Ideal gases irreversible isothermal expansion

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) ... 

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. 

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. 

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. 

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... 

Figure 4.14. An irreversible expansion of an ideal gas into a vacuum is isothermal (Joule s law). Therefore, for this process, AS is the same as for the reversible, isothermal expansion described in figure 4.13.

Expansion of an ideal gas at a specific temperature. The curve is the reversible path for the given temperature. Expansion from to V2 follows this path only if equilibrium is maintained throughout. Then, the area under the segment from V- to V2 is iPdV, which is the work associated with the reversible process. Several irreversible expansions are indicated. A one-step irreversible process is for the external pressure to go instantaneously from Pi to 2, and then the gas expands from to V2 against a pressure of P2. The amount of work expended by the gas in this irreversible process is the area of the shaded rectangle. It is clear that the area under any of the irreversible paths is less than the area under the path that follows the isotherm where the gas pressure is always equal to the external pressure. The amount of reversible work is approached through a sequence of irreversible steps as the steps are made smaller and smaller. 

SOLUTION In this example, we must consider both the entropy change of the system and the entropy change of the surroundings. We outline the solution methodology with process A, the isothermal, irreversible expansion from state 1 at 2 bar and 0.04 m /mol to state 2 at 1 bar and 0.08 mVmol. Since the process is isothermal, we assume the temperature of the surroundings and that of the system are identical. Their values can be found using the ideal gas law ... 

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