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Expansion Work without the External Pressure and Thermodynamics in Terms of Quasistatic Irreversible Processes Klaus Schmidt-Rohr* Department of Chemistry, Iowa State University, Ames, Iowa 50011, United States

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S Supporting Information *

ABSTRACT: We demonstrate that the formula for irreversible expansion work in most chemical thermodynamics textbooks does not apply during the expansion process. Instead of the “external pressure” Pext, the pressure Psys,mb on the piston or other moving boundary (hence the subscript mb), which is nearly equal to the system pressure Psys, should be used in the integral over volume. This formula only requires that Psys(V) and T are well defined, that is, a system of uniform P and T (“uPT”) undergoing a “uPT process”, which may be irreversible. An instructive example is an expanding gas accelerating a bullet horizontally and performing work without a conventional external pressure. We emphasize that δw = −Psys,mb dV ≈ −Psys dV is the only useful formula for infinitesimal PV work during a uPT process. The quasistatic approximation Psys,mb = Psys and δw = −Psys dV is usually excellent and enables analyses of irreversible uPT processes, for example, in heat engines; friction in the surroundings and a large piston mass improve the approximation. Slow chemical reactions at constant T and P are quasistatic, and many equations in advanced chemical thermodynamics apply specifically to uPT or quasistatic processes. We show that the equality dS = δqirr/T applies in irreversible quasistatic processes without composition change. In short, with well-defined P and T at constant composition, the simple equations for reversible processes are usually excellent approximations even when the process is irreversible. KEYWORDS: Upper-Division Undergraduate, Physical Chemistry, Misconceptions/Discrepant Events, Thermodynamics

E

xpansion work plays an important role in basic discussions of the first law of thermodynamics and serves as a starting point for more advanced differential expressions (e.g., dG < 0 for a spontaneous process at constant T and P) in almost all courses and textbooks on chemical thermodynamics. It is also of interest in the analysis of heat engines that convert chemical into mechanical energy. All books agree that w=−

∫V

V2

Psys dV

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1

Figure 1. (A) Gas under a movable piston of mass m and area A, with an “external pressure” Pext = mg/A, in an otherwise evacuated cylinder. (B) Gas accelerating a bullet-piston in a horizontal, otherwise evacuated, cylinder, with vanishing conventional external pressure but still performing expansion work.

describes the work done in a reversible process that changes the volume of a closed system from V1 to V2 when the system pressure Psys(V) is equal to, or only infinitesimally different from, the external pressure Pext. However, for an irreversible expansion or compression resulting from Pext ≠ Psys, most undergraduate physical-chemistry and chemical-engineering books,1−9 as well as a physics text,10 insist that the external pressure needs to be used in the equation for work done by or on the system w=−

∫V

V2

Pext dV

irreversible processes and what happens when Pext ≠ Psys. Others,18 including an advanced thermodynamics text,19 do not present a simple equation for PV work in slow irreversible processes. Some progress in the analysis of expansion work for irreversible processes has been made fairly recently.20−22 Important results have been obtained, such as that (i) kinetic energy must be considered,20,23 (ii) the work done by the system and the work done on the surroundings is not the same, |wsys| ≠ |wsurr|, in the presence of friction,20,21 and (iii) the

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1

For a piston of constant mass m and area A being raised or lowered by a gas (see Figure 1A), the external pressure is given as Pext = mg/A.1,2,4−6 Fermi11 and Pauli,12 as well as a few other authors,13−17 use eq 1 for all processes with well-defined Psys(V), but they do not explain why eq 1 is correct for © 2013 American Chemical Society and Division of Chemical Education, Inc.

Published: December 16, 2013 402

dx.doi.org/10.1021/ed3008704 | J. Chem. Educ. 2014, 91, 402−409

Journal of Chemical Education

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pressure exerted by the piston on the gas is close to Pgas,20−23 not mg/A. Careful reading of this literature (e.g., eq 8 of ref 20, eq 7 of ref 21, and the last two equations of ref 22) shows that, contrary to the more widely held opinion in the textbooks1−10 and in earlier publications,24 eq 1 is usually the more correct expression for the work done by or on a system of well-defined Psys in slow irreversible expansions and compressions. It has also been pointed out18,21,22,25,26 that the pressure Psys,mb (often called Pop)15,21,22 exerted by the system on a piston or other moving boundary deviates from Psys if the piston motion is not slow compared to the speed of the molecules in the gas or fluid (see below). The most accurate expression for system-based PV work is therefore15,21,22 w=−

∫V

V2

1

Psys,mb dV

derived. Finally, we summarize the resulting simple approach to teaching expansion work.

