Derivation of Van der Waals Critical Constants

Derivation of Critical Constants: Relationship between Critical Constants and Van der Waals Constants

The Van der Waals equation of state for one mole of a real gas is given by:

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

Expanding the algebraic terms yields:

$$PV - bP + \frac{a}{V} - \frac{ab}{V^2} = RT$$

Multiplying the entire equation by \(\frac{V^2}{P}\), we obtain a cubic expression in terms of volume:

$$V^3 - bV^2 + \frac{aV}{P} - \frac{ab}{P} - \frac{RTV^2}{P} = 0$$

Grouping identical powers of \(V\) results in:

$$V^3 - \left(b + \frac{RT}{P}\right)V^2 + \frac{a}{P}V - \frac{ab}{P} = 0$$

At the critical point, the temperature and pressure are equal to their critical values (\(T = T_c\) and \(P = P_c\)). Substituting these boundary conditions into the cubic framework gives:

$$V^3 - \left(b + \frac{RT_c}{P_c}\right)V^2 + \frac{a}{P_c}V - \frac{ab}{P_c} = 0 \quad \text{--- (Equation 1)}$$

Concurrently, at the critical point, the three roots of the cubic equation merge into a single, identical value, meaning:

$$V = V_c \implies V - V_c = 0$$

Cubing this expression defines the path of perfect confluence at the critical state:

$$(V - V_c)^3 = 0$$

Expanding this binomial expression yields:

$$V^3 - 3V_cV^2 + 3V_c^2V - V_c^3 = 0 \quad \text{--- (Equation 2)}$$

Because Equation 1 and Equation 2 describe the identical state of the system, we can equate the corresponding coefficients for each power of \(V\):

$$3V_c = b + \frac{RT_c}{P_c} \quad \text{--- (Equation 3)}$$ $$3V_c^2 = \frac{a}{P_c} \quad \text{--- (Equation 4)}$$ $$V_c^3 = \frac{ab}{P_c} \quad \text{--- (Equation 5)}$$

Step 1: Determining Critical Volume (\(V_c\))

Dividing Equation 5 by Equation 4 yields:

$$\frac{V_c^3}{3V_c^2} = \frac{\frac{ab}{P_c}}{\frac{a}{P_c}}$$ $$\frac{V_c}{3} = b$$

$$V_c = 3b$$

Step 2: Determining Critical Pressure (\(P_c\))

Substituting the derived value of \(V_c\) into Equation 4:

$$3(3b)^2 = \frac{a}{P_c}$$ $$27b^2 = \frac{a}{P_c}$$

$$P_c = \frac{a}{27b^2}$$

Step 3: Determining Critical Temperature (\(T_c\))

Substituting the expressions for both \(V_c\) and \(P_c\) back into Equation 3:

$$3(3b) - b = \frac{R T_c}{\frac{a}{27b^2}}$$ $$8b = \frac{27b^2 R T_c}{a}$$

Isolating \(T_c\) via algebraic simplification recovers the final critical parameter:

$$8a = 27b R T_c$$

$$T_c = \frac{8a}{27Rb}$$

Related Topic: Critical Phenomenon and Critical Constants

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