Ideal gases irreversible expansion

It suffices to carry out one such experiment, such as the expansion or compression of a gas, to establish that there are states inaccessible by adiabatic reversible paths, indeed even by any adiabatic irreversible path. For example, if one takes one mole of N2 gas in a volume of 24 litres at a pressure of 1.00 atm (i.e. at 25 °C), there is no combination of adiabatic reversible paths that can bring the system to a final state with the same volume and a different temperature. A higher temperature (on the ideal-gas scale Oj ) can be reached by an adiabatic irreversible path, e.g. by doing electrical work on the system, but a state with the same volume and a lower temperature Oj is inaccessible by any adiabatic path. 

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. 

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. 

Points a and b in Figure 6.7 represent the initial and final states of an irreversible adiabatic expansion of an ideal gas. The path between is not represented because the temperature has no well-defined value in such a change different parts of the system may have different temperatures. The inhomogeneities in the system that develop during the irreversible change do not disappear until a new equilibrium is reached at b. 

A reversible adiabatic expansion of an ideal gas has a zero entropy change, and an irreversible adiabatic expansion of the same gas from the same initial state to the same final volume has a positive entropy change. This statement may seem to be inconsistent with the statement that 5 is a thermodynamic property. The resolution of the discrepancy is that the two changes do not constitute the same change of state the final temperature of the reversible adiabatic expansion is lower than the final temperature of the irreversible adiabatic expansion (as in path 2 in Fig. 6.7). 

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... 

From the base equation (5.46d), the AG variations can also be evaluated for a wide variety of other changes. Analogous variations of other thermodynamic potentials (U, H, A) are easily derived from the corresponding differential expressions (5.46a-c). Sidebar 5.14 illustrates such evaluations for reversible volume expansion of an ideal gas, and Sidebar 5.15 shows how to carry out the analogous evaluations for irreversible conditions. Once the power of the differential expressions (5.46a-d) is understood, they will be found sufficient for the solution of nearly any practical problem that the student may encounter. 

Determine AS, ASsurr, and AStot for (a) the isothermal reversible expansion and (b) the isothermal irreversible free expansion (expansion against zero pressure) of LOO mol of ideal gas molecules from 8.00 L to 20.00 L atm at 292 K. Explain any differences between the two paths. 

Initially, a sample of ideal gas at 323 K has a volume of 2.59 L and exerts a pressure of 3.67 atm. The gas is allowed to expand to a final volume of 8.89 L via two pathways (a) isothermal, reversible expansion and (b) isothermal, irreversible free expansion. Calculate AStot, AS, and ASsurr for both pathways. 

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).

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

Consider the free, isothermal (constant T) expansion of an ideal gas. Free means that the external force is zero, perhaps because a stopcock has been opened and the gas is allowed to expand into a vacuum. Calculate AU for this irreversible process. Show that q = 0,so that the expansion is also adiabatic [q = 0) for an ideal gas. This is analogous to a classic experiment first performed by Joule. 

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. 

Comparison of a reversible and an irreversible isothermal expansion of an ideal gas for the same initial and final states (see text Example 7.12) ... 

In the isothermal expansion of a gas, the final volume V2 is greater than the initial volume Vi and, consequently, by equation (19.26), AS is positive that is to say, the expansion is accompanied by an increase of entropy of the system. Incidentally, when an ideal gas expands (irreversibly) into a vacuum, no heat is taken up from the surroundings ( 9d), and so the entropy of the latter remains unchanged. In this case the net entropy increase is equal to the increase in entropy of the system, i.e., of the gas, alone. 

Now let s consider another example, the expansion of an ideal gas at constant temperature (referred to as an isothermal process). In the cylinder-piston arrangement of Figure 19.5, when the partition is removed, the gas expands spontaneously to fill the evacuated space. Can we determine whether this particular isothermal expansion is reversible or irreversible Because the gas expands into a vacuum with no external... 

A Figure 19.5 An irreversible process. Initially an ideal gas is confined to the right half of a cylinder. When the partition is removed, the gas spontaneously expands to fill the whole cylinder. No work is done by the system during this expansion. Using the piston to compress the gas back to its original state requires the surroundings to do work on the system. 

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