Calculating and Visualizing Thermodynamic Equilibrium - Journal of

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Calculating and Visualizing Thermodynamic Equilibrium A Tutorial on the Isolated System with an Internal Adiabatic Piston João Paulo M. Ferreira Escola Superior de Biotecnologia–UCP, 4200-072 Porto, Portugal; [email protected]

Problems dealing with the prediction of the equilibrium state of a system under certain constraints can be quite instructive in introductory courses on thermodynamics. Generally, they involve the application of the first and second laws, establishment of appropriate hypotheses and approximations, and evaluation of variations in thermodynamic properties with changes in pressure, temperature, and volume. A typical problem involves a rigid and adiabatic container divided by a frictionless piston and containing certain amounts of ideal gases in both compartments (Figure 1). Different problems can be proposed concerning the evolution towards mechanical or thermal equilibrium of the system. Recently, Gislason and Craig (1) showed how such a system can be used to demonstrate the principle of entropy maximization, assuming a diathermic piston. A quite challenging situation is that of a similar system with an adiabatic piston. Several analyses of this problem have appeared in textbooks and articles, because it constitutes an example of the impossibility of predicting the final state based only on classical equilibrium thermodynamics principles. We can trace back this problem at least to Callen (2). We will discus that impossibility and review solutions proposed so far. Thermodynamic Analysis: Predicting Equilibrium State Let us consider the system in Figure 1 and assume that the initial pressures in the two compartments are different (P1i P2i ) or their temperatures are different (T1i T2i ). We want to predict the equilibrium position of the piston and the final pressures and temperatures of the two gases, once the piston stops are removed. Students have an intuitive feeling that the piston is likely to have an oscillatory motion before it stops. During the process, the two gases interchange work. If the piston is diathermic, then they will

also exchange heat. As the composite system is isolated, the equilibrium state will be that of maximum entropy under the prevalent conditions. For an ideal gas, we can calculate the entropy variation between an initial state at volume and temperature (Vi , Ti ) and a final state at (Vf , Tf ) as (3) ∆S = n CV ln

Tf Ti

+ nR ln

Vf Vi

(1a)

= n CP ln

Tf Ti

− nR ln

Pf Pi

(1b)

where n is the number of moles and CV and CP are the constant-volume and constant-pressure molar heat capacities of the gas, respectively, assumed not to vary with temperature. Therefore, for the two gases in Figure 1, and assuming they are chemically identical, ∆S = ∆S1 + ∆S 2 = CV n1 ln

T1f T1i

+ n2 ln

T2 f T2 i

+ R n1 ln

V1f V1i

+ n2 ln

V2f V2i

= n CV ln

T1f T2 f T1iT2i

+ nR ln

V1f V2f V1i V2 i

(2b)

= n CP ln

T1f T2f T1iT2i

− nR ln

P1f P2 f P1i P2 i

(2c)

(2a)

In case n1 n2 n, ∆S = ∆S1 + ∆SS2

The first law tells us that the final energy is the same as the initial one: n1CV T1f + n2CV T2 f = n1CV T1i + n2 CV T2 i Figure 1. A rigid and adiabatic container divided by a movable, frictionless piston and containing ideal gases in both compartments.

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(3)

Obviously Vtotal V1 V2 constant. Regarding pressure, mechanical equilibrium requires that P1f P2f .

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Considering a diathermic piston, then there are no internal constraints for energy exchange between the two gases and the equilibrium can be predicted by maximizing the entropy variation. With the volume and pressure conditions, plus the energy balance (eq 3), one can express the change of entropy of the whole system as a function of just one variable. Setting the derivative of this function to zero, one can calculate the equilibrium value (1). As expected, this equilibrium yields T1f T2f (1). If the piston is adiabatic, then the composite system is not able to attain this same equilibrium point, because restrictions exist on the way the energy (or temperature) of the gases vary. In this case, all of the above equations still apply. However, the mathematical maximization of entropy based solely on eqs 2 and 3 is not applicable because the restrictions above are not taken into account in these equations. As we discuss later, with an internal adiabatic piston the maximization of entropy is physically limited. Callen terms the equilibrium state attained in this situation as “constrained” (4), postulating that There exists a function (called the entropy S ) of the extensive parameters of any composite system for all equilibrium states and having the following property: The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy….. The single, all-encompassing problem of thermodynamics is the determination of the equilibrium state that eventually results after the removal of internal constraints in a closed, composite system.

