Irreversible expansion

As for a perfect gas the initial and final states must satisfy the perfect-gas equation of state 

The change in internal energy for an adiabatic expansion of one mole of gas from Vx to V2 is calculated using the relation derived in Section 2.7  

The initial and final temperatures are related to the volumes by the equation derived above. 

If the pressure on a gas is suddenly lowered from a value Px to P2 (Fig. 8.1) and the gas is allowed to expand against this new pressure to a volume V2 it does work 

If this expansion is carried out adiabatically so that q — 0, then, as AU q + w, 

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At the instant a pressure vessel ruptures, pressure at the contact surface is given by Eq. (6.3.22). The further development of pressure at the contact surface can only be evaluated numerically. However, the actual p-V process can be adequately approximated by the dashed curve in Figure 6.12. In this process, the constant-pressure segment represents irreversible expansion against an equilibrium counterpressure P3 until the gas reaches a volume V3. This is followed by an isentropic expansion to the end-state pressure Pq. For this process, the point (p, V3) is not on the isentrope which emanates from point (p, V,), since the first phase of the expansion process is irreversible. Adamczyk calculates point (p, V3) from the conservation of energy law and finds... 

The first term is due to the irreversible expansion from V, to Vj, and the second term to the isentropic expansion from Vj to Vj. Adamczyk does not actually say how p3 should be chosen. A reasonable choice for seems to be the initial-peak shock overpressure, as calculated from Eq. (6.3.22). The equation presented above can be compared to the results of Guirao et al. (1979). They numerically evaluated the work done by the expanding contact surface. When the difference between... 

In this discussion, we will limit our writing of the Pfaffian differential expression bq, for the differential element of heat flow in thermodynamic systems, to reversible processes. It is not possible, generally, to write an expression for bq for an irreversible process in terms of state variables. The irreversible process may involve passage through conditions that are not true states" of the system. For example, in an irreversible expansion of a gas, the values of p. V, and T may not correspond to those dictated by the equation of state of the gas. 

Because entropy is a state function, the change in entropy of a system is independent of the path between its initial and final states. This independence means that, if we want to calculate the entropy difference between a pair of states joined by an irreversible path, we can look for a reversible path between the same two states and then use Eq. 1 for that path. For example, suppose an ideal gas undergoes free (irreversible) expansion at constant temperature. To calculate the change in entropy, we allow the gas to undergo reversible, isothermal expansion between the same initial and final volumes, calculate the heat absorbed in this process, and use it in Eq.l. Because entropy is a state function, the change in entropy calculated for this reversible path is also the change in entropy for the free expansion between the same two states. 

Because A.Ssllll = —AS, AStor = 0. This value is in accord with the statement that the process is reversible, (b) For the irreversible process, AS is the same, at +7.6 J K 1. No work is done in free expansion (Section 6.3), and so w = 0. Because AU = 0, it follows that q = 0. Therefore, no heat is transferred into the surroundings, and their entropy is unchanged ASslirr = 0. The total change in entropy is therefore ASt()t = +7.6 J-K. The positive value is consistent with an irreversible expansion. 

In an intermediate irreversible expansion in which the external pressure is not zero but is less than the pressure of the gas by a finite amount, some work would be... 

Derive an explicit expression for the work performed in the irreversible expansion of a gas from volume Vi to volume V2 against a constant external pressure P that is less than the pressure of the gas throughout the expansion. 

In other words, for the system plus surroundings, this irreversible expansion has been accompanied by an increase in entropy. 

We may contrast this result for A totai with that for Al/totai for an ideal gas, as mentioned in Section 5.1. In the irreversible expansion of an ideal gas, Allgys = 0 the surroundings undergo no change of state (Q and W are both equal to zero), and hence, A /total = 0- ff we consider the reversible expansion of the ideal gas, AUsys is also equal to zero and AUsun is equal to zero because Q = —W, so again A /total = 0- Clearly, in contrast to AS, AU does not discriminate between a reversible and an irreversible transformation. 

Shown mathematically that the magnitude of the work involved in a reversible expansion of an ideal gas from volume Vj to V2 is larger than the corresponding work involved in an irreversible expansion... 

Figure 3.10 Schematic of Joule s experiment for irreversible expansion of a gas into an evacuated chamber in a water bath.

The expansion of an ideal gas in the Joule experiment will be used as a simple example. Consider a quantity of an ideal gas confined in a flask at a given temperature and pressure. This flask is connected through a valve to another flask, which is evacuated. The two flasks are surrounded by an adiabatic envelope and, because the walls of the flasks are rigid, the system is isolated. We now allow the gas to expand irreversibly into the evacuated flask. For an ideal gas the temperature remains the same. Thus, the expansion is isothermal as well as adiabatic. We can return the system to its original state by carrying out an isothermal reversible compression. Here we use a work reservoir to compress the gas and a heat reservoir to remove heat from the gas. As we have seen before, a quantity of heat equal to the work done on the gas must be transferred from the gas to the heat reservoir. In so doing, the value of the entropy function of the heat reservoir is increased. Consequently, the value of the entropy function of the gas increased during the adiabatic irreversible expansion of gas. 

