Why a reversible process does the maximum amount of work



Concepts in Thermodynamics

Many of the concepts and equations in chapter 18 are simply given to you and you are asked to accept them on faith. Here I try to fill in many of the blanks so you can acquire a deeper understanding of thermodynamics.

The second law of thermodynamics states that a process will occur if it results in an increase in the entropy of the universe. Your textbook defines entropy as a measure of energy dispersal and they point out that dispersed energy represents a more probable state than concentrated energy. (That’s why high energy molecules are very valuable: they are rare!) Thus, the second law simply says that processes occur spontaneously in the direction that corresponds to going from a low probability state to a high probability state. Most textbooks also equate entropy with “disorder” since a highly ordered system is also highly improbable. Your text does not give the general equation for ΔS because it involves a difficult technical concept: “reversible process”. The equation is:

[pic] where qrev represents the heat transfer due to a “reversible” process

The origin of this equation is explained in sections I and II below.

In this context, “reversible” means a process that is carried out infinitely slowly. Your textbook gives one special case in which this equation can be used to calculate ΔSsurr:

[pic]

A chemical reaction can transfer only a tiny amount of heat to or from the surroundings (compared to the total heat content of the surroundings). Thus, as far as the surroundings are concerned, a reaction transfers a negligible amount of heat. This is equivalent to transferring the heat reversibly. Thus the general equation can be used to calculate ΔSsurr. However, ΔSrxn can NOT be calculated using this equation because ΔHrxn is NOT negligible with respect to the system.

Section I tries to clarify the meaning of a reversible process and demonstrates that doing a process “reversibly” (very slowly) results in the maximum possible work. This idea is used in section II to show how the equation for entropy comes from the analysis of the efficiency of a steam engine that is run reversibly.

In section III, I expand on how one uses the third law of thermodynamics to measure the absolute entropy of a pure substance in a given state. Recall that the third law states that a perfect crystal at absolute zero has zero entropy (perfect order).

In section IV, I show how the analysis of the efficiency of a steam engine led to the second law of thermodynamics.

In section V, I use the expansion of an ideal gas to illustrate the connection between entropy, energy dispersal, disorder, and probability. It is not at all obvious how the equation for entropy is related to the probability of a system being in a particular state. This section tries to help you see that connection.

Without proof, your textbook makes the following statement: “ΔG represents the maximum useful work that can be done by a product-favored system on its surroundings under conditions of constant temperature and pressure.” In section VI, I show why this is true. (As you will see, the statement in your textbook is not exactly correct. ΔG represents the maximum amount of “non-PV work” that a reaction can do.)

Some of the explanations below require the use of simple calculus. I don’t think this will cause you any problems.

I. Reversible processes and reversible chemical reactions

In thermodynamics, “reversible” has a technical meaning. A reversible process is one that is infinitely slow! Obviously, no real process is reversible but you can approximate a reversible process by doing it as slowly as possible. For example, you can think of a reaction that is at equilibrium as occurring infinitely slowly (nothing is happening!). Imagine that the forward reaction occurs just barely faster than the reverse reaction so, for all practical purposes, the reaction is at equilibrium. You can make the reaction reverse by adding a tiny amount of product or by removing a tiny amount of reactant. Thus, a reaction at equilibrium is reversible! Conversely, a reaction that is not at equilibrium is irreversible. All real processes are irreversible.

Reversible processes produce the maximum amount of work

If a process does work on the surroundings, you get more work out of the process if it is done slowly. This is because less heat is lost to the surroundings. So a reversible process (infinitely slow) does the maximum work. There is one example of this that is easy to understand: the work done by a gas expanding inside a piston. We will see that the maximum amount of work is obtained when the gas is allowed to expand very slowly (reversibly).

[pic]

Let’s compare a reversible vs. an irreversible expansion of the gas. Initially, let’s assume that the internal and external pressures are the same (Pint = Pext). To expand the gas reversibly, we reduce the external pressure very slowly (technically, infinitely slowly) so the system never strays significantly from equilibrium. This is a reversible process because a very small increase in the external pressure would cause the gas to be compressed. To expand the gas irreversibly, we reduce the pressure very quickly. As the gas expands it does work on the surroundings (by moving the piston against the external pressure). The plots of pressure vs. volume for these two processes are shown below. Figure (a) represents the reversible expansion. Figure (b) is the irreversible expansion.

