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

The three master results you will lean on the whole page:

Level 1 — Recognition

Exercise 1.1

State the two mathematical conditions that pin down the critical point on a $P$ – $V$ isotherm, and say in plain words what each one means geometrically.

Recall Solution

$(\frac{\partial P}{\partial V})_{T} = 0, (\frac{\partial ^{2} P}{\partial V ^{2}})_{T} = 0.$

First derivative $= 0$ : the isotherm has a horizontal tangent — the curve is momentarily flat, so a tiny change in volume causes no change in pressure. This is the last trace of the "liquefaction plateau."
Second derivative $= 0$ : zero curvature — the curve neither smiles nor frowns; it is an inflection point where it switches from concave to convex.

The figure below draws three isotherms from the reduced equation of state, one for each temperature regime, so you can see both conditions being born at once.

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

Reading the figure. The magenta curve ( $T < T_{c}$ ) has a genuine flat segment — a whole stretch of constant pressure where liquid and gas coexist (the liquefaction plateau). As temperature rises, that flat segment shrinks. On the violet curve ( $T = T_{c}$ ) it has collapsed to the single navy dot at $(V_{r} = 1, P_{r} = 1)$ : there the dashed navy line shows the tangent is exactly horizontal (first condition) and the curve stops curving one way and starts curving the other (second condition) — flatness and straightness meet at one point. The orange curve ( $T > T_{c}$ ) has no flat part at all: raising pressure just smoothly compresses the fluid, so no liquid can ever separate out. That is why $T_{c}$ is a hard ceiling for liquefaction.

Exercise 1.2

Write $T_{c}$ , $P_{c}$ , $V_{c}$ in terms of the van der Waals constants $a$ , $b$ (and $R$ ). Then compute the critical compressibility factor $Z_{c}$ these predict.

Recall Solution

$V_{c} = 3 b, T_{c} = \frac{8 a}{27 R b}, P_{c} = \frac{a}{27 b ^{2}} .$ $Z_{c} = \frac{P _{c} V _{c}}{R T _{c}} = \frac{\frac{a}{27 b ^{2}} \cdot 3 b}{R \cdot \frac{8 a}{27 R b}} = \frac{\frac{3 a}{27 b}}{\frac{8 a}{27 b}} = \frac{3}{8} = 0.375.$ Notice every $a$ , $b$ , $R$ cancels — the answer is a pure number, the same for every gas vdW describes.

Level 2 — Application

Exercise 2.1

For CO₂, $T_{c} = 304 K$ and $P_{c} = 73 atm$ . Find the van der Waals constants $a$ and $b$ .

Recall Solution

Step 1 — get $b$ from the pair $(T_{c}, P_{c})$ . Why this combination? We have two experimental numbers, $T_{c}$ and $P_{c}$ , and two unknowns, $a$ and $b$ . If we divide $T_{c}$ by $P_{c}$ , the unknown $a$ appears in both numerator and denominator and cancels, leaving an equation in $b$ alone — so we can solve for $b$ directly without ever knowing $a$ : $\frac{T _{c}}{P _{c}} = \frac{8 a /27 R b}{a /27 b ^{2}} = \frac{8 b}{R} \Rightarrow b = \frac{R T _{c}}{8 P _{c}} .$ (Bonus: this also dodges the poorly-measured $V_{c}$ .) $b = \frac{0.0821 \times 304}{8 \times 73} = \frac{24.96}{584} = 0.0427 L mol^{- 1} .$ Step 2 — get $a$ from $P_{c} = \frac{a}{27 b ^{2}}$ now that $b$ is known: $a = 27 P_{c} b^{2} = 27 \times 73 \times (0.0427)^{2} = 27 \times 73 \times 0.001824 = 3.59 atm L^{2} mol^{- 2} .$

Exercise 2.2

Using the constants from 2.1, predict CO₂'s critical molar volume $V_{c}$ and verify $Z_{c}$ .

Recall Solution

$V_{c} = 3 b = 3 \times 0.0427 = 0.128 L mol^{- 1} .$ $Z_{c} = \frac{P _{c} V _{c}}{R T _{c}} = \frac{73 \times 0.128}{0.0821 \times 304} = \frac{9.34}{24.96} = 0.374 \approx \frac{3}{8} . ✓$ The near-exact 0.375 confirms our arithmetic reproduces the universal vdW value.

