Damped Oscillations of a Frictionless Piston in an Adiabatic Cylinder Enclosing an Ideal Gas

European Journal of Physics Eur. J. Phys. 38 (2017) 035102 (8pp) https://doi.org/10.1088/1361-6404/aa6054

Damped oscillations of a frictionless piston in an adiabatic cylinder enclosing an ideal gas

Carl E Mungan

Physics Department, US Naval Academy, Annapolis, MD 21402-1363, United States of America

Received 7 December 2016, revised 23 January 2017 Accepted for publication 14 February 2017 Published 1 March 2017

Abstract Suppose that a piston of nonzero mass encloses an ideal gas in a vertical cylinder. The piston and cylinder are thermally insulated so that no energy is lost by heat transfer to the surroundings. If the piston is set into motion, it oscillates like a block on a spring. However, the motion is damped because the gas density and pressure immediately adjacent to the piston are higher or lower than the bulk values, depending on whether the piston is compressing or expanding the gas, respectively. That automatically gives rise to a drag force linear in the velocity of the piston. It is not necessary to add extra dissipative terms such as kinetic friction between the piston and cylinder, or viscosity of the gas, to damp out the motion of the piston. The ideas should thus be helpful to undergraduate students performing a Rüchardt experiment to measure the ratio of the heat capacities of a gas.

Keywords: underdamped oscillations, piston, ideal gas, adiabatic, entropy, dynamic pressure

(Some ﬁgures may appear in colour only in the online journal)

1. Introduction

A cylinder enclosing an ideal gas having a piston that can slide frictionlessly is a standard setup for discussing the first and second laws of thermodynamics in textbooks [1, 2]. Such a configuration is often used to distinguish between state and process variables, and in appli- cations such as Carnot heat engines. If the piston is restricted to move infinitesimally slowly, then the process is quasistatic because the gas is always nearly in equilibrium so that it is has a single overall pressure and temperature [3]. On the other hand, if the piston (possibly

Figure 1. The bulk of the gas has time-varying pressure P, absolute temperature T, and volume V = Ax. However, the gas atoms next to the piston (occupying the hatched slice of negligible volume compared to V ) exert dynamic pressure P˜ on the piston. including an extra weight attached to or dropped onto the piston) is set into ﬁnite-amplitude motion, it can oscillate like a mass on a spring and a full analysis may require dividing the gas into a spatial grid of hydrodynamic zones [4]. It is often believed that an explicit dissipative term such as piston friction or gas turbulence has to be introduced in order to damp the motion of a massive piston enclosing a gas [5–7]. However, kinetic theory for the collisions between the gas atoms and the moving piston leads to a dynamically corrected pressure and density of the gas next to the piston that is different than the static pressure and density of the rest of the gas. Since the pressure is not uniform throughout the gas, the overall motion of the system cannot be described as quasistatic even at low piston speeds [8, 9]. This irreversibility leads to a monotonic increase in the entropy S of the universe. The ﬁnal equilibrium state corresponds to a maximum [10, 11] in S. Thus the dissipation is entropically driven. In this article, the dynamics of the system are analysed in terms of Newton’s second law and energy conservation using straightforward ideas and notation, so that they are accessible to undergraduate physics students. A numerical differential equation solver is then used to plot quantities of interest such as the position and velocity of the piston, the temperature of the gas, and the entropy of the universe.

2. Coupled mechanical and thermodynamical analysis of the gas and piston

Consider the setup in figure 1. A vertical cylinder encloses an ideal monatomic gas via a piston of cross-sectional area A and mass m that slides without friction. The height x of the piston above the bottom of the cylinder is initially xi. The weight mg of the piston keeps it from flying out of the cylinder, with earth’s surface gravitational field strength equal to g = 9.80 N kg-1 . The cylinder and piston have negligible heat capacity (compared to the gas)