[pic]

The work done in these two processes is the area under the curves. (Remember that PV has units of energy or work). Clearly, more work is done by the reversible expansion. The reversible expansion does the maximum amount of work because the gas is pushing against the maximum possible external pressure. (Since the process remains at equilibrium throughout the reversible expansion, Pext = Pint.) If the external pressure was any higher, the process would reverse and the gas would be compressed.

In general, energy released by a reversible process can do the maximum amount of work because less of the energy is lost as heat. A process that is done quickly (irreversibly) tends to generate turbulence and friction resulting in heat loss to the surroundings.

II. Entropy and the efficiency of a steam engine: The origin of the equation ΔS = qrev/T

Thermodynamics was developed by people who wanted to maximize the efficiency of a steam engine. A steam engine operates much like an internal combustion engine of a car except high pressure steam is used to move a piston rather than the combustion of gasoline. The movement of the piston is then used to do useful work. After the piston has moved, the remaining steam is ejected from the engine and the piston returns to its original position and is ready for another cycle. This cycle is represented schematically below.

[pic]

In essence, a steam engine is a machine for converting heat into work. Heat is extracted from a high temperature reservoir (hot steam at Th). Some of the heat is converted to work (w) by allowing the steam to expand. After expansion, the remaining heat (cooler steam at Tl) is ejected into a low temperature reservoir. The goal is to convert as much of the extracted heat into useful work as is possible. The efficiency of a steam engine is:

[pic]

Work (w) is a negative number because work is done by the system (steam) on the surroundings. qh is positive because it is heat added to the system from the surroundings. Thus, the negative sign makes the efficiency a positive number.

Since the engine operates in a cycle (returns to its initial state): ΔEsys = 0 (E is a state function.)

From the first law of thermodynamics: ΔEsys = qh + ql + w = 0 (note that ql is negative with respect to the system)

Thus: qh + ql = -w

[pic]

[Notice that efficiency = 1 (100%) only if ql = 0 which means that all of the heat has been converted to work!]

The maximum efficiency results if the engine is operated reversibly (see section I).

If the steam behaves as an ideal gas and the engine is operated reversibly, then it can be shown (I won’t derive it here) that:

[pic] (If the engine is run irreversibly, then[pic])

(Notice that efficiency = 1 only if Tl = 0 which means that all of the heat has been converted to work so the “steam” is now at absolute zero! To maximize the efficiency of an internal combustion engine, the engine is designed to operate at very high temperature and to discharge the products at the lowest possible temperature!)

Thus:

[pic]

Rearranging gives:

[pic]

This result shows that the function q/T is a state function (that we call entropy!) since it sums to zero for a cyclic process. (The system returns to its original state so all state functions describing the system must not change). But understand that q/T is a state function only if the engine is run reversibly so that q = qrev. If the engine is run irreversibly, then the sum above would be less than zero! Recall that q is not a state function so for q/T to be a state function we must choose a particular path: a reversible path!

At constant temperature: [pic]

If T is allowed to vary, ΔS =[pic] integrated between the initial and final temperatures.

III. Measuring absolute entropy (where did the numbers in appendix 3 come from?)

From the third law of thermodynamics, we know that the entropy of a perfect crystalline substance at

0 K is zero. The existence of a zero point on the entropy scale allows us to assign an absolute entropy value (as opposed to a relative entropy value) to a pure substance (see appendix 3). Notice that the unit for entropy (J/K) is the same as that for heat capacity. Absolute entropy is measured by determining how the heat capacity of a substance changes with temperature. (The temperature is changed VERY slowly to approximate a reversible process.) A substance with a higher heat capacity also has more entropy (see section IV)!

Remember that heat capacity (C) is the amount of heat required to raise the temperature of a substance 1 K (or 1 oC).