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ANALYSIS AND DISCUSSION

Uniform P and T Processes

In order for eq 1 to be applicable at all, the pressure Psys(V) (also simply called P in the following) during the process must be uniquely defined. Paraphrasing Fermi11 (whose text is still in print), it must be possible to represent the state of the system as a point in a P−V diagram.11 The pressure and temperature throughout the closed system must then be spatially uniform, so that P and T each has a single value. This requires that all changes occur slowly, so that pressure or temperature gradients are always smaller than unavoidable thermal fluctuations. We call this a uniform P and T (uPT) process, but for generality allow for a different pressure Psys,mb ≠ Psys at a moving boundary of the system. Most importantly, the fundamental relation dU = −P dV + T dS + ∑i μi dni holds for any uPT process thus defined (see derivation in the last section of the Supporting Information). The closely related but more restrictive term quasistatic, where even the pressure at all system boundaries is the same, is discussed next. Note that many inhomogeneous systems, including coexisting liquid and vapor, can also have the same P and T throughout and, therefore, fulfill the requirements of a uPT process. When uPT processes are analyzed, it is assumed that P and T equilibrate quickly27 and are, therefore, related by the same P(V, T) function as in equilibrium states of the same fixed composition.8,11,28 We may refer to this condition as internal pressure−temperature equilibrium. However, chemical equilibrium is not required; we can describe the irreversible uPT expansion of a (kinetically trapped) mixture of O2 and H2 of fixed composition, without requiring them to form H2O and reach chemical equilibrium. Even the chemical potentials μi do not have to be uniform throughout the system, so we can eventually predict slow movement of molecules from high to low chemical potential within a closed system at constant T and P. Nevertheless, spatially uniform chemical potentials are often assumed, for instance, in the analysis of slow irreversible homogeneous chemical reactions at constant T and P. We emphasize that dG < 0 for a spontaneous process at constant T and P, a central result of chemical thermodynamics, applies specifically to irreversible uPT processes.

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Nevertheless, these publications lack the simplicity and clarity to really settle the issue and result in corrections of the textbooks. For instance, a recent paper by Gislason and Craig21 also presented eq 2 as a valid formula for both system-based and surroundings-based work. For system-based work, eq 2 was obtained only by redefining Pext = Psys,mb (see the Supporting Information for details),21 which is contrary to the definition of Pext elsewhere in ref 21 and contrary to the convention for Pext in all textbooks.1−10 Even for surroundings-based work,20,21,23 which is often of limited interest for the thermodynamics of the system, eq 2 applies only under restrictive conditions not met in most heat engines (see below). Several authors20,23,24 have struggled to come up with simple examples of processes where the two equations, eqs 1 and 2, produce dramatically different results. By considering an expanding gas accelerating a heavy bullet in a horizontal cylinder, Figure 1B, we give a simple example that shows very clearly that only eq 1 gives a good approximation for the work done by the system in slow irreversible expansion processes, whereas eq 2 leaves out kinetic energy and work done by the system against friction in the surroundings. This example highlights that work can be performed by a gas even in the absence of an external pressure as defined above. We introduce the term “uniform P and T (uPT) process” for a slow process in a closed system where P and T are uniform throughout the system11 (allowing for a moving system boundary subject to a slightly different pressure Psys,mb) and discuss that work and heat for such processes are particularly simple to calculate in the “quasistatic” approximation, where simply δw = −P dV. This approach is similar to that taken by Fermi11 but contrasts with the surroundings-based considerations championed by Gislason and Craig,20,21,23 which cannot provide a correspondingly simple and useful formula for infinitesimal work δw associated with the expansion from V to V + dV. We note that uPT processes can be quasistatic and irreversible, for instance with friction, with large pressure or temperature differences between system and surroundings, or with spontaneous chemical reactions, and thus approximate real processes much better than do the reversible processes that have been emphasized in most thermodynamics textbooks. On this basis, we introduce a new type of homework problem calculating changes in a gas performing expansion work under realistic conditions, with friction in the surroundings. We point out that any equation in advanced thermodynamics containing T and P assumes a uPT process. Some implications of δw = −P dV for heat and entropy change in quasistatic processes are also