Exercise 2.3

Gas A has $T_{c} = 150 K$ , gas B has $T_{c} = 300 K$ . At what actual temperatures are they in corresponding states with $T_{r} = 1.2$ ?

Recall Solution

Reduced temperature $T_{r} = T / T_{c}$ , so $T = T_{r} T_{c}$ : $T_{A} = 1.2 \times 150 = 180 K, T_{B} = 1.2 \times 300 = 360 K .$ Different real temperatures, but the same fraction of their critical temperatures — so (at equal $P_{r}$ ) they share the same $Z$ and the same reduced volume $V_{r}$ .

Level 3 — Analysis

Exercise 3.1

A real gas is measured to have $Z_{c} = 0.28$ . Van der Waals predicts $0.375$ . Compute the percentage error of the vdW prediction and state what the mismatch tells you physically.

Recall Solution

$error = \frac{0.375 - 0.28}{0.28} \times 100% = \frac{0.095}{0.28} \times 100% = 33.9%.$ vdW over-predicts $Z_{c}$ by about $34%$ . Physically: the vdW $a / V^{2}$ and $b$ corrections are too crude near the critical point (attractions actually vary with distance and molecules are soft, not hard spheres). So vdW is only qualitatively right there — better equations like Redlich–Kwong push $Z_{c}$ down toward the observed $0.27$ – $0.31$ .

Exercise 3.2

Show that $P_{c} V_{c} = \frac{3}{8} R T_{c}$ directly, and use it to find $P_{c}$ for a gas with $T_{c} = 126 K$ and $V_{c} = 0.090 L mol^{- 1}$ (this is N₂).

Recall Solution

From $Z_{c} = P_{c} V_{c} / (R T_{c}) = 3/8$ , multiply both sides by $R T_{c}$ : $P_{c} V_{c} = \frac{3}{8} R T_{c} .$ Solve for $P_{c}$ : $P_{c} = \frac{3 R T _{c}}{8 V _{c}} = \frac{3 \times 0.0821 \times 126}{8 \times 0.090} = \frac{31.03}{0.72} = 43.1 atm .$ (Experimental N₂ $P_{c} \approx 33.5 atm$ — the vdW estimate is the right order and a bit high, exactly as the imperfect $Z_{c}$ warns.)

Exercise 3.3

Two gases are each brought to $T_{r} = 1.0$ and $P_{r} = 1.0$ . Without knowing which gases they are, what is their common reduced volume $V_{r}$ ? Justify using the reduced equation of state.

Recall Solution

Plug $P_{r} = 1$ , $T_{r} = 1$ into the reduced equation: $(1 + \frac{3}{V _{r}^{2}}) (3 V_{r} - 1) = 8.$ By definition the critical point is $T_{r} = P_{r} = V_{r} = 1$ . Check: $(1 + 3) (3 - 1) = 4 \times 2 = 8.$ ✓ So $V_{r} = 1$ : all gases at their critical point sit at reduced volume 1 — that is precisely what "corresponding states" guarantees. No gas-specific data was needed.

Level 4 — Synthesis

Exercise 4.1

Starting from $P = \frac{R T}{V - b} - \frac{a}{V ^{2}}$ , derive $V_{c} = 3 b$ from scratch using the two critical conditions. (Reproduce the full inflection argument.)

Recall Solution

First derivative $= 0$ (horizontal tangent): $\frac{\partial P}{\partial V} = - \frac{R T}{( V - b ) ^{2}} + \frac{2 a}{V ^{3}} = 0 \Rightarrow \frac{R T}{( V - b ) ^{2}} = \frac{2 a}{V ^{3}} (1)$ Second derivative $= 0$ (zero curvature): $\frac{\partial ^{2} P}{\partial V ^{2}} = \frac{2 R T}{( V - b ) ^{3}} - \frac{6 a}{V ^{4}} = 0 \Rightarrow \frac{2 R T}{( V - b ) ^{3}} = \frac{6 a}{V ^{4}} (2)$ Divide (2) by (1) (kills $R, T, a$ , leaving pure geometry): $\frac{2/ ( V - b ) ^{3}}{1/ ( V - b ) ^{2}} = \frac{6 a / V ^{4}}{2 a / V ^{3}} \Rightarrow \frac{2}{V - b} = \frac{3}{V} .$ Cross-multiply: $2 V = 3 (V - b) = 3 V - 3 b \Rightarrow V = 3 b$ . Hence $V_{c} = 3 b$ .

