Q.Derive the values of critical constants in terms of van der Waals constants.

Step 1. Start from the one-mole van der Waals equation, $\left(P+\dfrac{a}{V^2}\right)(V-b)=RT$. Clearing the $V^2$ denominator gives a cubic in V: $V^3-\left(b+\dfrac{RT}{P}\right)V^2+\dfrac{a}{P}V-\dfrac{ab}{P}=0$.

Step 2. At the critical point, all three roots of this cubic coincide at $V=V_c$, so the cubic must be identical to $(V-V_c)^3=0$, which expands to $V^3-3V_cV^2+3V_c^2V-V_c^3=0$.

Step 3. Matching coefficients of $V^2$, $V^1$ and $V^0$ between the two cubics (at $T=T_c$, $P=P_c$) gives three equations: $b+\dfrac{RT_c}{P_c}=3V_c$, $\dfrac{a}{P_c}=3V_c^2$, $\dfrac{ab}{P_c}=V_c^3$.

Step 4. Dividing the third equation by the second cancels $a/P_c$, leaving $b=\dfrac{V_c}{3}$, i.e. $V_c=3b$. …

Turn the 1-mole van der Waals equation into a cubic in V, use the triple-root condition at the critical point to match it against (V - Vc)^3 = 0, and solve the resulti …

• Trying to solve for Tc, Pc and Vc simultaneously rather than in the sequence the coefficient-matching naturally allows (Vc from the ratio of the constant and linear-coefficient equations …