Entropy:

Efficiency of the engine working on the principle of Carnot cycle is of huge use to engineers; however, its main use in physics and chemistry is in the discussion and understanding of the second law of thermodynamics. This leads to the given comparison which might assist you in understanding the transformation of equation:

[(q_H + q_C)/q_H] = [(T_H -T_C)/T_H]

To

[(q₂ + q₁)/q₂] = [(T₂ - T₁)/T₂]

If in a Carnot engine, heat q₂ is absorbed at the higher temperature T₂ and q_l is absorbed at the lower temperature T_l (in reality q₁ will be a negative quantity since heat is discarded) then, according to the above equation,

[(q₂ + q₁)/q₂] = [(T₂ - T₁)/T₂]

Or 1 + (q₁/q₂) = T₁/T₂

That is, q₁/T₁ = - q₂/T₂

Therefore, q₁/T₁ + q₂/T₂ = 0

Therefore, the sum of such quantities as obtained by dividing the heat absorbed reversibly via the temperature is zero over a complete Carnot Cycle.

Any reversible cyclic process can be broken to a big number of infinitesimal Carnot cycles (as shown in the figure below). If in each such small Carnot cycle, heat dq₁ is absorbed at temperature T₁'and dq₂ absorbed at T₂', then for each and every small Carnot cycle,

dq₁/T₁' + dq₂/T₂' = 0

Summing these over all the cycles we can represent in general, sum of dq_i/T_i terms

Over all the cycles = 0

Here, 'i' stands for an individual process of expansion or contraction 'm', each cycle. As the summation is continuous, we can substitute the summation by integration and encompass

Δdq_rev/T = 0

Here, dq_rev is the heat absorbed reversibly at a temperature in an infinitesimal step in the cyclic process and Δ is the integral over a whole cycle.

Fig:  Infinitesimal-Carnot cycles

Now consider a system going in the reversible manner from an initial state 'A' to an intermediate state 'B' and then back to 'A' through the other path (figure shown).  This cyclic process can be broken up to a large number of Carnot cycles. Beginning from 'A' and following all these cycles we might reach 'A' once again. The paths within the figure cancel out each other and just a zigzag path is left). The bigger the number of Carnot cycles, the closer will be the resemblance between this zigzag path and the overall path ABA.

Fig: The cyclic change ABA

Sum of dq_rev/T terms = Σ dq_rev/T = 0

Here, once again dq_rev is the heat absorbed reversibly at temperature 'T' in an infinitesimal procedure. We can break this up into two parts, that is, one in which we go from 'A' to 'B' and the other in which we go from 'B' to 'A'. Therefore,

Σ_Cycle dq_rev/T = Σ_A→B dq_rev/T = Σ_B→A dq_rev/T = 0

Or in terms of integrals,

∫dq_rev/T = _A∫^B dq_rev/T + _B∫^A dq_rev/T = 0

_A∫^B dq_rev/T = - _B∫^A dq_rev/T

Therefore, the quantity _A∫^B dq_rev/T is not based on the path selected and is only dependent on the initial and final states of the system. This signifies that it symbolizes a change in some thermodynamic property. This property is known as entropy (S), and we represent,

dq_rev/T = dS

Therefore ∫dS = 0

As well if we symbolize the entropy of the initial state A as S_A and that of the final state B as S_B, then,

ΔS = S_B - S_A = _A∫^B dq_rev/T

Let us associate the changes in internal energy and enthalpy to entropy change.

If we now place dq = TdS, dU = nC_VdT and dw = - pdV in dU = dq + dw

Therefore,

dU = nC_V‾dT  = TdS - pdV

As per the equation H = U + pV

On differentiating, it gives:

dH = dU + pdV + Vdp

By using the equation, dH = TdS - pdV + pdV + Vdp = TdS + VdP

Equation dU = nCv‾dT = TdS - pdV and dH = TdS - pdV + pdV + Vdp = TdS + VdP are the joint mathematical statements of the first and second laws of thermodynamics. The first law of thermodynamics is mainly concerned by the conservation of energy and the second law of thermodynamics introduces the theory of entropy.

