Yes.. In an isothermal transformation it follows that Q = W, since ΔU = 0. This means that in an isothermal transformation all the energy that the system can be exchanged into heat, becomes work.
If the transformation is reversible isothermal addition, the work done by the system will be the maximum possible between the initial and final states of the transformation.
How many constraint conditions there are, how many “the maximum work” would be there, correspond to dU, dH, dF, dG, dΩ, etc.
“the work done” correspond to a state change Ydx, and is the process of the free energy transportation. “pV work” correspond to the conversion between heat and the real free energy “ψ”.
The internal energy U may be reclassified into “the internal heat energy” + “the real free energy”, i.e. U=q+ψ. Then we will find the thermodynamic potentials and “the maximum work” quite easy to understand, moreover, we don’t need the concept of “the maximum work” for explaining “the thermodynamic potentials”.
consider a system, thermally insulated comprised of several bodies that are not thermally equlibrio together. During the process establishing the equlibrio, the system can perform a job. However, the way to balance diferetes shapes can be made, therefore, you can reach different final states. (in particular, different energy, different entropy).
Accordingly, the amount of work, which can realize a system imbalance, depending on the shape, which is reached equlibrio. Now, as the maximum work is done? ... should be taken into account, only work because of the imbalance, regardless, due to dilatation of the system, therefore is considered a change of zero volume.
E0 the initial energy of the system, and E (S) the steady-state energy as a function of the entropy. If the work done thermally isolated, the work is equal to the energy variation. (R=energy variation)
R=E0-E(S)
Differentiating R with respect to the entropy of the final state
(dR/dS)=(dE/dS)=-T (constant volume)
The derivative is negative, ie, R decreases with increasing S.However, S does not change into a thermally isolated system. Therefore, reached a maximum value of R, if S remains constant during the process.
Concludes: maximum working ----> entropy remains constant, that is, when the transition to equilibrium takes place in a reversible manner.
Reversible work is nothing but the maximum amount of work that can be obtained as a system undergoes a process between to specified end states.This maximum amount of work potential that is available for doing some useful work is the availability or exergy of the system. The part of the maximum work potential that diappears into the suurounding and becomes unavilable is the unavailability or exergy destroyed or irreversibility . for example, a red hot iron block (500K) left in the atmosphere (290K) has maximum exergy in terms of its heat energy ( m.Cp. dT). This energy could be utilized in doing some useful work. But if it is allowed to cool down in the atmosphere, its available energy disappears into the atmosphere and which could not be regained back from the atmosphere. This means the process is irreversible. In this case the entire exergy is equal to irreversibility.
Theoretically, during a reversible process, we don't dissipate energy, so the created entropy (S irr) equals to zero (S system=S reservoir) and system performs maximum work:
w max = q rev (the reversible heat)
but in the case of irreversible process, we have energy dissipation and thus the work is less than q rev. In this case the created entropy (S irr) is positive (>0).
The value of the work is between zero and W max, depending on irreversibility of the process.
maximum work done by a system is calculated between initial and dead state on the other hand reversible process dont always end up at the dead state .actually i think you are confusing with different type of reversible process but when they taken to dead state they all give same work. if u consider final state = dead state ,both are same
But a process will occur only if it result in the increase of entropy of universe, isn't it? @ Mr.Yousef Payandeh and Mr.Daniel Ricardo Delgado
Yes, the maximum work that is capable of performing a system is subject to the proximity of a reversible process
In both the case, i.e that may be reversible or irreversible the change in volume is same, W= Pdv, if dv is same P should be different so that work done be different. If in reversible process p is more than that of irreversible process, then reversible work done is greater than than that of irreversible work done. But my question is that how P is different in both the case.
Entropy: A concept that is not a physical quantity
https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity
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