Exercise 4.2

Using $V_{c} = 3 b$ and condition (1), derive $T_{c} = \frac{8 a}{27 R b}$ , then obtain $P_{c} = \frac{a}{27 b ^{2}}$ .

Recall Solution

With $V_{c} = 3 b$ , we have $V_{c} - b = 2 b$ . Put both into (1): $\frac{R T _{c}}{( 2 b ) ^{2}} = \frac{2 a}{( 3 b ) ^{3}} = \frac{2 a}{27 b ^{3}} .$ Now solve for $R T_{c}$ by multiplying both sides by $(2 b)^{2} = 4 b^{2}$ : $R T_{c} = \frac{2 a}{27 b ^{3}} \cdot 4 b^{2} .$ Multiply the numbers ( $2 \times 4 = 8$ ) and cancel the powers of $b$ ( $b^{2} / b^{3} = 1/ b$ ): $R T_{c} = \frac{8 a b ^{2}}{27 b ^{3}} = \frac{8 a}{27 b} \Rightarrow T_{c} = \frac{8 a}{27 R b} .$ Now plug $V_{c} = 3 b$ and $T_{c}$ back into the vdW equation for $P$ : $P_{c} = \frac{R T _{c}}{V _{c} - b} - \frac{a}{V _{c}^{2}} = \frac{R \cdot \frac{8 a}{27 R b}}{2 b} - \frac{a}{9 b ^{2}} = \frac{8 a}{54 b ^{2}} - \frac{a}{9 b ^{2}} = \frac{4 a}{27 b ^{2}} - \frac{3 a}{27 b ^{2}} = \frac{a}{27 b ^{2}} .$

Exercise 4.3

Derive the reduced (universal) equation of state by substituting $P = P_{r} P_{c}$ , $V = V_{r} V_{c}$ , $T = T_{r} T_{c}$ into the vdW equation, showing every cancellation.

Recall Solution

Start: $(P + \frac{a}{V ^{2}}) (V - b) = R T$ . Substitute $P_{c} = \frac{a}{27 b ^{2}}$ , $V_{c} = 3 b$ , $T_{c} = \frac{8 a}{27 R b}$ : $(P_{r} \frac{a}{27 b ^{2}} + \frac{a}{( 3 b V _{r} ) ^{2}}) (3 b V_{r} - b) = R T_{r} \frac{8 a}{27 R b} .$ Factor $\frac{a}{27 b ^{2}}$ from the first bracket and $b$ from the second: $\frac{a}{27 b ^{2}} (P_{r} + \frac{27 b ^{2}}{9 b ^{2} V _{r}^{2}}) \cdot b (3 V_{r} - 1) = \frac{8 a}{27 b} T_{r} .$ Since $\frac{27 b ^{2}}{9 b ^{2}} = 3$ : $\frac{a}{27 b ^{2}} \cdot b (P_{r} + \frac{3}{V _{r}^{2}}) (3 V_{r} - 1) = \frac{8 a}{27 b} T_{r} \Rightarrow \frac{a}{27 b} (P_{r} + \frac{3}{V _{r}^{2}}) (3 V_{r} - 1) = \frac{8 a}{27 b} T_{r} .$ Cancel $\frac{a}{27 b}$ on both sides: $(P_{r} + \frac{3}{V _{r}^{2}}) (3 V_{r} - 1) = 8 T_{r} .$ No $a$ , $b$ , or $R$ survives — the equation is identical for every gas.

Level 5 — Mastery

Exercise 5.1

A gas obeys vdW with $a = 1.36 atm L^{2} mol^{- 2}$ and $b = 0.0318 L mol^{- 1}$ (this is O₂). Compute $T_{c}$ , $P_{c}$ , $V_{c}$ , and confirm $Z_{c} = 3/8$ numerically.