Thus the heat absorbed during a temperature change is:

q = CΔT

We have to use a little calculus here:

dq = CdT

The entropy change caused by a change in temperature is:

ΔS = [pic] (note: C depends on temperature)

To determine the absolute entropy at temperature T, the integral is evaluated from 0 K to T. This is equivalent to determining the area under the curve produced by plotting C/T vs. T (see graph below). From the third law, we know that the entropy at 0 K is zero so the area under the curve is the absolute entropy at temperature T!! If phase changes occur, then ΔHfus/T and ΔHvap/T must be added to obtain the total entropy.

Assuming no phase changes, the area under the curve is the absolute entropy of this substance at 300 K. The unit for heat capacity is J/K so the area would also have this unit! If phase changes occur, then ΔHfus/T and ΔHvap/T must be added to obtain the total entropy. The numbers in appendix 3 were measured using one mole of each substance at a pressure of one bar so they are standard molar entropies (So) whose unit is J/mol-K.

IV. What does any of this have to do with the second law?

Based on the previous argument about the efficiency of a steam engine, we will show that the entropy of the universe increases whenever an irreversible process occurs (the second law!). It is easy to show that no steam engine can have an efficiency greater than that calculated above for a reversible engine. If there was such an engine, one could transfer heat from a colder object to a hotter object without doing any work. From our common every day experience, we know that this never happens (heat only flows spontaneously from hot to cold)! Pretend that you had two reversible steam engines and that engine 1 is more efficient than engine 2. Hook them up as shown below. Engine 2 uses the work produced by engine 1 to pump heat from the low temperature reservoir to the high temperature reservoir. For the overall process w = 0. However, since engine 1 is more efficient than engine 2, q1 0

Here are 5 different ways to state the second law of thermodynamic:

1. For any spontaneous process the entropy of the universe increases.

2. It is impossible for heat to spontaneously flow from a cold object to a hot object.

3. No steam engine can be more efficient than a reversible steam engine.

4. It is impossible to build a perfectly efficient steam engine.

5. It is impossible to build a perpetual motion machine of the second kind.

The second law also explains why gases always expand to fill the available space (see section V). You never see a gas spontaneously contract!

(Note: It can be shown that the efficiency of a heat engine is independent of the working material. So the second law applies to all systems, not just those consisting of an ideal gas!)

V. Relationship between entropy, probability, and disorder

Another interpretation of the second law is that a spontaneous process corresponds to a system changing from a less probable state to a more probable state. It should be clear that a disordered state is more probable than an ordered state since there are many more ways for a system to be disordered than ordered.

Let’s use the expansion of an ideal gas to see the connection between entropy and probability.

Consider two connected vessels filled with one mole of an ideal gas as shown below:

[pic]

It should be obvious that it is very unlikely that all of the gas will be in just one of the vessels. Instead, it will spread out into both vessels until the pressure is uniform throughout the system. If V = ½ V’, then the right-hand vessel will contain twice as much gas as the left-hand vessel (2/3 of the gas will in the right-hand vessel and 1/3 in the left-hand vessel). In general, the most probable distribution of the gas is for the left hand vessel to contain

[pic] molecules of gas (NA is Avogadro’s number)

Let’s start with all of the gas (one mole) in the left-hand vessel and calculate the entropy change that occurs when this ideal gas expands reversibly at constant temperature to fill the entire system. So the gas is expanding from an initial volume of V to a final volume of V + V’.

Since there are no attractive or repulsive forces between the molecules of an ideal gas, its internal energy depends only on the temperature (see note below for further discussion of this fact). Since T is constant, E does not change upon expansion:

ΔE = q + w = 0 so:

q = -w

Recall that for the expansion of a gas (see section I):

w = -[pic](The area under the PV curve, the negative sign produces the correct sign for w.)

For one mole of an ideal gas:

[pic] so:

w = -RT[pic]

If the gas expands from V1 to V2, then integrating gives.

w = -RTln[pic]

In our example, V1 = V and V2 = V + V’ so:

w = -RTln[pic]

Since q = -w and the gas was expanded reversibly, and [pic]

ΔS = Rln[pic]

Now we can see the connection to probability:

The probability (P1) of one molecule of the gas being within a volume, V, is:

[pic]

The probability (P) of all NA molecules being in a volume, V, is:

[pic] (If V < V + V’, then this probability is VERY low as it should be!)