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Figure 2. A gas atom of mass μ and upward velocity u makes an elastic collision with the piston of mass m and upward velocity υ, where m  m and u  u. and the gas is thermally isolated from the surroundings. There is no viscosity or turbulence of the enclosed gas. The gas is initially in mechanical and internal thermal equilibrium at pressure PmgAi = and temperature Ti = 300 K. The piston is now given an impulsive downward push, so that its initial velocity is ui < 0. The objective is to describe the sub- sequent evolution of the gas and piston. The pressure of the gas immediately adjacent to the moving piston is known as the dynamic pressure1 P˜ and is in general different than the bulk pressure P of the remainder of the gas [9, 12]. The dynamic pressure is smaller than P if the piston is moving upward in figure 1 so that it leaves a rarefied zone in its wake, and conversely P˜ > P if the piston is moving downward and pushing gas atoms together in front of it like a snowplow. To find an expression for the dynamic pressure, suppose that a gas atom of mass μ and initial vertical velocity u makes an elastic collision with the piston of mass m moving with velocity u = dd,xtas sketched in figure 2. Since m  m, the atom’s initial relative velocity of u - u is reversed by the collision, whereas the piston’s velocity is almost unchanged. Thus the final velocity of the atom in the cylinder’s frame of reference is 2u - u for a net change in momentum of Dpu=-mu()22.The atom travels back and forth across the height of the 2 gas cylinder in a time of Dtxu= 2 ∣∣avg where the denominator is the atom’s average speed of (uu+-22.u) Consequently the average force exerted by all N gas atoms on the piston is ⎡ Dp ⎤ ⎡ Numu()- 2 ⎤ ⎡ Nu m (2 - 2 u u )⎤ FN=-⎢ ⎥ = ⎢ ⎥ » ⎢ ⎥ ()1 avg ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Dt avg x avg x avg

1 This deﬁnition of ‘dynamic pressure’ is related to but not identical with the term of the same name often invoked in Bernoulli’s principle. 2 Technically one should add together x/u and xu( - 2u) to get the total travel time, but that sum is equal to Δt to within a negligible factor of ()u u 2.

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Figure 3. Graph of the height xt()of the piston by simultaneously solving equations (4) and (5) using the parameters speciﬁed in the text. since u  u, so that the dynamic pressure is ⎡ ⎤ Favg N ⎛ 2u ⎞ ˜ ==⎢m 2 ⎜⎟ -⎥ P ⎣ u ⎝1 ⎠⎦ .2() A Vuavg The average must be performed over all directions of the atomic velocity υ having a positive upward component and over atomic speeds having a Maxwell–Boltzmann distribution appropriate to gas temperature T and molar mass M. Equation (2) is the sum of a term independent of the piston speed, which must represent the bulk pressure P, and a damping term linear in the piston velocity which Bauman has calculated in detail [13], with the result that ⎛ ⎞ 8M PP˜ =-⎜1 u ⎟,3() ⎝ pRT ⎠ where R = 8.314 J K--11 mol is the gas constant. Using this dynamic pressure, Newton’s second law applied to the piston gives its acceleration a as ⎛ ⎞ d2x nRT dx 8M PA˜ -= mg ma =⎜1 -⎟ -g,4() dt 2 mx ⎝ dt pRT ⎠ where the ideal gas law was used to replace P with nRT Ax. In addition, the work the piston does on the monatomic ideal gas is equal to the increase in internal energy U of the gas [14], ⎛ ⎞ 3 dT 2T dx 8M dx -=PA˜ d x nRd T =⎜ -1⎟ .5() 2 dt 3x ⎝ dt pRT ⎠ dt Equations (4) and (5) are two coupled differential equations for xt()and Tt()that can be solved numerically, here done using NDSolve in Mathematica. (Alternatively, students could execute the Euler–Cromer algorithm in a spreadsheet [15].) Suppose that the piston has cross- 2 sectional area A = 10 cm , mass m = 10 kg, initial height xi = 1m, and initial velocity -1 ui =-2ms .In that case, there are n =»mgxii RT 0.0393 mol of ideal gas (taken to be krypton with a molar mass of M = 83.8 g mol-1) in the cylinder at an initial pressure of

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Figure 4. Graph of the entropy S ()t of the universe from equation (6) using the solution of equations (4) and (5) with the parameters speciﬁed in the text. There are weak oscillations in this curve at all times, most noticeably below 20 s.

PmgAi ==/ 9.8 kPa. Thus the damping term in equations (4) and (5) has an initial relative 2 12 value of (8MRTupi i ) on the order of 2%, so that the oscillations substantially decay away after a few dozen cycles. The resulting position x of the piston as a function of time t is plotted in ﬁgure 3. Similar oscillations are found for Tt(), Pt(), and u (t) which damp out within a few minutes. The entropy change of the gas (and hence of the universe) can be calculated as [9]

3 T x TSdd=+ U PV d = S nRln +nR ln() 6 2 Tiix starting from an initial value of zero. The result is plotted in ﬁgure 4. As expected for an irreversible process, it increases monotonically and levels off to the ﬁnal equilibrium value calculated next. The square of the piston speed can be expressed in terms of the values of x and T at that instant in time (without needing to know their full time-dependent solutions) using conservation of the sum of the kinetic energy of the piston, the internal energy of the gas, and the gravitational potential energy of the piston to obtain3

2 2 3nR u =+u ()()TTii -+2. gxx - () 7 i m

Furthermore the initial and ﬁnal equilibrium pressures of the gas are equal, which implies

nRTi nRTf Tf ==xf xi.8() Axi Axf Ti

Substituting this result into equation (7), along with the fact that the piston ends up at rest, leads to

3 Equation (7) could be used in place of equation (4) in the simultaneous solution with equation (5), provided one accounts for the sign ﬂip of υ at each turning point.

