# 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:

21 dS
= – P2 P1 (∂V / ∂T) P) dP ……. (3)

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

Equation 3 will become:
(Sr) 1 – (Sr) P
= – 1P (∂V / ∂T) P) dP ……………………………………………………….. (4)

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

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

Where So —–> standard entropy
S —————> entropy of real gas
Both entropies are determined at 1 atm. We have to calculate So. Here S is given as:
S = 1/3 a (T*) 3 + Tf T* (CP, S / T) dT + ∆Hf / Tf + Tb Tf (CP, L / T) dT + ∆Hv / Tb + T Tb (CP, 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) Pc V2c Tc
b = V c / 4
R = (32/9) Pc Vc / Tc ………………… (10)

Hence, equation (9) will become:
PV = RT [1 + 9/128 (PTc / Pc T) (1 – (T2c /
6 T2)]
Dividing the above equation by P, we will have:
V = RT / P + 9/128 (RTc / Pc) – 27 / 64 (R T3c / Pc T2) …… (11)
Therefore, (∂V / ∂T) P) = R / P + 27 / 32 (R T3c / Pc T3) …….. (12)
Substituting for (∂V / ∂T) P) in equation (6), we will get:
So = S + 27 / 32 (R T3c / Pc T3) 1P dP
= S + 27 / 32 (R T3c / Pc T3) (1 – P)
As P << 1 Therefore, So = S + 27 / 32 (R T3c / Pc T3)
Hence standard entropy So for a real gas has been obtained

Category: Third Law of Thermodynamics