This is worth mentioning that the entropy change in a system is represented by:

dS = dq_rev/T

This signifies that the entropy change in a system is to be computed by supposing the process to be reversible, irrespective; of the fact that the process is reversible or not.

Entropy changes in Isolated Systems:

We are now interested in determining the entropy change in an isolated system where cyclic processes of isothermal expansion and compression occur. These cyclic processes can take place in two ways; one in which both the expansion and compression are reversible and the other, in which one is irreversible whereas the other is reversible. Let us take an isolated system comprising of a cylinder that contains a gas between it and a smooth air tight piston and is put in a heat reservoir.

Isothermal Reversible Expansion and Reversible Compression:

Assume that the gas (system) undergo isothermal reversible expansion from volume V₁ to V₂ at a temperature 'T'. In this reversible process, the gas absorbs heat, qrev, from the reservoir; the entropy change of the system, ΔS₁, is represented by,

ΔS₁ = q_rev/T

As the reservoir too loses heat q_rev in a reversible manner, the entropy change, ΔS₂, of the reservoir is represented by,

ΔS₂ = - q_rev/T

The net entropy change of the isolated system, ΔS_a, in this reversible expansion process is, represented by, 

ΔS_a = ΔS₁ + ΔS₂ = q_rev/T - q_rev/T = 0

Assume that the gas undergo isothermal reversible compression back to its original state. Suppose that throughout this compression, heat lost from the system and the heat gained via the reservoir are both reversible. Then, the total entropy change (ΔS_b) of the isolated system illustrated above, throughout reversible compression, is as well equivalent to zero.

ΔS_b = 0

Therefore, ΔS in this cyclic process = ΔS_a + ΔS_b = 0 

This signifies that the total entropy change in the Reversible cycle is zero. Let us now observe how the entropy changes in the cyclic process comprising an irreversible phase.

Isothermal Irreversible Expansion and Reversible Compression:

Assume that the gas undergo isothermal irreversible expansion from a volume V₁ to V₂ at a temperature T. In this process, let us suppose that the gas absorbs heat 'q' irreversibly while the reservoir loses the similar heat reversely. Though, the entropy change of the system (ΔS₁) is still represented by the equation per definition.

ΔS₁ = q_rev/T

However, as the reservoir loses heat 'q' reversibly, the entropy change of the reservoir, ΔS₂, is represented by,

ΔS₂ = - q/T

Therefore, the net entropy change of the isolated system, ΔS_a, in this irreversible expansion process is represented by,

ΔS_a = ΔS₁ + ΔS₂ = q_rev/T - q/T > 0

As q_rev > q

Suppose the gas now experience isothermal reversible compression in such a way that the heat loss via the system and the heat gain by the surroundings are both reversible. The net entropy change of the isolated system, ΔS_b, in this reversible compression process is represented by,

ΔS_b = 0

Therefore, the net entropy change of the isolated system over the whole cycle is:

ΔS_a + ΔS_b > 0

Therefore for any reversible process or cycle ΔS_total = 0

For any irreversible process or cycle ΔS_total > 0

In another words, the second law of thermodynamics recommends that the entropy should increase in an irreversible or a spontaneous process. As all the natural processes are irreversible, the entropy of the universe is constantly increasing. The first and second laws of thermodynamics can be summed up as:

The first law: The energy of the universe is constant.

The second law: Entropy of the universe is tending to maximum.

Statements of the second law of Thermodynamics:

There are three generalized statements which are used to obtain the Statement of the second law of thermodynamics. These statements are illustrated below:

1) The entropy of an isolated system tends to rise and reaches a maximum. This means that the most stable state of an isolated system is the state of maximum entropy. As the universe might be considered as an isolated system, it follows that the entropy of the universe for all time rises.  