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⎛ 2 ⎞ mui TTfi=+⎜1 ⎟.9() ⎝ 32nRTii+ mgx ⎠ Alternatively, one could derive this result by equating the change in internal energy of the gas to the sum of the impulsive applied work done to initially impart kinetic energy to the piston, plus the work done by gravity to change the height of the piston. That gravitational work is equal to the constant external pressure mg/A on the gas [16] multiplied by the negative of the volume change of the gas. Likewise, substituting equation (8) into (6) implies that the overall increase in the entropy of the universe due to this dissipative process is

5 Tf Sf = nR ln .() 10 2 Ti -1 Equations (8)–(10) result in xf » 1.08 m, Tf » 324 K, and Sf » 64.1 mJ K in agreement with ﬁgures 3 and 4. If, in contrast, the entire process were treated reversibly, then the entropy of the thermally insulated gas could not change. In that case, equation (6) would become nR nRT ddUPV=-  dd,T =- V () 11 g - 1 V where γ is the adiabatic exponent equal to the ratio of the isobaric and isochoric heat capacities of the ideal gas. Equation (11) can be separated and integrated to obtain the adiabatic gas law, TV g-1 = constant. Since reversibility implies the absence of dissipation, the piston would exhibit undamped oscillations after being initially set into motion. Hence no ﬁnal state of equilibrium would be attained.

3. Fit to underdamped harmonic oscillations of the piston

Figure 3 resembles standard underdamped oscillations about the ﬁnal equilibrium position, as described by the equation

-bt2 m xt()=- xf Xesin, (wf t + ) ( 12 ) where X is the initial amplitude, ω is the angular frequency, b is the damping coefﬁcient in the equation Fbd =- u for the drag force, and f is the phase constant. The angular frequency is expected to match the prediction from Rüchardt’s experiment for determining the adiabatic exponent g = 5/3 for a monatomic ideal gas [17], gPA2 5nRT w ==»f f -1 2 3.89 rad s ,() 13 mVf 3mxf where the ﬁnal values of the time-dependent parameters are used, because the oscillations are about those values and are thus on average equal to them. Equation (3) implies that the drag force on the piston is -bPPAu =-( ˜ ) so that

88M n MRTf -1 b ==PAf »0.872 kg s ,() 14 ppRTffx where the ﬁnal values have again been used. Finally, take the time derivative of equation (12) and equate the result at t = 0 to υi. Also evaluate equation (12) at t = 0 and equate it to xi. That gives two equations in the two remaining unknowns whose simultaneous solution gives

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Figure 5. Graph of the height xt() of the piston predicted by the underdamped oscillator model of equation (12) using the parameters speciﬁed in the text.

X » 0.522 m and f » 0.157 rad. Substituting these numerical values for ω, b, X, and f into equation (12) results in the graph in ﬁgure 5, which is in excellent agreement with ﬁgure 3.

4. Conclusion

When a piston of mass m moves at velocity υ relative to the bulk of an ideal gas it encloses in a cylinder, a drag force Fbd =- u on the piston arises. The power transfer to the gas from the 2 piston is -Fbd uu= which is positive regardless of the sign of υ. Although a small value of b will result in a small exponential decay rate b 2m of the piston’s amplitude of oscillations, the dissipation will eventually prove signiﬁcant no matter how slow the piston moves. Starting from any initial speed, it will take approximately the same amount of time for the piston to lose half its amplitude of motion, because the model of equation (12) with a constant value of b accurately describes the actual dynamics. The common rule of thumb that one can neglect this damping for piston speeds much less than the speed of sound in the gas is thus wrong. The piston will always progressively lose kinetic energy on average and the gas will overall warm up. This irreversible rise in the average temperature of the gas could equivalently have been accomplished by a reversible heat transfer, which explains the increase in the entropy of this adiabatic system. Remarkably, this effect arises from the purely elastic collisions on which the kinetic theory of ideal gases is based. I thank Paul Mikulski, whose question about entropic damping of an adiabatic system motivated this article.

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