Question

Derive the relationship between critical constants and van der Waals constants.

Answer 1

The Relationship between the van der Waals Parameters and the Critical Constants

If

$$\left(p + \frac {a}{V ^ {2}}\right) (V - b) = R T$$

then

$$p = \frac {R T}{V - b} - \frac {a}{V ^ {2}}$$

At the critical point,

$$\left(\frac {\partial p}{\partial V}\right) _ {T} = 0 \quad \text{and} \quad \left(\frac {\partial^ {2} p}{\partial V ^ {2}}\right) _ {T} = 0$$

So, differentiating our expression for pressure with respect to volume at constant temperature

$$\left(\frac {\partial p}{\partial V}\right) _ {T} = - \frac {R T _ {c}}{\left(V _ {c} - b\right) ^ {2}} + \frac {2 a}{V _ {c} ^ {3}} = 0$$

so that

$$\frac {R T _ {c}}{\left(V _ {c} - b\right) ^ {2}} = \frac {2 a}{V _ {c} ^ {3}}$$

Differentiating again gives

$$\left(\frac {\partial^ {2} p}{\partial V ^ {2}}\right) _ {T} = + \frac {2 R T _ {c}}{\left(V _ {c} - b\right) ^ {3}} - \frac {6 a}{V _ {c} ^ {4}} = 0$$

so that

$$\frac {2 R T _ {c}}{\left(V _ {c} - b\right) ^ {3}} = \frac {6 a}{V _ {c} ^ {4}}$$

Dividing these two expressions gives

$$\frac {(V _ {c} - b)}{2} = \frac {V _ {c}}{3}$$

and so

$$V _ {c} = 3 b$$

Substitution of this result into the expression that we obtained by differentiating once

$$\frac {R T _ {c}}{(V _ {c} - b) ^ {2}} = \frac {2 a}{V _ {c} ^ {3}}$$

gives

$$\frac {R T _ {c}}{(3 b - b) ^ {2}} = \frac {2 a}{(3 b) ^ {3}}$$

which, when rearranged, gives

$$T _ {c} = \frac {8 a}{2 7 b R}$$

Substitution of the expressions for the critical volume $V_{c}$ and critical temperature $T_{c}$ into the van der Waals equation gives

$$\begin{array}{l} p _ {c} = \frac {R T _ {c}}{V _ {c} - b} - \frac {a}{V _ {c} ^ {2}} \\ = R \frac {1}{(3 b - b)} \frac {8 a}{2 7 b R} - \frac {a}{9 b ^ {2}} \\ = \frac {8 a}{5 4 b ^ {2}} - \frac {a}{9 b ^ {2}} \\ = \frac {8 a}{5 4 b ^ {2}} - \frac {6 a}{5 4 b ^ {2}} \\ \end{array}$$

and so

$$p _ {c} = \frac {a}{2 7 b ^ {2}}$$

Given these expressions for the values of the critical constants

$$\frac {T _ {c}}{p _ {c}} = \frac {8 a}{2 7 b R} \frac {2 7 b ^ {2}}{a} = \frac {8 b}{R}$$

so that

$$b = \frac {R T _ {c}}{8 p _ {c}}$$

Similarly

$$T _ {c} ^ {2} = \frac {6 4 a ^ {2}}{7 2 9 b ^ {2} R ^ {2}}$$ $$\frac{T_{C}^{2}}{P_{C}} = \frac{64a^{2}}{729b^{2}R^{2}} \frac{27b^{2}}{a} = \frac{64a}{27R^{2}}$$

Hence

$$a = \frac{27R^{2}T_{C}^{2}}{64P_{C}}$$