2) It is not possible to transfer heat from a cold body to a hotter body without doing some work. This was hypothesized by Clausius. 

 3) According to Kelvin, it is not possible to take heat from a source (that is, a hot reservoir) and convert all of it into work via a cyclic process devoid of losing some of it to a colder reservoir.

Entropy changes during expansion and Compression:

In common, the entropy change of the system is stated by the entropy of the final state (B) minus the entropy of the initial state (A). This is equivalent to _A∫^B dq_rev/T which at constant temperature can be represented as 1/T _A∫^B dq_rev = q_rev/T, here q_rev is the net amount of the heat absorbed reversibly in the process.

Entropy Change in the Isothermal Expansion of an Ideal Gas:

If n mol of an ideal gas is isothermally and reversibly expanded  from the initial state in which it has pressure p₁ and volume V₁ to the final state of volume V₂ and pressure p₂, then as illustrated earlier,

q_rev = w = nRT ln V₂/V₁ = nRT ln p₁/p₂  

And therefore, ΔS = q_rev/T = nR ln V₂/V₁ = nR ln p₁/p₂

                            = 2.303 nR log V₂/V₁ = 2.303 nR log p_l/p₂

Therefore, to compute the entropy change of an ideal gas throughout isothermal expansion or compression, n, V₁ and V₂ or p₁ and p₂ should be known.

Entropy Change during Adiabatic Expansion:

The adiabatic expansion comprises no heat change that signifies ΔS = 0 for the system.

Entropy Change of an Idea Gas if it is expanded under conditions which are not isothermal:

Let n mol of the ideal gas is expanded from an initial state of V₁ and T₁ to a final state of V₂ and T₂. Then according to the first law of thermodynamics, we have:

d_qrev = dU + pdV 

Using the equation dU = nCv‾dT

Here, C_V‾ is the molar heat capacity of the gas under constant volume conditions. 

According to the ideal gas equation, p = nRT/V

By using the above two expressions we encompass,

dq_rev = nC_V‾dT + nRT dV/V 

Or TdS = nC_V‾dT + nRT dV/V

Therefore, dS = 1/T [nC_V‾dT + nRT dV/V]

                     = nC_V‾dT/T + nRdV/V

On integration between the limits T₁ → T₂, V₁ → V₂ and S₁ → S₂

_S1∫^S2 dS = _T1∫^T2 nC_V‾dT/T + _V1∫^V2 nRdV/V

Supposing that C_V‾ is independent of temperature, we encompass

S₂ - S₁ = ΔS = nC_V‾ ln (T₂/T₁) + nR ln (V₂/V₁)

          = 2.303 n [C_V‾ log (T₂/T₁) + R log (V₂/V₁)]

This is the case when we take T and V to be the variable quantities. As for an ideal gas p, y and T are associated by the ideal gas equation, only two of these are independent Variables.

Now assume that we consider T and p as the variables. Let them change from T_l to T₂ and p₁ to p₂ throughout the process. Then these are associated to the initial and final volumes V₁ and V₂ as:

V₂/V₁ = T₂P₁/T₁p₂

Therefore, from the equation:

ΔS = nC_V‾ ln T₂/T₁ + nR ln T₂P₁/T₁p₂  

ΔS = nC_V‾ In T₂/T₁ + nR In T₂T₁+ nR ln = p₁/p₂

     = (C_V‾ + R)n In T₂/T₁ + nR ln p₁/p₂ 

As C_p‾ - C_V‾ = R, C_p‾ = R + C_V‾

ΔS = nC_p‾ ln(T₂/T₁) + nR ln p₁/p₂

    = 2.303 n [C_p‾ log (T₂/T₁) + R log (p₁/p₂)]

Therefore, by using the above equations, we can compute the entropy change of the ideal gas whenever its pressure or volume changes due to the temperature change.

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