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## Entropy of Change, Chemistry tutorial

:EntropyEfficiency 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

[(q

_{2}+ q_{1})/q_{2}] = [(T_{2}- T_{1})/T_{2}]If in a Carnot engine, heat q

_{2}is absorbed at the higher temperature T_{2}and q_{l}is absorbed at the lower temperature T_{l}(in reality q_{1 }will be a negative quantity since heat is discarded) then, according to the above equation,[(q

_{2}+ q_{1})/q_{2}] = [(T_{2}- T_{1})/T_{2}]Or 1 + (q

_{1}/q_{2}) = T_{1}/T_{2}That is, q

_{1}/T_{1}= - q_{2}/T_{2}Therefore, q

_{1}/T_{1}+ q_{2}/T_{2}= 0Therefore, 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 dq

_{1}is absorbed at temperature T_{1}'and dq_{2}absorbed at T_{2}', then for each and every small Carnot cycle,dq

_{1}/T_{1}' + dq_{2}/T_{2}' = 0Summing these over all the cycles we can represent in general, sum of dq

_{i}/T_{i}termsOver 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 = 0Here, 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 = 0Here, 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 = 0Or 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}/TTherefore, 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 = dSTherefore ∫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}/TLet us associate the changes in internal energy and enthalpy to entropy change.

If we now place dq = TdS, dU = nC

_{V}dT and dw = - pdV in dU = dq + dwTherefore,

dU = nC

_{V}‾dT = TdS - pdVAs 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}/TThis 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 SystemsWe 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 V

_{1}to V_{2}at a temperature 'T'. In this reversible process, the gas absorbs heat, qrev, from the reservoir; the entropy change of the system, ΔS_{1}, is represented by,ΔS

_{1}= q_{rev}/TAs the reservoir too loses heat q

_{rev}in a reversible manner, the entropy change, ΔS_{2}, of the reservoir is represented by,ΔS

_{2}= - q_{rev}/TThe net entropy change of the isolated system, ΔS

_{a}, in this reversible expansion process is, represented by,ΔS

_{a}= ΔS_{1}+ ΔS_{2}= q_{rev}/T - q_{rev}/T = 0Assume 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}= 0Therefore, ΔS in this cyclic process = ΔS

_{a }+ ΔS_{b }= 0This 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 V

_{1}to V_{2}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 (ΔS_{1}) is still represented by the equation per definition.ΔS

_{1}= q_{rev}/THowever, as the reservoir loses heat 'q' reversibly, the entropy change of the reservoir, ΔS

_{2}, is represented by,ΔS

_{2}= - q/TTherefore, the net entropy change of the isolated system, ΔS

_{a}, in this irreversible expansion process is represented by,ΔS

_{a}= ΔS_{1}+ ΔS_{2}= q_{rev}/T - q/T > 0As q

_{rev }> qSuppose 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}= 0Therefore, the net entropy change of the isolated system over the whole cycle is:

ΔS

_{a}+ ΔS_{b}> 0Therefore for any reversible process or cycle ΔS

_{total}= 0For any irreversible process or cycle ΔS

_{total}> 0In 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 ThermodynamicsThere 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 CompressionIn 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 p

_{1}and volume V_{1}to the final state of volume V_{2}and pressure p_{2}, then as illustrated earlier,q

_{rev}= w = nRT ln V_{2}/V_{1}= nRT ln p_{1}/p_{2}And therefore, ΔS = q

_{rev}/T = nR ln V_{2}/V_{1}= nR ln p_{1}/p_{2}= 2.303 nR log V

_{2}/V_{1}= 2.303 nR log p_{l}/p_{2}Therefore, to compute the entropy change of an ideal gas throughout isothermal expansion or compression, n, V

_{1}and V_{2 }or p_{1}and p_{2}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 V

_{1}and T_{1}to a final state of V_{2}and T_{2}. Then according to the first law of thermodynamics, we have:d

_{qrev}= dU + pdVUsing 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/VOr TdS = nC

_{V}‾dT + nRT dV/VTherefore, dS = 1/T [nC

_{V}‾dT + nRT dV/V]= nC

_{V}‾dT/T + nRdV/VOn integration between the limits T

_{1}→ T_{2}, V_{1}→ V_{2}and S_{1}→ S_{2 }_{S1}∫^{S2 }dS =_{T1}∫^{T2}nC_{V}‾dT/T +_{V1}∫^{V2}nRdV/VSupposing that C

_{V}‾ is independent of temperature, we encompassS

_{2}- S_{1}= ΔS = nC_{V}‾ ln (T_{2}/T_{1}) + nR ln (V_{2}/V_{1})= 2.303 n [C

_{V}‾ log (T_{2}/T_{1}) + R log (V_{2}/V_{1})]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 T_{2}and p_{1}to p_{2}throughout the process. Then these are associated to the initial and final volumes V_{1}and V_{2}as:V

_{2}/V_{1}= T_{2}P_{1}/T_{1}p_{2}Therefore, from the equation:

ΔS = nC

_{V}‾ ln T_{2}/T_{1}+ nR ln T_{2}P_{1}/T_{1}p_{2}ΔS = nC

_{V}‾ In T_{2}/T_{1}+ nR In T_{2}T_{1}+ nR ln = p_{1}/p_{2}= (C

_{V}‾ + R)n In T_{2}/T_{1}+ nR ln p_{1}/p_{2}As C

_{p}‾ - C_{V}‾ = R, C_{p}‾ = R + C_{V}‾ΔS = nC

_{p‾}ln(T_{2}/T_{1}) + nR ln p_{1}/p_{2}= 2.303 n [C

_{p}‾ log (T_{2}/T_{1}) + R log (p_{1}/p_{2})]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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