Maximum Work
For a change of state from A to B, the First Law gives
UB - UA = q − w
Is there a definite limit to the amount of work which the system can do during its change of state?
The First Law gives no indication of the allowed magnitudes of q and w in the given process.
We have seen in the preceding discussion that, although the values of q and w can vary depending on the degree of irreversibility of the path taken between the state A and B , the Second Law sets a definite limit on the maximum amount of work which can be obtained from the system during a given change of state and, hence, sets a limit on the quantity of thermal energy which the system may absorb. For an infinitesimal change of state
dSsys = δq/T + dSirr
and, from the First Law,
δq = dUsys + δw
Thus
dSsys = (dUsys + dw)/T + dSirr
δw = TdSsys - dUsys - dSirr
and finally,
δw ≤ TdSsys − dUsys (3.1)
If the temperature remains constant throughout the process (and equal to the temperature of the reservoir supplying heat to the system), then the integration of Equation 3.1 from state A to state B gives
w ≤ T (SB − SA) − ( UB − UA)
and since U and S are functions of state, then w cannot be greater than a certain amount, wmax, the work which is obtained from the system when the process is conducted reversibly; that is,
wmax = T ( SB − SA) − ( UB − UA)
This work, wmax , corresponds to the absorption of the maximum heat, qrev, and is the most work that can be performed during the change of state. Since entropy is a state function, then, in undergoing any specific change of state from A to B,
The change in the entropy of the system is the same whether the process is conducted reversibly or irreversibly.
The preceding discussion indicates that it is the heat effect which is different in the two cases; that is, if the process involves the absorption of thermal energy and is conducted reversibly, then the thermal energy absorbed, qrev, is greater than the thermal energy which would have been absorbed if the process had been conducted irreversibly. As has been seen, when an ideal gas is isothermally and reversibly expanded from state A to state B , energy q, where
q = n RT ln VB / VA
is reversibly transferred from the heat reservoir to the gas, and the increase in the entropy of the gas, SB − SA, equals n R ln VB /VA. The entropy of the thermal reservoir decreases by an equal amount and, therefore, entropy is not created; that is, ΔSirr = 0. However, if the gas is allowed to expand freely from VA to VB, then, since the gas performs no work, no thermal energy is transferred from the reservoir to the gas, and there is no change in the entropy of the reservoir. Since entropy is a state function, the value of SB − SA is independent of the process path, and hence, the entropy created, ΔSirr , equals SB − SA, which equals n R ln VB / VA. This entropy is created as a result of the degradation of the work which would have been performed by the gas had the expansion not been carried out against zero pressure. This degraded work equals wmax as well as qrev. The free expansion therefore represents the limit of complete irreversibility, during which all of the potential work is degraded because of the increase in volume of the gas. The degraded potential work in the gas accounts for the increase in the entropy of the gas. For the isothermal expansion of an ideal gas from the state A to the state B, the value of ΔSirr is
0 ≤ ΔSirr ≤ n R ln VB /VA
ΔSirr = 0 for a reversible isothermal expansion and ΔSirr = n R ln VB/VA for a free expansion. The value of ΔSirr is thus shown to depend on the degree of irreversibility of the process.