The key to this solution lies in the fact that the operation involved is an irreversible expansion. Taking Cv as constant between P i and T2, AU = — W = nCv(T2 — Pi) where n is the kmol of gas and T2 and Pi are the final and initial temperatures, then for a constant pressure process, the work done, assuming the ideal gas laws apply, is given by ... 

Figure 9.1 Work of (a) reversible expansion and (b) irreversible expansion of an ideal gas from (V P<) to (l/f, Pf).

If an amount n of gas, at pressure, Pi, having volume, V, at constant temperature, T, is expanded in such a way that the opposing pressure is always made just infinitesimally smaller than P, at any snapshot instant (i.e. = Pi- d P) and this process is continued (always maintaining the current pressure to be only marginally less than that of the gas) until the same final state as in the irreversible expansion discussed in the section above (that of Pf and Vf) is reached, then this expansion is termed an expansion under reversible conditions. It is the case that if, at any moment, the opposing pressure to the gas pressure, P were made equal to P + dP then the expansion would be reversed and become a contraction. Thus during the course of this particular mode of (reversible) expansion we proceed through a succession of equilibrium states. 

In contrast to the case of the irreversible expansion, P is not constant during the reversible expansion but varies along the isotherm (equation (4.1), Frame 4) ... 

However the change in entropy, AS, will be independent of whether the gas is expanded reversibly or irreversibly since the initial (Vi, Pi, T) and final (Vf, Pf, T) states of the system are identical in all three experiments (reversible and irreversible expansions) and the entropy is given, in each case, by ... 

This relationship is correct for both reversible and irreversible expansion processes. Also since for an amount n of an ideal gas (Frame 9) ... 

For an irreversible expansion of a real gas at constant temperature due to a heat reservoir, the change of entropy flow is d,.S = 8q/T, where 8q is the heat flow between the gas and the reservoir to maintain the constant temperature. The increase of entropy during the expansion is... 

A certain gas obeys the equation of state P(V -nb) - nRT and has a constant volume heat capacity, Cv, which is independent of temperature. The parameter b is a constant. For 1 mol, find W, AE, Q, and AH for the following processes (a) Isothermal reversible expansion. (b) Isobaric reversible expansion. (c) Isochoric reversible process, (d) Adiabatic reversible expansion in terms of Tlf Vlt V2, Cp, and Cv subscripts of 1 and 2 denote initial and final states, respectively. (c) Adiabatic irreversible expansion against a constant external pressure P2, in terms of Plf P2, Tj, and 7 = (Cp/Cy). 

For given values of Tc and Tu, the highest possible value of co is attained for Carnot-cycle refrigeration. The lower values for the vapor-eompression eycle result from irreversible expansion in a tlrrottle valve and irreversible compression. The following example provides an indication of typical values for eoefficients of perfomianee. 

Irreversible swelling is a phenomenon principally observed with acrylic strong base anion exchange resins whereby upon undergoing their first few aqueous ion exchange cycles an irreversible expansion occurs of around 7—10% over and above the reversible volume changes which thereafter apply. 

FIGURE 12.2 Stages in an irreversible expansion of a gas from an initial state (a) of volume V to a final state (c) of volume Vj. In the intermediate stage shown (b), the gas is not in equilibrium because of turbulence, pressure and temperature cannot be defined. 

The tools are now in place for using the second law to predict whether specific processes will be spontaneous. We illustrate the procedure first for spontaneous cooling of a hot body, and then for irreversible expansion of an ideal gas. 

IRREVERSIBLE EXPANSION OF AN IDEAL GAS Consider a gas confined within a piston-cylinder arrangement and held at constant temperature in a heat bath. Suppose the external pressure is abruptly reduced and held constant at the new lower value. The gas immediately expands against the piston until its internal pressure declines to match the new external pressure. The total entropy of system plus surroundings will increase during this expansion. In preparation for a quantitative example, a general comparison of irreversible and reversible processes connecting the same initial and final states provides insight into why the total entropy increases in a spontaneous process. 

FIGURE 13.7 Work done by a system in reversible and irreversible expansions between the same initial and final states. The work performed is greater for the reversible process. 

During an expansion, must be less than F, the pressure of the gas. For a reversible expansion, Fg t is only infinitesimally smaller (so the system is always close to equilibrium) but for an irreversible expansion, F xt is measurably smaller. Therefore, the area under a graph of F xt plotted against V is less than that of a graph of F against V (Fig. 13.7), so... 

How can the heat from Example 13.7, which is irreversible from the perspective of the system, be reversible from the perspective of the surroundings This can be accomplished by enclosing the gas in a material (such as a metal) that can efficiently transfer heat to and from the surroundings, and thus remain close to equilibrium, at the same time that the gas itself is far from equilibrium due to the gas currents that occur during the irreversible expansion. 

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