Quasistatic Processes and the Quasistatic Approximation

We use the term quasistatic to describe a uPT process with fixed or infinitely slowly moving system boundaries.6,10,15 Clearly, all reversible processes are quasistatic. Although some texts12,18 also claim that all quasistatic processes are reversible (possibly based on different definitions of the terms), various valid examples of simple quasistatic and irreversible processes have been provided. They include slow expansion against a nearly matched friction force,6 slow heat flow into a constantvolume system from surroundings of significantly higher temperature,10 or a slow spontaneous chemical reaction at constant T and P, as discussed further below. A reversible process, by contrast, would have to occur without friction, with the temperatures or pressures of system and surroundings only infinitesimally different, and with the system remaining in chemical equilibrium.15,21,22 The need for a distinction between quasistatic and other uPT processes becomes apparent after considering that the pressure Psys,mb of a gas on a moving piston depends on the piston speed 403



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The quantity W in eq 6 is positive for expansion (V2 > V1) because Psys > 0. However, the sign convention in physical chemistry, which is very sensible in giving ΔU = q + w rather than the more complicated q − W, requires work done by the system to be negative. Therefore, we set w = −W and obtain

u. For a gas exerting a pressure Psys on its static boundaries, Bauman and Cockerham have shown25 that because of decreased (u > 0) or increased (u < 0) momentum transfer from the gas molecules, the pressure on the moving piston is 2 ⎛ 8⎛u⎞ 8⎛u⎞ ⎞ Psys,mb = Psys⎜1 − ⎜ ⎟ + ⎜ ⎟ ⎟ π⎝v⎠ π⎝v⎠ ⎠ ⎝

w = −W = −

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∫x

x2

Fsys,mb dx = −

1

where u > 0 for expansion and u < 0 for compression and v is the average speed of the gas molecules.15,21,22,25,26 For some uPT processes with volume changes that occur without friction for extended time periods, the difference between Psys,mb and Psys can be crucial.27,28 On the other hand, the u-dependent terms in eq 4 often produce only a fairly small correction to the system pressure in practical situations of interest in chemical thermodynamics and piston engines. For a typical maximum engine piston speed of u = 10 m/s and with v = 700 m/s for N2 at 600 K, the correction

≈−

−w =

(5b)

The process usually involves a volume change of the system. Specifically, a slowly expanding fluid of pressure Psys exerting a pressure Psys,mb ≈ Psys and a force Fsys,mb = Psys,mb A on a piston of area A, produces work W=

∫x

x2

1

Fsys,mb dx =

∫V

V2

1

Psys,mb dV ≈

∫V

V2

1

Psys dV

x1+Δx

Fgas dx =

∫V

V1+ΔV

Psys,mb dV

1

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This is equivalent to eq 3. In particular, at the beginning of the process, the piston velocity is small so that the quasistatic approximation Psys,mb ≈ Psys and eq 1 holds very well. The work done by the gas imparts kinetic energy to the bullet; this is missing in eq 2, which corresponds to a potential-energy change only (see below). In the present example, the gas is not subject to an external pressure in the sense of physical chemistry texts1−9 because the weight force of the bullet is orthogonal to the force applied by the gas pressure. Generally, Pext = mg |cosθ|/A, where θ is the angle between the external force vector and the gas-pressure force. For the horizontal cylinder, θ = 90° and Pext = mg |cos(90°)|/A = 0. Therefore, eq 2 would have predicted w = 0. This prediction corresponds to considering work only in terms of lifting of a weight,1,2,4−6,8 and in the horizontal cylinder, the piston is indeed not lifted. But the result w = 0 is clearly incorrect for the system, which has lost energy as work; if the expansion is adiabatic, qsys = 0 and wsys = ΔUsys < 0. We know that the energy loss ΔUsys is nonzero, since this energy appears in the surroundings as kinetic energy of the piston and frictioninduced thermal energy.20,21,23 It is also unquestionable that the expanding gas (the system) was performing mechanical work on the bullet (which is part of the surroundings) because the gas was applying a force (Psys,mb A) on the bullet through a distance (Δx = ΔV/A). All of these considerations apply even with friction in the surroundings, so eq 3 still gives the work done by the system under those conditions. Equation 1 remains a good approximation as long as the expansion is slow, and friction will indeed slow the bullet down.