Taking the natural log of both sides and multiplying by “Boltzmann’s constant” (k) gives:

klnP = kNAln[pic] = -Rln[pic] = -ΔS (k = R/NA)

We have shown that: ΔS = -klnP (where P is the probability of the gas occupying volume V)

As the gas expands form V to V + V’, P increases from a small fraction to 1, and ΔS decreases from a large positive number to zero. That is, initially, all of the gas is in the left-hand vessel which represents a very low probability, highly ordered state with very low entropy. Thus, the gas expands spontaneously until the probability, disorder, and entropy are as high as possible. When the gas has filled the entire volume, the probability is one, the system is maximally disordered which means that the system is at equilibrium!

Note: There is a subtle but important point that I did not emphasize above. If we let the gas expand freely from the left hand vessel until it fills the entire system, then no work will be done and no heat will be absorbed. No work is done because the gas is expanding into a vacuum so it is not pushing anything out of the way. Since no work is done and the internal energy doesn’t change (constant T), then no heat is absorbed and q/T = 0. Does this mean the ΔS = 0? NO! The free expansion of a gas is irreversible so q is NOT qrev and q/T is NOT ΔS! In order to calculate ΔS, we have to imagine a reversible pathway and determine how much heat is absorbed. So we imagined that the gas expanded exactly as was done in section I! That is, we allowed it to expand (at constant temperature) against an external pressure that was reduced VERY slowly. Under these conditions, the gas does work as it expands and therefore it must absorb an equivalent amount of heat from the surroundings in order for the internal energy to remain constant.

Imagine the apparatus is submerged in a very large water bath at temperature T. Initially, the pressure of the gas is equal to the external pressure and the temperature of the gas is the same as the temperature of the water bath. Now imagine the expansion occurring in a series of tiny steps:

1. The external pressure is reduced by a tiny bit.

2. Since the pressure of the gas is now a little bit higher than the external pressure, the gas expands a little bit and does a little bit of work in the process.

3. Recall that the internal energy (E) of a system is equal to the sum of the potential and kinetic energy of all of the particles in the system. Since an ideal gas is free of attractive or repulsive forces, its potential energy is zero. Therefore, its internal energy is equal to its kinetic energy. Because the gas does a little bit of work, its internal energy and, therefore, its kinetic energy drops a little bit. This was a long-winded way of saying that the temperature of the gas drops a little bit.

4. Now the temperature of the gas is a little bit lower than the temperature of the water bath so heat flows into the gas from the water bath until their temperatures are the same again. Since the water bath is very large, its temperature does not change so the final temperature of both the water bath and the gas are the same as the original temperature. (If you prefer, the water bath can be relatively small but maintained at a constant temperature by an electrical heater equipped with a thermostat. So when heat flows from the water bath into the gas, the temperature of the water bath starts to drop but then the heater clicks on and heats the bath back to its original temperature. In other words the heat source is actually some huge power plant somewhere!) The point is that the gas must absorb heat from the surroundings in order to maintain a constant temperature as it expands.

5. Repeat steps 1-4 until the gas has reached the desired final volume.

We can then calculate ΔS as we did above. Since S is a state function this is also ΔS for the free expansion of the gas! (ΔS for an isothermal expansion of a gas depends only on the initial and final volumes!)

