**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 P_{1} and P_{2} 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} + ^{Tf}∫_{ T*} (C_{P, S} / T) dT + ∆H_{f} / T_{f} + ^{Tb}∫_{ 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 / TV^{2}) (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 / TV^{2} ……………………. (8)

The term ab / TV^{2} 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 / R^{2} T^{2}] …………………………………… (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} V^{2}_{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 – (T^{2}_{c} /

6 T^{2})]

Dividing the above equation by P, we will have:

V = RT / P + 9/128 (RT_{c} / P_{c}) – 27 / 64 (R T^{3}_{c} / P_{c} T^{2}) …… (11)

Therefore, (∂V / ∂T) _{P}) = R / P + 27 / 32 (R T^{3}_{c} / P_{c} T^{3}) …….. (12)

Substituting for (∂V / ∂T) _{P}) in equation (6), we will get:

S^{o} = S + 27 / 32 (R T^{3}_{c} / P_{c} T^{3}) ^{1}∫_{P} dP

= S + 27 / 32 (R T^{3}_{c} / P_{c} T^{3}) (1 – P)

As P << 1
Therefore,
S^{o} = S + 27 / 32 (R T^{3}_{c} / P_{c} T^{3})

Hence standard entropy S^{o} for a real gas has been obtained