and therefore Fsys,mb dx

∫x

1

(5a)

1

(7)

As a particularly simple and instructive example, we consider an expanding gas accelerating a heavy bullet or piston in an evacuated horizontal gun barrel or cylinder (see Figure 1B). The gas exerts a force Fgas = Psys,mb/A on the bullet−piston of area A and pushes it through a distance Δx, increasing the volume by ΔV = AΔx and performing work

In thermodynamics, work is energy transferred between system and surroundings. Combining this with the basic definition of mechanical work, the infinitesimal PV work done by or on a closed system is the force Fsys,mb applied by the system via a moving boundary to the surroundings times the distance dx through which the force acts11,20,22

x2

Psys dV

Expansion Work without an External Pressure: Gas Accelerating a Bullet

PV Work in Thermodynamics

∫x

Psys,mb dV

1

which confirms eq 3 and, in the quasistatic approximation, eq 1.

is less than 4%, and it is even smaller averaged throughout the engine cycle. In the quasistatic limit, where u approaches zero, we obtain Psys,mb = Psys. Many real processes with sufficiently small piston speed u and friction or stored kinetic energy in the surroundings can be approximated as quasistatic,15,22 which means that we ignore the difference between Psys,mb and Psys and use eq 1 rather than eq 3. This greatly simplifies the calculation of expansion work throughout a process; the approximation is further justified below.

W=

V2

1

2 ⎛ 8⎛u⎞ 8⎛u⎞ ⎞ ⎜− ⎜ ⎟ + ⎜ ⎟ ⎟ ⎝ π⎝v⎠ π⎝v⎠ ⎠

dW = Fsys,mb dx

V2

∫V

∫V

(6)

Work During an Expansion Process

Note that this definition does not require an applied external pressure Pext or a force opposing motion;3,10,22,29 this is confirmed below by a simple example, gas performing work while accelerating a bullet in a horizontal gun barrel, and by an analysis in terms of potential and kinetic energy. Equation 6 is also valid when a spontaneous chemical reaction occurs uniformly within the system. In a pistonless expansion into a vacuum, we obtain W = 0 because no force on the surroundings acts through a distance (see the Supporting Information for details).

The simple example of the gas pushing a bullet highlights the limitations of calculating work using eq 2, which gives surroundings-based work only for a process that starts and ends with the piston at rest.20,21,23 As a result of this limitation, the surroundings-based approach using eq 2 does not provide a simple expression for the work at an arbitrary point during the process, and therefore does not yield a useful formula for the infinitesimal work δw for a volume change from V to V + dV at the system pressure P(V). In particular, 404



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δw ≠ −Pext dV

Gas Raising a Weight: Oscillations and Kinetic Energy

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In many textbooks treating an expanding gas raising a weight, it is assumed that the height of the frictionless piston of mass m somehow changes monotonically from its initial to its final height.1−4,6 Due to inertia, this description is incorrect, except for constant pressure processes. A massless, frictionless piston,3 while free of inertia, would be subject to an infinite acceleration a = F/m = (Psys,mb − Pext)A/0 and reach infinite or at least molecular speed within a short time, which makes its motion quite intractable. A piston of finite mass m is accelerated by the net force Psys,mb A − mg, and the piston will not stop at the height heq at which the forces are balanced, Psys,mb = Pext. Instead, in a system with ideally insulating walls and no friction or other dissipation, a heavy piston will oscillate (Figure 2A).20,30 In a uPT process with small piston speeds u < 1 m/s,

because the right-hand side would give zero work for the gas accelerating the bullet, which is incorrect even for surroundings-based work (which would need to include the change in kinetic energy).20,21,23 By contrast, the system-based approach associated with eqs 1 and 3 provides the desired correct answer, that the system-based work done by or on the system is δw = −Psys,mb dV at any point during a uPT process, and even simpler that δw = −P(V ) dV