VI. Relationship between ΔG and work

From the first law:

ΔE = q + w

w = ΔE – q

We showed in section IV that for any process at constant temperature:

q ≤ TΔS (The equality applies only to a reversible process.)

so:

w ≥ ΔE – TΔS

at constant pressure:

w = -PΔV + w’ (w’ represents all non-PV work such as electrical work. The negative sign is due to our convention that positive work means work done on the system which corresponds to a decrease in volume)

-PΔV + w’ ≥ ΔE – TΔS

w’ ≥ ΔE – TΔS + PΔV

By definition, ΔH = ΔE + PΔV

w’ ≥ ΔH – TΔS

w’ ≥ ΔG

Remember that only reactions with negative ΔG can occur and produce work. For example, the last inequality says that if ΔG = -10 kJ/mol, then the reaction is spontaneous and w’ can be -10, -9, -8 etc. Thus, -10 kJ/mol is the maximum amount of work that can be done by the reaction. (Negative work is work done by the system on the surroundings.) If ΔG = +10 kJ/mol, the reaction is not spontaneous and the number represents the minimum amount of work that must be done on the system to force the reaction to occur.

We have shown that the change in Gibbs Free Energy represents the maximum non-PV work obtainable from a spontaneous chemical reaction at constant T and P. The maximum is obtained only if the reaction is done reversibly (otherwise some of the released energy is lost as heat). Remember that a large ΔG means that the reaction is far from equilibrium. Any system (e.g. a spring) that is far from equilibrium is capable of doing a lot of work.

Here is another way to look at this:

ΔG = ΔH – TΔS

ΔH = q (heat actually produced or absorbed by the reaction)

TΔS = qrev [heat that would be transferred if the reaction was done reversibly (very slowly)]

ΔG = q - qrev

This form of the equation say that for a spontaneous process (negative ΔG), the amount of heat transferred is always less than the amount of heat that would be transferred by a reversible process. The difference is the amount of energy available to do work. If a reaction is at equilibrium, then it cannot do any work (ΔG = 0) and all of the energy produced by the reaction is lost as heat (q = qrev).

Let’s again put in some numbers to make it easier to understand this:

Let’s first consider a spontaneous reaction (ΔG < 0).

If q = -10 kJ/mol (exothermic), then qrev could be -9, -8, -7, etc. The differences ( q – qrev) would be negative (-1, -2, -3, etc.). This represents the portion of q that can be converted into (non-PV) work. The remainder of the heat is dispersed into the surrounding and is wasted. Remember that the second law says that it is impossible to convert heat into work with 100% efficiency.

If q = +10 kJ/mol (endothermic), then qrev could be 11, 12, 13, etc. The differences ( q – qrev) would again be negative (-1, -2, -3, etc.) and the same argument applies. Notice in this case, ΔS would have to be positive in order for the reaction to overcome the unfavorable ΔH and allow the reaction to occur. For an exothermic reaction, ΔS can be negative as long as it isn’t too negative!

Now let’s consider a non-spontaneous reaction (ΔG > 0)

If q = -10 kJ/mol (exothermic), then qrev could be -11, -12, -13, etc. The differences ( q – qrev) would be positive (1, 2, 3, etc.). This represents the amount of work that must supplied by the surrounding to force the non-spontaneous reaction to occur.

If q = +10 kJ/mol (endothermic), then qrev could be 9, 8, 7, etc. The differences ( q – qrev) would again be positive (1, 2, 3, etc.) and the same argument applies.

If a reaction is at equilibrium (ΔG = 0) then it can do no work. In this case q = qrev and all of the heat produced is dispersed into the surroundings and none of it can be converted into work.

Note: You might have noticed a seeming contradiction. In part I, we showed that a reversible mechanical process produces the most work. Here we seem to be saying that a reversible reaction (one at equilibrium) produces the least amount of work (zero!). What’s really happening here is that the reaction is proceeding at equal rates in both directions. The heat produced by the exothermic direction is absorbed by the reverse reaction so no heat is available to do work. A reaction at equilibrium does no work because there is no net reaction! If a spontaneous reaction could be carried out reversibly, then more of the heat produced could be converted into work. In some sense, you could say that the heat produced by the exothermic reaction is driving the endothermic reaction with 100% efficiency! Thus, the reaction IS doing the greatest amount of work when it is at equilibrium and no heat is lost to the surroundings! (I admit that this last argument is a bit shaky!)

-----------------------

Reservoir at Th

Reservoir at Tl

qh

ql

Engine

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Work (w)

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Engine 1

(more efficient)

Reservoir at Tl

Reservoir at Th

q1

q2

Engine 2

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q4

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