Therefore, the problem proposed can be considered beyond the scope of classical thermodynamics. In order to find the (constrained) equilibrium point, we need a kinetic model of the process (5, 7).

on physical chemistry or thermodynamics that lead to reasonable estimates of the equilibrium point and that require only a pocket calculator. Activities

The Ideal Nondissipative Case Let us suppose that the initial piston position is Xi , with X x兾L, that is, the position of the piston normalized by the container length L. The two chambers contain n1 and n2 moles of the same ideal gas, at pressures P1i , P2i , and temperatures T1i , T2i (Figure 1). The piston stops are released. In the limit of a quite heavy piston, it will move sufficiently slowly that one can consider the process occurring in each gas as reversible. Therefore, the two gases undergo reversible adiabatic expansions–compressions. They exchange energy with each other via work, but they also exchange energy with the piston. While the piston moves, albeit slowly, it has kinetic energy that was imparted to it by the gases. This point should be made clear to students early on. For an adiabatic change of volume, P1V1γ constant (3), or equivalently, P1 X γ constant, where γ CP 兾CV. Thus,

P1 = P1i

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γ

(4)

Similarly, for gas 2, P2 = P2 i

1 − Xi 1 − X

γ

(5)

The variation of temperature with volume follows T1V1γ−1 constant, or in this situation, T1 X γ−1 constant. So,

Previous Solutions The analysis is facilitated by assuming a sufficiently heavy piston, so that the time scale for equilibration in the gas is much smaller than the time scale of the piston motion (5). Based on this hypothesis, there are at least two reports in the literature using different methods for finding a predictive equilibrium point. Crosignani et al. (5) considered the effect of a finite piston velocity u (with u P1, so the piston will be pushed back. Then reset initial conditions with those of the turning point and proceed with the calculations moving the piston backwards. One finds the same XP on the way back, and a stop point at X 0.1, which is the initial starting position (Table

1). All the properties of the gases are also the same as the initial ones. Therefore, one expects that the system will undergo another identical cycle. In conclusion, in the ideal limiting case of reversible adiabatic volume changes of the gases, with no dissipations, the piston will undergo an endless oscillatory motion. Some additional observations can still be made, based on the information in Table 1. First, the position with maximum kinetic energy of the piston is XP , as expected from physical principles (in all other positions the piston has a net force accelerating it). A mathematical derivation can be done to show that XP is indeed the position corresponding to maximum kinetic energy. Furthermore, in the first course of the piston to the right, from Xi to XP , gas 1 loses energy, that is used to compress gas 2 and to accelerate the piston; from XP to XT , the drop in energy of gas 1 is much smaller than from X i to XP , and it is mainly the kinetic energy of the piston that is responsible for the increase in energy of gas 2 (Table 1).

Equilibrium Point Estimations When Dissipations Exist Students have the intuition that in a real situation the piston will have a damped oscillatory motion, ending by stopping. This outcome is due to the fact that the piston will have significant speeds, leading to irreversible compressions and expansions of the gases. These irreversibilities lead to temperatures of the gases higher than the ones expected for reversible processes (we neglect possible heating of the piston and cylinder due to friction between them). How can we use the background of first- or second-year students to estimate equilibrium positions in this situation? This exercise can constitute a follow-up of the previous one. We propose here some methods, but clearly other ones might be suggested in class. The basic idea in these methods is that the kinetic energy of the piston—as calculated for the reversible compression– expansion—is converted into thermal energy of the gases. The irreversible back-and-forth motion of the piston has an effect

Table 1 . Comparative Piston Movement Simulation Data with Reversible Adiabatic Volume Changes of Two Gases X Values

P1/bar

P2/bar

T1/K

T2/K

(E1 E2)/Ei

K/Ei

Initial conditions at X 0.1 and corresponding values. 0.1000

9.000

1.000

300.0

300.0

1.0000

0.0000

0.2000

2.835

1.217

189.0

324.5

0.8560

0.1440

0.2934

1.497

1.497

146.4

352.5

0.8315

0.1685

0.5000

0.615

2.660

103.0

444.0

0.9110

0.0890

0.6000

0.454

3.860

90.9

515.0

1.0100

0.0100

0.5920

0.464

3.738

91.7

508.4

1.0000

0.0000

Reset initial conditions at X 0.592 and corresponding values. 0.2934

1.497

1.497

146.4

352.5

0.8315

0.1685

0.1000

9.000

1.000

300.0

300.0

1.0000

0.0000

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on the gases equivalent to the action of stirrers and this leads to the dissipation of the kinetic energy of the piston into thermal energy of the gases. Therefore, we will consider two types of steps: reversible adiabatic volume changes (RAVC) and selective and controlled heating of the gases at constant volume, restoring the total initial energy of the gases.

P2 n T XP = 2 2 P1 n1 T1 1 − X P

(12)

In our example, two rounds lead to equilibrium conditions, with Xf 0.310, T1f 186 K and T2f 414 K. Steps for Method 3

Steps for Method 1

Step 1: RAVC from position Xi to XP ; calculate T1 and T2 according to eqs 6 and 7 with X XP .