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applies at any point during a quasistatic process. System- versus surroundings-based work20,21,23 and their confusion in ref 24 is presented in the Supporting Information, and so is the notion of an external pressure exerted on the gas, which is misleading; according to Newton’s third law, the pressure exerted on the gas by the piston is always equal to the pressure exerted on the piston by the gas.21 Gas Raising a Massive Piston: Kinetic and Potential Energy

The following brief analysis, analogous to that in ref 20, justifies the use of eq 3 in the classical case of an expanding gas lifting a piston. We consider an expanding gas exerting a force Fgas = Psys,mb Apiston on a frictionless piston of mass m in a vertical cylinder. The piston is accelerated by the sum of Fgas and the weight Fext = −mg Ftot = Fgas − mg = (Psys,mb − Pext)A

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(Note that though all pressures are positive, it is convenient to distinguish forces pointing in opposite directions by opposite signs.) The sum of the change in kinetic and potential energy reproduces eq 3. First, for a piston vertically displaced by a distance h, the change in potential energy is obtained as ΔEpot = −Fexth = mgh = PextAh = PextΔV =

∫V

V2

Pext dV

1

Figure 2. (A) Schematic showing the oscillation of a frictionless, massive piston accelerated and raised by an expanding gas. At the end of each cycle, the initial compressed state is nearly recovered. (B) Schematic P(V) curve between the initial volume and the maximum volume Vmax reached by the oscillating piston. The colored area between the P(V) curve and P = Pext is the kinetic energy of the piston at that gas volume, which is zero at V1 and at Vmax = V1 + Ahmax.

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According to the work−energy theorem of basic mechanics, the change in kinetic energy is the integral of the total (or net) force acting on the piston ΔE kin =

∫0

∫0

=

h

Ftot dx =

∫0

h

(Fgas + Fext) dx

h

(Psys,mb − Pext)A dx=

∫V

V2

achievable with a piston of large mass, according to eq 4 we have Psys,mb ≈ Psys to a very good approximation. Setting Psys,mb = Psys, according to eq 13, the kinetic energy of the piston is equal to the area between the Psys,mb = Psys = P(V) curve and the constant P = Pext line (see Figure 2B). In particular, at the equilibrium height heq where the forces due to gas pressure and piston weight cancel, the piston has significant kinetic energy

(Psys,mb − Pext) dV

1

(13)

The magnitude of the total work done by the gas on the piston is the sum of these changes in potential and kinetic energy of the piston (in the absence of friction) −w = W = ΔEpot + ΔE kin =

∫V

V2

1

Psys,mb dV

heq

1 m[u(heq )]2 (15) 2 and, therefore, a finite speed u(heq). It will keep moving further up (see Figure 2A), but now the net force points downward (mg/A > Psys). This decelerates the piston up to the height hmax at which the areas under the P(h) curve and under the line P = Pext are equal, so that Ekin = 0 according to eq 13. Next, the piston is accelerated downward by the net force, passes heq again with kinetic energy, and is then decelerated by the net upward force until it reaches the initial h = 0, where Ekin and the E kin(heq ) =

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which reproduces eq 3 and shows that the formula in eq 2 disregards work performed to increase the kinetic energy of the surroundings (as well as work against friction;20,21 see below). We note that expansion work can be calculated using eq 14 or eq 3 even in the presence of temperature or chemical-potential gradients (though it is unusual to have uniform pressure in the presence of a temperature gradient). 405

∫0

(P(h) − Pext)A dh =



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Damping of the Frictionless Piston

integral in eq 13 vanishes again. Then, the next oscillation cycle starts. The case of compression by Pext > P(V1) is discussed in the Supporting Information. For small amplitudes, the oscillation period can be expressed in terms of the piston mass, area, the equilibrium gas volume, and the compressibility of the gas, which is exploited in Rüchardt’s experiment for determining the ratio of the specific heats, Cp/CV, of the gas.30

Even a frictionless piston will eventually come to rest, after many oscillation cycles,30 because of the accumulating effect of the difference between Psys,mb (see eq 4) and Psys.27,28 This is an interesting but rather unusual example of a uPT process that cannot be approximated as quasistatic. The damping of the oscillations is explained in Supporting Information Figure S2D. Because the piston has come to a rest at the end of process, eq 2 can be applied to give the surroundings-based work20,21,23 −wsurr (see the Supporting Information for a discussion of the sign of wsurr), which in this special frictionless case is also equal to the system-based work28 calculated by integrating eq 3 over many oscillation cycles. For an adiabatic expansion of an ideal gas with CV = (3/2)nR,

