Q1(b). Andrew's isotherms and the critical-point conditions
Draw Andrew's plot of P vs. V isotherms of a real gas at different temperatures. At the critical point why (i) ((∂ P)/(∂ V))_(T=T_c)=0 and (ii) ((∂²P)/(∂ V²))_(T=T_c)=0?
Answer in brief
On the critical isotherm the flat liquid–vapour line has shrunk to a point which is a horizontal point of inflexion: slope zero (i) and curvature zero (ii). From these, V_c=3b, P_c=a/27b², T_c=8a/27Rb.
Key points
- Below T_c the isotherm has a horizontal part (liquid and vapour coexist). Above T_c it is smooth and falls with V. At T_c the horizontal part shrinks to one point.
- (i) The critical isotherm is tangent to the horizontal there: (∂ P/∂ V)_(T_c)=0. (ii) The van der Waals loop below T_c has a maximum and a minimum; as T→ T_c they merge, which needs the curvature to vanish too: (∂²P/∂ V²)_(T_c)=0.
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
- Below T_c the isotherm has a horizontal part (liquid and vapour coexist). Above T_c it is smooth and falls with V. At T_c the horizontal part shrinks to one point.
- (i) The critical isotherm is tangent to the horizontal there: (∂ P/∂ V)_(T_c)=0. (ii) The van der Waals loop below T_c has a maximum and a minimum; as T→ T_c they merge, which needs the curvature to vanish too: (∂²P/∂ V²)_(T_c)=0.
For P=(RT)/(V-b)-(a)/(V²): -(RT)/((V-b)²)+(2a)/(V³)=0 and (2RT)/((V-b)³)-(6a)/(V⁴)=0. Dividing the first by the second gives (V-b)/(2)=V/3, i.e. V_c=3b; then T_c=(8a)/(27Rb), P_c=(a)/(27b²).

Coverscritical phenomenaAndrew's isotherms and the critical-point conditions
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