Question

Write the van der Waals equation and explain the van der Waals isotherm (P–V curve).

Hint:

The S-shaped region in van der Waals isotherm indicates phase transition. At the critical point, the curve has an inflection point.

Answer 1

Step 1: Van der Waals Equation.
The ideal gas equation is:
\[ PV = nRT. \]
Van der Waals modified it to account for:
1. Finite volume of gas molecules
2. Intermolecular attraction
The van der Waals equation is:
\[ \left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT. \]
Where:
$a$ = measure of intermolecular attraction
$b$ = excluded volume of gas molecules
Step 2: Explanation of Corrections.
Pressure correction:
\[ P_{real} = P_{ideal} - \frac{a n^2}{V^2}. \]
The term $\frac{a n^2}{V^2}$ corrects for intermolecular attraction.
Volume correction:
\[ V_{real} = V_{container} - nb. \]
The term $nb$ accounts for finite molecular volume.
Step 3: Van der Waals Isotherm (P–V Curve).
When plotting pressure vs volume at constant temperature:
1. At high temperature ($T>T_c$):
Curve resembles ideal gas behaviour (smooth decrease).
2. At critical temperature ($T = T_c$):
Curve shows a point of inflection (critical point).
3. At low temperature ($T<T_c$):
Curve shows a characteristic S-shaped region.
This region represents liquid–gas phase transition.
The flat portion after Maxwell correction represents coexistence of liquid and vapour phases.
Step 4: Critical Constants.
At critical point:
\[ \left(\frac{\partial P}{\partial V}\right)_T = 0 \] \[ \left(\frac{\partial^2 P}{\partial V^2}\right)_T = 0. \]
Critical constants are:
\[ V_c = 3nb, \] \[ P_c = \frac{a}{27b^2}, \] \[ T_c = \frac{8a}{27Rb}. \]
Final Answer:
The van der Waals equation is:
\[ \boxed{ \left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT } \] and its P–V curve shows deviation from ideal behaviour with a characteristic S-shaped isotherm below critical temperature.