Damping of the Oscillations: Different Mechanisms

Eventually, the piston will come to rest at the height where Psys = Pext. This can be achieved in two important ways (two more, inductive braking and a friction pad in the system, are discussed in the Supporting Information): (i) the kinetic energy of the piston passing the force equilibrium height can be converted into friction heat in the surroundings (see the Supporting Information) or (ii) the difference between Psys,mb and Psys (see eq 4) can result in damping of the frictionless piston (without energy dissipation as heat). It is important to note that the change in the system properties V, T, and U will usually be different for the two damping mechanisms; this is shown in the following. Although most textbooks tacitly assume the completely frictionless case (ii), we argue that friction in the surroundings, case (i), is more realistic and typical of heat engines; it is also more compatible with a development of chemical thermodynamics in the differential formalism.

C V(T2 − T1) = ΔUsys = −ΔUsurr = −wsurr + 0 = −PextΔV = −Pext(V2 − V1)

We use Pext = P2 and the ideal gas law P2V2 = nRT2, as well as P1V1 = nRT1, to eliminate to eliminate T2 and T1, which yields V2 P = 0.6 1 + 0.4 V1 Pext

Friction in the surroundings slows the piston down, so the quasistatic approximation and eq 1 can be applied to the uPT process. This is in agreement with the conclusions of Gislason and Craig20,21 and confirmed by a detailed analysis given in the Supporting Information. We can use eq 1 and δw = −P(V) dV as an excellent approximation, which makes the calculations and results formally equivalent to those of reversible expansion. For instance, during an adiabatic expansion of a monomolecular ideal gas with CV = (3/2)nR, in the quasistatic approximation (16)

This problem is treated in all thermodynamics texts under the rubric of ‘reversible adiabatic expansion’. Starting from an initial pressure P1 and volume V1, the final volume V2 = Veq, with pressure balance P2 = Pext, is obtained from eq 16 as ⎛ P ⎞ V2 =⎜ 1⎟ V1 ⎝ Pext ⎠

(19)

This result eq 19 differs from that of eq 17 (see also Supporting Information Figure S3), confirming that during an irreversible adiabatic expansion, friction in the surroundings affects the changes in the expanding gas. Interestingly, eq 19 predicts that a frictionless piston, even of very large mass (i.e., with very large Pext), cannot adiabatically compress a monatomic ideal gas to less than 40% of its original volume. The reader can confirm that in the limit V2 = 0.4V1, the final temperature becomes infinite. Although a gas lifting a frictionless piston is often the only irreversible expansion discussed in chemical thermodynamics textbooks, it is actually of little use in the further development of the topic. It is neither reversible nor approximately quasistatic, the two conditions under which the simple differential expressions of advanced thermodynamics can be derived. The fact that it is not useful is also reflected in the fact that Pext does not show up in the formulas of advanced thermodynamics.

Expansion against Friction

C V dT = dU = δw = −P(V , T ) dV

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Relevance of Kinetic Energy and Work against Friction

0.6

Lifting of a weight, that is, work to change potential energy, is the only type of PV work described explicitly in most physical chemistry texts.1,2,4−6,8 The complete analysis outlined above has shown that kinetic energy and work against friction in the surroundings must also be considered.23 Are these other forms of work of interest in heat engines that extract mechanical work from an expanding gas? The answer is clearly yes. In the case of heat engines propelling cars and airplanes, as well as in electric power plants, work by expanding gas increases the kinetic energy of the vehicle, of a turbine, or of a flywheel. Therefore, the restriction to processes that start and end without kinetic energy21,24 prevents the application of the surroundings-based approach with eq 2 to heat engines. When the vehicle is cruising at constant speed, work is done by the engine, that is, the expanding gas, against friction in the surroundings.