Step 1: RAVC from position Xi to XP.

Step 2: Heating of the cooler gas only, until the initial energy is reset. In case the temperature of that gas reaches the value of the other one, then start heating both, maintaining equality of temperatures. This heating leads to a new pressure imbalance, so calculate the new pressure ratio P2/P1 (eq 12).

Step 2: Heating of the two gases at constant volume and maintaining equal pressures in both chambers, until the initial energy of the system is reset.

The rationale for this method is that, in many situations, the piston is likely to oscillate around XP, eventually stopping close to this point. The back-and-forth movements cause viscous dissipation in both gases. The calculations for the first step were already carried out. For the second step, the imposition of equal pressures and the ideal gas law yield n1 R T1f n RT nT XP = 2 2 f ⇒ 1 1f = V1P V2 P n2T2 f 1 − XP

(10)

With this relation and the balance of energy, eq 3, it is straightforward to calculate the final temperature values. For the example under consideration, T1f 176 K and T2f 424 K. It is interesting to have these results in a dimensionless form, defined as the ratio of the thermal energy of one gas to the total thermal energy: θ1 =

n1CV T1 n1CV T1 + n2 CV T2

θ2 =

n2 CV T2 n1CV T1 + n2 CV T2

(11)

From eq 10, it is easy to conclude that, at equilibrium, θ1f XP . Therefore, θ1f 0.293. Steps for Method 2 Step 1: RAVC from position Xi to XP ; calculate T1 and T2 according to eqs 6 and 7 with X XP . Step 2: Heating of both gases at constant volume, partitioning energy proportionally to n1 and n2, until the initial total energy is restored. This heating yields a new pressure imbalance, so calculate the new pressure ratio, P2/P1. Step 3: Repeat steps 1 and 2, until the pressures are equal and energy is in balance.

The rationale for this method is that the viscous dissipation affects the temperature of both gases to the same extent. Regarding the pressure ratio after the heating step, the gas law yields

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Step 3: Repeat steps 1 and 2, until the pressures are equal and energy is in balance.

This process, although essentially unrealistic, is interesting from a thermodynamic point of view. By allowing the dissipations to take place only in the cooler gas (or in both, only if they reach the same temperature), this process—under the restriction that the volume changes are adiabatic and reversible—maximizes the total entropy. In the example above, equilibrium is achieved in three rounds, with Xf 0.367, T1f 220.5 K and T2f 379.5 K. Besides this numerical example, which we call Example A, the three methods were tested with other initial conditions. In particular, we used the conditions worked by Hoover and Mooran (7), and by Crosignani et al. (5), here called Examples B and C, respectively. Furthermore, different amounts of the gas in the two compartments were also tested (Examples D and E). Table 2 summarizes the information for all the examples, indicating the initial pressure ratio, initial and equilibrium positions of the piston, and the fractional thermal energies of gas 1 (eq 11). The total entropy changes for the processes, calculated according to eq 2, are also presented. It is interesting to observe that in all examples, except B, the three methods gave rather close estimates for Xf and θ1f . The differences in Example B are attributable to the “drastic” initial value of Xi . The results of Examples B2 and C2 are also in close agreement with the reported ones (Table 2). Method 3 leads to the higher possible change of entropy for a given set of initial conditions, as we considered that the dissipations take place only on the cooler gas. In all examples except Example C, the equilibrium position Xf of this method is beyond those of Methods A and B, because the gas heated in step 2 of both methods was always gas 1; in Example C, both gases were heated in the first round to θ 0.5, but in the second round only gas 2 was heated, pushing then the piston to the left (Table 2). It should be emphasized that, under the restrictions of step 1, the final equilibrium values of X and θ are not 0.5. Comparing Examples A, D, and E, one concludes that, using Method 1 or 3, the number of moles in each compartment does not influence the final position, as long as the initial position, fractional thermal energies, and pressure ratio are maintained.


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In the above, the values of the pressure at equilibrium are not evident. In fact, for ideal gases, the equilibrium pressure is set only by the initial conditions and is independent of the process that took place. This can be concluded using the ideal gas law and the observation that θ1f Xf :

The Diathermic Piston Case

P1iV1i P V P X P X = 1f 1f ⇒ 1i i = 1f f T1i T1f θ1i θ1f X ⇒ P1f = P1i i θ1i

(13)

Analogously, for gas 2: P2f = P2 i

For example A, taking P1i 9 bar and P2i 1 bar, the equilibrium pressure is P1f P2f 1.8 bar. This same equilibrium value applies for Examples D and E if we consider the same initial pressures, even though the quantities of gas in the two compartments are different.

