$V_{r}$ grows huge? Take the reduced equation of state and let $V_{r} \to \infty$ (gas spread out very thin). Show it collapses to the ideal-gas relation, and confirm numerically at $V_{r} = 100$ , $T_{r} = 1.0$ .

Forecast: Very dilute gas → molecules rarely meet → attractions ( $a$ ) and finite size ( $b$ ) both irrelevant → must become the ideal gas.

Step 1 — In $(P_{r} + \frac{3}{V _{r}^{2}}) (3 V_{r} - 1) = 8 T_{r}$ , drop the small terms. Why this step? As $V_{r} \to \infty$ : the attraction term $\frac{3}{V _{r}^{2}} \to 0$ (spacing kills attraction), and the $- 1$ next to $3 V_{r}$ is negligible (finite size vanishes vs huge volume). $P_{r} \cdot 3 V_{r} \approx 8 T_{r}$

Step 2 — Rearrange to see the ideal form. $P_{r} V_{r} \approx \frac{8}{3} T_{r}$ This is $P V \propto T$ in reduced units — the ideal-gas law, since $\frac{8}{3} = \frac{1}{Z _{c}}$ ties it back to $Z_{c} = 3/8$ .

Step 3 — Numerical spot-check at $V_{r} = 100, T_{r} = 1.0$ . Exact: $(P_{r} + \frac{3}{10 0 ^{2}}) (3 \times 100 - 1) = 8 \Rightarrow (P_{r} + 0.0003) (299) = 8 \Rightarrow P_{r} = 0.026455$ Ideal approximation: $P_{r} = \frac{8}{3 \times 100} = 0.026667$ .

Verify: Exact $0.02646$ vs ideal $0.02667$ — agree to within $0.8%$ , and the gap shrinks as $V_{r}$ grows. Degenerate limit confirmed. ✓

Example 7 — Cell G (above $T_{c}$ : liquefaction ceiling) — with figure

Worked example Can you liquefy CO₂ at room temperature by squeezing?

CO₂ has $T_{c} = 304 K (\approx 3 1^{\circ} C)$ . A student at $T = 298 K$ (25 °C) compresses CO₂ to 200 atm. Do they get liquid CO₂? What about at $T = 290 K$ ?

Forecast: 298 K is below 304 K, so liquefaction should be possible with enough pressure; but only just. 290 K is comfortably below → definitely liquefiable.

Step 1 — Compare $T$ to $T_{c}$ . Why this step? $T_{c}$ is the hard ceiling: only below it does the isotherm have a flat plateau (the coexistence region) where liquid can form. $T = 298 K < 304 K = T_{c} \Rightarrow plateau exists \Rightarrow liquefiable ✓$

Step 2 — Reduced temperature check. $T_{r} = \frac{298}{304} = 0.980 < 1 (and \frac{290}{304} = 0.954 < 1)$ Both are below $T_{r} = 1$ , so both can be liquefied by pressure.

Step 3 — Contrast the failure case. Why this step? If instead $T = 310 K$ , then $T_{r} = 1.02 > 1$ : no plateau, so no amount of pressure makes a distinct liquid — you only get a dense supercritical fluid.

Verify: $298 < 304$ and $290 < 304$ → both liquefiable; $310 > 304$ → not. The figure shows the plateau present below $T_{c}$ and gone above it. ✓

Example 8 — Cell H (real-world word problem)

Worked example Designing a storage cylinder

An engineer wants to store methane (CH₄) as a liquid at $T_{c}$ -limited conditions. Given $a = 2.28 atm L^{2} mol^{- 2}$ , $b = 0.0428 L mol^{- 1}$ , what critical pressure must the cylinder at least withstand, and what temperature must it be kept below?

Forecast: Methane is a light gas, so $T_{c}$ should be low (~190 K), needing refrigeration; $P_{c}$ around 45 atm.

Step 1 — Translate "kept below" → $T_{c}$ . Why this step? Liquid CH₄ exists only below $T_{c}$ ; that is the temperature the design must stay under. $T_{c} = \frac{8 a}{27 R b} = \frac{8 \times 2.28}{27 \times 0.0821 \times 0.0428} = \frac{18.24}{0.09487} = 192.3 K$

Step 2 — Translate "withstand" → $P_{c}$ (worst-case liquefaction pressure at $T_{c}$ ). $P_{c} = \frac{a}{27 b ^{2}} = \frac{2.28}{27 \times ( 0.0428 ) ^{2}} = \frac{2.28}{27 \times 0.001832} = \frac{2.28}{0.04946} = 46.1 atm$

Verify: Experimental CH₄: $T_{c} \approx 190.6$ K, $P_{c} \approx 45.8$ atm — our design targets ( $192.3$ K, $46.1$ atm) match. So: keep below $\sim 192$ K, rate the cylinder above $\sim 46$ atm. ✓

Example 9 — Cell I (exam twist: "which liquefies most easily?")

Worked example Rank three gases by ease of liquefaction

Given $T_{c}$ : He $= 5.2$ K, CO₂ $= 304$ K, H₂O $= 647$ K. Which is hardest to liquefy, and how much stronger (via $a / b^{2}$ intuition) are water's attractions reflected in $T_{c}$ ? Also compute the ratio $T_{c} (H_{2} O) / T_{c} (He)$ .

Forecast: Ease of liquefaction tracks higher $T_{c}$ (stronger attractions win at higher temperatures). Water (H-bonding) should be easiest, helium (feeble dispersion only) hardest.

Step 1 — Order by $T_{c}$ . Why this step? $T_{c} = \frac{8 a}{27 R b}$ rises with the attraction constant $a$ ; a bigger $T_{c}$ means you don't have to cool as far, i.e. easier to liquefy. $Easiest \to H_{2} O (647) > CO_{2} (304) > He (5.2) \to hardest$

Step 2 — Quantify the extreme spread. $\frac{T _{c} ( H _{2} O )}{T _{c} ( He )} = \frac{647}{5.2} = 124.4$

Verify: Helium's $T_{c} = 5.2$ K explains why liquid helium needs extreme cryogenics, while water is liquid at room temperature — a factor of $\sim 124$ in $T_{c}$ , driven by hydrogen bonding vs weak dispersion. ✓

Recall Quick self-test across the matrix

Which inversion do you use to find $b$ when only $T_{c}, P_{c}$ are given (never $V_{c}$ )? ::: $b = R T_{c} / (8 P_{c})$ — dividing $T_{c} / P_{c}$ cancels $a$ . As $V_{r} \to \infty$ , the reduced EOS becomes what? ::: $P_{r} V_{r} = \frac{8}{3} T_{r}$ , i.e. the ideal gas law. Two gases at equal $T_{r}$ AND equal $P_{r}$ share what property? ::: the same $V_{r}$ and the same $Z$ (corresponding states). A gas at $T_{r} = 1.02$ — can pressure liquefy it? ::: No; above $T_{c}$ there is no plateau, only a supercritical fluid.