Derive the Entropy of Real Gases

Entropy of real gases:
We know that:
(∂S / ∂P)_T = – (∂V / ∂T)_P ……………………. (1)
is one of the Maxwell relations which can also be written as:
dS – (∂V / ∂T) _P) dP …………………. (2)

Integrating between pressure P₁ and P₂ at constant tempearute T , we will get:

²∫₁ dS
= – ^P₂ ∫ _P1 (∂V / ∂T) _P) dP ……. (3)

At low temperature, a real gas behaves ideally. Let the pressure be P. let (S_r) ₁ be the entropy of the real gas at 1 atm pressure and (S_r) _P be the entropy at pressure P, at constant temperature. Then,

Equation 3 will become:
(S_r) ₁ – (S_r) _P
= – ¹∫_P (∂V / ∂T) _P) dP ……………………………………………………….. (4)

For an ideal gas, (∂V / ∂T) _P) = R / P. if (S_i) ₁ and (S_i) _P are the entropies of an ideal gas at 1 atm and P atm, respectively, then eq. 4 will be:
(S_i) _P – (S_i) ₁
= ¹∫_P (R / P) dP ……………………….. (5)

As At low temperature, a real gas behaves ideally so we can equate (S_r) _P with (S_i) _P
Adding equations 4 and 5, we will get:
(S_i) _P – (S_r) ₁ = S^o – S
= ¹∫_P [(∂V / ∂T) _P) – R / P] dP ………………….. (6)

Where S^o —–> standard entropy
S —————> entropy of real gas
Both entropies are determined at 1 atm. We have to calculate S^o. Here S is given as:
S = 1/3 a (T^*) ³ + ^T_f∫ _T^* (C_{P, S} / T) dT + ∆H_f / T_f + ^T_b∫ _Tf (C_{P, L} / T) dT + ∆H_v / T_b + ^T∫ _Tb (C_{P, g} / T) dT ………………………………………… (7)

Berthelot equation of state is:
(P + a / TV²) (V – b) = RT
is more appropriate to use than the van der Waals equation of state.
Multiplying and rearranging, we will get:
PV = RT + Pb – a / TV + ab / TV² ……………………. (8)

The term ab / TV² in the above equation is negligible as compared to other terms since the Berthelot constants a and b are small.

Thus, Eq. (8) becomes:
PV = RT + Pb – a / TV
As V = RT / P, therefore
PV = RT + Pb – a / TV = [1 + (Pb / RT) – aP / R² T²] …………………………………… (9)

For Berthelot’s equation of state, the constants a, b and R can be written in terms of the critical constants. Accordingly,
a = (16/3) P_c V²_c T_c
b = V _c / 4
R = (32/9) P_c V_c / T_c ………………… (10)

Hence, equation (9) will become:
PV = RT [1 + 9/128 (PT_c / P_c T) (1 – (T²_c /
6 T²)]
Dividing the above equation by P, we will have:
V = RT / P + 9/128 (RT_c / P_c) – 27 / 64 (R T³_c / P_c T²) …… (11)
Therefore, (∂V / ∂T) _P) = R / P + 27 / 32 (R T³_c / P_c T³) …….. (12)
Substituting for (∂V / ∂T) _P) in equation (6), we will get:
S^o = S + 27 / 32 (R T³_c / P_c T³) ¹∫_P dP
= S + 27 / 32 (R T³_c / P_c T³) (1 – P)
As P << 1 Therefore, S^o = S + 27 / 32 (R T³_c / P_c T³)
Hence standard entropy S^o for a real gas has been obtained