Question

At T=Tc, (∂P/∂V)V T=Tc = 0 and (∂2P/∂V2)T=TC = 0. Use this information to derive expressions for a and b in the van der Waals equation of state in terms of experimentally determined Pc and Tc.

Answer 1

Vanderwaal equation can be written as

(P+a/V²)*(V-b)= RT   (1)

P= RT/(V-b)- a/V2 (1A)

Differentiating Eq.1A with respect to V at constant temperature gives

(dp/dV)_T= -RT/(V-b)²+2a/V³     (2)

At critical point T= Tc and (dp/dV)_{T =Tc} =0

RT_C/(Vc-b)² =2a/V_C³      (2A)

Differentiation Eq.2 again

Eq.2 becomes (d²P/dV²) _{T= Tc} gives 2RT_C/(V_C-b)³- 6a/V_C⁴ =0        (3)

Or 2RT_C/(Vc-b)³= 6a/V_c⁴     (4)

Dividing Eq.2A and Eq.4 gives

V_C =3b   (5)

But RT_C/(Vc-b)²= 2a/VC³ (Eq.2A)

substituting Eq.5 in this equation gives

RT_C/(3b-b)²= 2a/(3b)³

RT_C/4b²= 2a/27b³

T_C= 8a/27bR    (6)

But at critical point from Eq.1

P_C= RT_C/(V_C-b)- a/V_C²     (7)

Substitution of TC from Eq.6 and Vc=3b from eq.5 gives

PC= (8a/27bR)*R/(2b)- a/9b²

Pc= 8a/54b² -a/9b²   = a/27b²   (8)

Eq.6/Eq.8 gives

T_C/P_C= (8a/27bR*(27b²/a) =8b/R

Hence b= RT_C/8P_C    (9)

from Eq.6, squaring it gives

T_C²= 64a²/ 729b²R²

Tc²/PC= (64a²/729b²R²)* 27b²/a = 64a/27R²

a= 27R²T_C²/ 64Pc

These are the expressions for a and b in terms of P_Cand T_C