Recall Solution

$V_{c} = 3 b = 3 \times 0.0318 = 0.0954 L mol^{- 1} .$ $T_{c} = \frac{8 a}{27 R b} = \frac{8 \times 1.36}{27 \times 0.0821 \times 0.0318} = \frac{10.88}{0.07049} = 154.3 K .$ $P_{c} = \frac{a}{27 b ^{2}} = \frac{1.36}{27 \times ( 0.0318 ) ^{2}} = \frac{1.36}{0.02731} = 49.8 atm .$ Check: $Z_{c} = \frac{P _{c} V _{c}}{R T _{c}} = \frac{49.8 \times 0.0954}{0.0821 \times 154.3} = \frac{4.751}{12.67} = 0.375.$ ✓ (Experimental O₂: $T_{c} \approx 154.6$ K, $P_{c} \approx 49.8$ atm — a genuinely excellent match here.)

Exercise 5.2

Prove that the ratio $\frac{R T _{c}}{P _{c} V _{c}}$ equals $\frac{8}{3}$ for any vdW gas, purely symbolically, and explain why this is the reciprocal of $Z_{c}$ .

Recall Solution

$\frac{R T _{c}}{P _{c} V _{c}} = \frac{R \cdot \frac{8 a}{27 R b}}{\frac{a}{27 b ^{2}} \cdot 3 b} = \frac{\frac{8 a}{27 b}}{\frac{3 a}{27 b}} = \frac{8}{3} .$ Every gas-specific quantity cancels, so the value is universal. It is the reciprocal of $Z_{c}$ because $Z_{c} = \frac{P _{c} V _{c}}{R T _{c}} \Rightarrow \frac{1}{Z _{c}} = \frac{R T _{c}}{P _{c} V _{c}} = \frac{8}{3},$ and indeed $\frac{1}{3/8} = \frac{8}{3}$ . ✓

Exercise 5.3

The reduced equation gives $Z$ as a function of $(T_{r}, V_{r})$ only. Using $Z = \frac{3 P _{r} V _{r}}{8 T _{r}}$ (the reduced form of $Z = P V / R T$ ), derive an expression for $Z$ in terms of $T_{r}$ and $V_{r}$ alone, and evaluate it at $T_{r} = 1.5$ , $V_{r} = 2$ .

Recall Solution

From the reduced equation solve for $P_{r}$ : $P_{r} = \frac{8 T _{r}}{3 V _{r} - 1} - \frac{3}{V _{r}^{2}} .$ Since $Z = \frac{P V}{R T}$ and $\frac{P _{c} V _{c}}{R T _{c}} = \frac{3}{8}$ , the reduced form is $Z = \frac{P _{r} V _{r}}{\frac{8}{3} T _{r}} = \frac{3 P _{r} V _{r}}{8 T _{r}}$ . Substitute $P_{r}$ and multiply through by $\frac{3 V _{r}}{8 T _{r}}$ — keep the factor of 3 on both terms: $Z = \frac{3 V _{r}}{8 T _{r}} (\frac{8 T _{r}}{3 V _{r} - 1} - \frac{3}{V _{r}^{2}}) = \frac{3 V _{r}}{3 V _{r} - 1} - \frac{9}{8 T _{r} V _{r}} .$ The first term came from $\frac{3 V _{r}}{8 T _{r}} \cdot \frac{8 T _{r}}{3 V _{r} - 1} = \frac{3 V _{r}}{3 V _{r} - 1}$ (the $8 T_{r}$ cancels, the $3$ stays). At $T_{r} = 1.5$ , $V_{r} = 2$ : $Z = \frac{3 \times 2}{3 \times 2 - 1} - \frac{9}{8 \times 1.5 \times 2} = \frac{6}{5} - \frac{9}{24} = 1.200 - 0.375 = 0.825.$ So $Z \approx 0.825$ : less than 1, telling us attractions still dominate at this moderately expanded state. The key deliverable is the universal formula $Z = \frac{3 V _{r}}{3 V _{r} - 1} - \frac{9}{8 T _{r} V _{r}}$ , which depends on nothing but reduced variables — the whole point of corresponding states. (See Real Gases and Compressibility Factor Z for the empirical chart version.)