1 − Xi 1 − θ1i

For comparison, we can calculate the equilibrium position and temperatures when starting with the same conditions, but with a diathermic piston. In this case, we will have equality of pressures and of temperatures at equilibrium (1). The calculations are straightforward. From eq 10 and the condition T1f T2f , Xf =

(14)

n1 n1 + n2

Table 2. Comparative Initial Conditions and Respective Equilibrium Values Estimated by Methods 1–3 for Examples A–E, in Adimensional Forms Xf

θ1f

∆S/(n1 n2)R

0.293

0.293

0.277

0.310

0.310

0.309

0.367

0.367

0.420

Method 1 (B1)

0.173

0.173

0.586

Method 2 (B2)

0.234

0.234

0.587

0.345

0.345

1.157

0.250

0.250

0.932

Method 1 (C1)

0.608

0.608

1.217

Method 2 (C2)

0.572

0.572

1.251

0.541

0.541

1.269

0.560

0.560

1.259

0.293

0.293

0.277

0.331

0.331

0.433

0.367

0.367

0.560

0.293

0.293

0.278

0.357

0.357

0.702

0.367

0.367

0.763

Method Used

n1 /n2

P1i /P2i

θ1i

Xi

Example A Method 1 (A1) Method 2 (A2)

01

09

0.5

0.1

Method 3 (A3) Example B

Method 3 (B3)

01

50

0.5

1/51

Blita Example C

Method 3 (C3)

01

81

0.9

0.1

C l itb Example D Method 1 (D1) Method 2 (D2)

02

09

0.5

0.1

Method 3 (D3) Example E Method 1 (E1) Method 2 (E2)

10

09

0.5

Method 3 (E3)

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Regarding temperatures, from eq 11, we see that, in this case, we still have θ1f X f , that is, the piston position continues to reflect the ratio of thermal energies of the two gases. Table 3 lists the equilibrium values of piston position and thermal energy for all examples and assuming a diathermic piston. The corresponding values of entropy changes for the overall processes are included. It must be pointed out that, for each example, this entropy change is higher than any of those in Table 2. Furthermore, this difference is accentuated by higher ratios of n1兾n2. Conclusions For the system and processes considered, there is spontaneous evolution towards mechanical equilibrium between the two compartments, but not necessarily towards temperature equilibration. The pressure equality is just the result of a balance of forces acting on the piston. However, when this is adiabatic, the final temperatures need not be the same. The system gains entropy in its way towards mechanical equilibrium, but that gain is less than a diathermic piston would permit. A limit on the entropy gain permitted by an adiabatic piston is evaluated by our Method 3. The activities proposed above constitute a good practice of fundamental principles of thermodynamics—first and second laws, gas relations, entropy changes—and they demand the visualization of physical processes. The students will also practice the ability to construct approximate solutions to complex or uncertain situations, when their knowledge does not permit a more scientifically elaborate approach. Conclusions about entropy changes of processes occurring under different constraints can be inferred. These problems are well suited for group work, as different teams can deal with different initial conditions and different methods. A joint session can follow, where the results of different groups are critically compared.

Table 3. Comparative Equilibrium Conditions in Adimensional Forms for Overall Entropy Change with a Diathermal Piston Example

n1 /n2

Xf

θ1f

∆S/(n1 n2)R

A

01

0.5

0.5

0.511

B

01

0.5

0.5

1.282

C

01

0.5

0.5

1.277

D

02

2/3

2/3

0.962

E

10

10/11

10/11

2.381

manuscript. I also thank the anonymous reviewers for their careful analyses that led to improvements in the article. Literature Cited 1. Gislason, E. A.; Craig, N. C. J. Chem. Educ. 2006, 86, 885– 889. 2. Callen, H. B. Thermodynamics; John Wiley and Sons: New York, 1963; p 321 and Appendix C. 3. Chang, R. Physical Chemistry for the Chemical and Biological Sciences; University Science Books: Sausalito, CA, 2000; pp 94–96, 156. 4. Callen, H. B. Thermodynamics and an Introduction to Thermostatistics, 2nd ed.; John Wiley and Sons: New York, 1985; pp 26–27 5. Crosignani, B.; Di Porto, P.; Segev, M. Am. J. Phys. 1996, 64, 610–613. 6. Bauman, R. P.; Cockerham H. L., III. Am. J. Phys. 1969, 37, 675–679.

Acknowledgments

7. Hoover, W. G.; Moran, B. Am. J. Phys. 1979, 47, 851–856.

I sincerely thank Eric Gislason for the exchange of information and ideas and for his comments on drafts of this

8. Crawford, F. S. Am. J. Phys. 1993, 61, 317–326.

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9. Leff, H. S. Am. J. Phys. 1994, 62, 120–129.


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