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This analysis of expansion against friction can serve as the basis for new, realistic homework problems, asking students to calculate the change in V, T, and U of an ideal gas lifting (or being compressed by) a piston of mass m and area A coupled to a friction pad in the surroundings under isothermal conditions or with thermally insulating walls. The results, calculated using eq 1 and the final pressure Pext = mg/A, are the same as for a reversible expansion. Strictly speaking, to solve this problem we need to evaluate the integral in eq 1 over a few oscillations (but not many, due to the frictional damping). While Gislason and Craig20 have claimed (p 964) that this is impossible, such an analysis is actually straightforward; to a good approximation, the system moves back and forth on one and the same P(V) curve and finally stops at Veq, so one can just integrate P(V) from V1 to Veq (see also Supporting Information Figure S2B).

Irreversible Quasistatic Processes

Our analysis above and in the following repeatedly refers to not just uPT but also specifically quasistatic processes that may be irreversible. But what irreversible and truly quasistatic processes 406



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(i.e., without a system boundary moving with finite speed) are there? We first note that every expansion starting from pressure−temperature equilibrium is initially quasistatic, as P and T are uniform and the initial piston velocity u = 0; the expansion is irreversible when the difference between Pext and Psys is finite or when friction occurs in the surroundings. As the expansion continues, the speed u of a piston of sufficiently large mass m will remain small because the acceleration a and speed u of the massive piston, which equal F/m, and (2Ekin/m)1/2, respectively, will be small. Even with a large mass m, the external pressure Pext = mg/A can still be kept small, by increasing the area A of the piston. These considerations suggest that the limit m → ∞ provides the rigorous quasistatic limit for an irreversible expansion with unbalanced forces (Pext < P) such as that shown in Figure 2. Indeed, it has been shown analytically that in the limit of infinite piston mass, a frictionless adiabatic piston separating two gases of different initial pressures undergoes quasistatic oscillations of finite amplitude, infinitely slowly and without damping.27,28 The conceptual analogy between truly quasistatic expansions and reversible expansions, both of which are infinitely slow, is pointed out in the Supporting Information. Friction in the surroundings, which makes a process irreversible, improves the quasistatic approximation because it reduces the piston speed.31 For instance, we can consider an irreversible quasistatic expansion of a gas pushing a massive piston in a horizontal cylinder coupled to a friction pad whose friction force is adjusted to be infinitesimally smaller than the force Psys,mbA of the gas on the piston. The piston velocity u will remain infinitesimal, making the process quasistatic. Other important examples of irreversible quasistatic processes involve pure heat flow, for example, from hot surroundings to a rigid cold system (with a finite temperature difference and therefore not reversible) through a nearly thermally insulating boundary or in the limit of infinite heat capacity of the system (constant T, i.e., like a heat bath). Finally and importantly, chemical reactions occurring slowly and homogeneously throughout a system at constant T and P or V are also quasistatic; at constant V, the system boundaries do not move, and at constant P, the difference between Psys and Pext remains infinitesimal, so forces on, as well as accelerations and velocities of the system boundaries do also.

moving system boundaries.22 Unlike approximately reversible processes, which require minimal friction and nearly matched Psys and Pext, approximately quasistatic expansion processes are quite easy to realize using a piston of large mass with significant friction in the surroundings. Implications for Quasistatic Entropy Change

Our analysis has shown that the equation for expansion work is the same for reversible and for quasistatic irreversible processes (see eq 10). Therefore, we can drop the subscript rev (reversible) in most equations (as Fermi and Pauli do11,12) and write δw = −P dV

for a quasistatic PV change

(20)

which includes δwirr = −P dV . With eq 20, we can write the first law for a quasistatic process in a closed system with only expansion work as dU = −P dV + δq

(21)

Combined with the fundamental relation for a uPT process in a closed system of constant composition dU = − P dV + T dS

(22)

(carefully derived in the Supporting Information), this gives dS =

δq T

(at const. composition)

(23)

for a quasistatic process without composition change, which includes13 dS =

δqirr T

(quasistatic at const. composition)

(24)

not just the accepted dS = δqrev/T . Equation 24 is in conflict with the inequality dS > δqirr/T claimed in many texts;1,2,4,6,10,16 in the next section, we provide a simple example confirming eq 24. (The inequality does hold with Tsurr instead of T 18 or if a composition change or chemical reaction occurs isothermally.7) Example of dS = δqirr/T

As an example of a process with dS = δqirr/T, consider irreversible quasistatic heat flow (through a nearly thermally insulating barrier to make it really slow) from hot surroundings of temperature Tsurr into a rigid system of well-defined temperature T ≪ Tsurr and constant composition. The first law and the fundamental relation combine to

