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

1. What are the critical constants?

WHY the meniscus disappears: below $T_c$, on a $P$–$V$ isotherm you see a flat "liquefaction plateau" (gas + liquid coexist). As $T \to T_c$ this plateau shrinks to a single point of inflection — the horizontal tangent and the curvature both vanish there.

2. Deriving $T_c, P_c, V_c$ from van der Waals (from scratch)

We start from the van der Waals equation for 1 mole:

$\left(P + \frac{a}{V^2}\right)(V - b) = RT$

Solve for $P$:

$P = \frac{RT}{V-b} - \frac{a}{V^2}$

WHAT defines the critical point mathematically? It is the inflection point of the isotherm, so both the first and second derivatives of $P$ w.r.t. $V$ vanish:

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

Step 1 — first derivative. Why? At the plateau's collapse the tangent is horizontal. $\frac{\partial P}{\partial V} = -\frac{RT}{(V-b)^2} + \frac{2a}{V^3} = 0 \Rightarrow \frac{RT}{(V-b)^2} = \frac{2a}{V^3}\quad (1)$

Step 2 — second derivative. Why? An inflection also needs zero curvature. $\frac{\partial^2 P}{\partial V^2} = \frac{2RT}{(V-b)^3} - \frac{6a}{V^4} = 0 \Rightarrow \frac{2RT}{(V-b)^3} = \frac{6a}{V^4}\quad (2)$

Step 3 — divide (2) by (1). Why? This kills $R$, $T$, $a$ and leaves pure geometry. $\frac{2/(V-b)}{1} = \frac{6a/V^4}{2a/V^3}\cdot\frac{(V-b)^2}{1}\;\Rightarrow\; \frac{2}{V-b} = \frac{3}{V}$ $2V = 3(V-b) \Rightarrow \boxed{V_c = 3b}$

Step 4 — get $T_c$ from (1). Substitute $V_c = 3b$, so $V_c - b = 2b$: $\frac{RT_c}{(2b)^2} = \frac{2a}{(3b)^3}=\frac{2a}{27b^3} \Rightarrow RT_c = \frac{2a}{27b^3}\cdot 4b^2 = \frac{8a}{27b}$ $\boxed{T_c = \frac{8a}{27Rb}}$

Step 5 — get $P_c$ by plugging $V_c$ and $T_c$ into the vdW equation: $P_c = \frac{RT_c}{V_c - b} - \frac{a}{V_c^2} = \frac{R\cdot \frac{8a}{27Rb}}{2b} - \frac{a}{9b^2} = \frac{8a}{54b^2} - \frac{a}{9b^2} = \frac{4a}{27b^2}-\frac{3a}{27b^2}$ $\boxed{P_c = \frac{a}{27b^2}}$

3. The Critical Compressibility Factor (a universal number)

WHY compute it? Form $Z_c = \dfrac{P_c V_c}{R T_c}$. For an ideal gas $Z=1$; the value here tests van der Waals against reality.

$Z_c = \frac{P_c V_c}{R T_c} = \frac{\left(\frac{a}{27b^2}\right)(3b)}{R\cdot\frac{8a}{27Rb}} = \frac{\frac{3a}{27b}}{\frac{8a}{27b}} = \boxed{\frac{3}{8} = 0.375}$

4. Law of Corresponding States (from scratch)

Idea (HOW): define reduced variables — each property divided by its critical value: $P_r = \frac{P}{P_c},\quad V_r = \frac{V}{V_c},\quad T_r = \frac{T}{T_c}$

Substitute $P = P_r P_c$, $V = V_r V_c = 3bV_r$, $T = T_r T_c$ into the vdW equation:

$\left(P_r\frac{a}{27b^2} + \frac{a}{(3bV_r)^2}\right)(3bV_r - b) = R\,T_r\frac{8a}{27Rb}$

Factor cleverly. Why? We want $a$ and $b$ to cancel completely. $\frac{a}{27b^2}\left(P_r + \frac{3}{V_r^2}\right)\cdot b(3V_r - 1) = \frac{8a}{27b}T_r$

The $\frac{a}{27b}$ cancels on both sides, leaving:

WHY it matters: you can predict an unknown gas's behaviour using a generalized compressibility chart ($Z$ vs $P_r$ for various $T_r$) that works for every gas — one chart, all gases.

5. Worked Examples

6. Common Mistakes (Steel-manned)

7. Flashcards

Connections

• Van der Waals Equation of State — the parent equation we differentiated.
• Real Gases and Compressibility Factor Z — $Z_c$ and generalized charts.
• Liquefaction of Gases and Andrews Isotherms — the $P$–$V$ isotherms that show the plateau.
• Ideal Gas Equation — the $a=b=0$ limit where $Z=1$ always.
• Intermolecular Forces — the $a$ term's physical origin (attractions setting $T_c$).
• Supercritical Fluids — the state reached beyond the critical point.

Concept Map

Hinglish (regional understanding)