Approximations in Thermodynamics

Although eq 3 with Psys,mb as given in eq 4 represents the most accurate expression of PV work, it is inconvenient in practice because the work done by the system depends not only on P(V) but also on the piston speed. It is therefore often useful to make the quasistatic approximation, which amounts to neglecting the difference between Psys,mb and Psys;15,22 as noted above, that difference amounts to only a few percent in real piston engines. Approximations like this are actually pervasive in basic thermodynamics of irreversible processes (i.e., of processes that can actually occur). For instance, the equation dG = V dP − S dT + ∑i μi dni cannot be applied, strictly speaking, as soon as the system has any gradients in T or P because then the right-hand side does not have a unique value. In addition, the P(V, T) relations used are often approximative. Following Jones and Dugan32 “(i)n solving physical problems, we usually focus our attention not on the actual system but rather on some idealized system that is similar to, but simpler than, the actual system” and use the quasistatic approximation Psys,mb = Psys even for most processes with

δwirr + δqirr = dU = −P dV + T dS

(25)

Because the system is rigid, δwirr = 0 and dV = 0, so eq 25 simplifies to δqirr = T dS, which is equivalent to eq 24.

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TEACHING EXPANSION WORK IN THERMODYNAMICS Implementing our results in the classroom (or in a textbook) is extremely simple: One can tell the students that δw = −P dV applies as an excellent approximation unless the piston is frictionless or moves very fast (u > 1 m/s). To explain these limitations, one could briefly discuss Psys,mb and show eq 4. One should further mention that like all equations containing P, δw = −P dV can be used only if P is well defined, that is, uniform throughout the system. The thermodynamics of quasistatic processes then follow naturally. A realistic new homework problem based on these results was described above (in the section on Expansion against Friction). 407



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For advanced students, one could mention that δw = −P dV is infinitesimal system-based work and that the concept of surroundings-based work20,21,23 enables analysis of some processes without well-defined P(V) or with a fast-moving frictionless piston. Here, the requirement is that at the end of the process, no macroscopic object moves as a result of the process; then, surroundings-based work is given by eq 2. However, this discussion is not necessary for developing the standard theory of chemical thermodynamics.

CONCLUSIONS The analysis presented here shows that work during expansion or compression of a uPT system needs to be calculated as the integral of δw = −Psys,mb dV. As long as the motion of the system boundary is slow compared to the molecular speed and the process does not continue for too long (e.g., when slowed down by friction), the quasistatic approximation δw = −P dV (i.e., eq 1) can be used. The widely copied eq 2 with the external pressure misses the work producing kinetic energy and the work against friction forces in the surroundings, which cannot be neglected in heat engines, and it does not provide the required general expression for infinitesimal work δw. We have introduced an expanding gas accelerating a bullet as a particularly simple system to identify the correct equations. Our results provide the basis for new, realistic homework problems analyzing changes in a gas lifting a piston of given large mass with friction, using eq 1. We have shown that quasistatic δw = −P dV leads to the equality dS = δqirr/T for irreversible quasistatic processes without composition change, which we have confirmed in an example. Our results show that in discussions of piston heat engines, one can analyze uPT expansions and compressions with friction in the surroundings and no force balance, rather than unrealistic frictionless, forcebalanced reversible processes, and still use the familiar eq 1 for calculating the work done by the system to an excellent approximation. ASSOCIATED CONTENT

S Supporting Information *

The following topics are discussed: expansion into or against vacuum; the signs of energy terms if the piston weight compresses the gas; heat due to friction; system- and surroundings-based work in the presence of friction; additional mechanisms for damping of piston oscillations; the external pressure and Newton’s laws; and a rigorous derivation of the fundamental relation for uPT processes. This material is available via the Internet at http://pubs.acs.org.

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REFERENCES

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AUTHOR INFORMATION

Corresponding Author

*E-mail: [email protected]. Notes

The authors declare the following competing financial interest(s): The author declares no competing financial interest.

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ACKNOWLEDGMENTS The author would like to thank Pat Thiel, Evgenii Levin, Mei Hong, and three of the anonymous reviewers for stimulating and insightful discussions, and Associate Editor A. M. Halpern for professionally handling the manuscript throughout the many rounds of review. 408



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