It is agreed that the calculation of work in processes such as the movement of a piston dividing two chambers of gas a t different pressures is tricky in that various different assumptions may be used in evaluating the path for the compression work in the system of interest where the pressure is non-constant.
To the Editor: The difficulty expressed by Professor Chesick in understanding "the pressure drop across the system boundary" points directly to the source of confusion. The system is the gas and there is no discontinuity of pressure a t or near the boundary of the gas, unless there is a shock wave within the gas. Straightforward application of Newton's third law of motion shows that in any system, static or dynamic, there is a balance of forces when inertial effects are properly considered. Thus it can make no difference whether the driving force or opposing force is employed in calculating work done, provided these are properly chosen. The preference indicated for looking a t the opposing force is based on the conviction that it is more often easier to recognize the opposing force correctly than the driving force. The new experiment suggested, in which a mass M rests on a frictionless, weightless piston, which moves between stops, is illustrative and may he considered in somewhat greater detail. I n the first place, if there were no gas in the container the mass, M, would be in free fall and would exert no force on the piston, or the vacuum, beneath it. It exerts a force only when it is prevented from free fall, and the force exerted is just equal to the reaction force provided by the gas. Therefore, as the piston falls the work done is JPdV where P is the restraining pressure exerted by the gas on the piston. The difference between the gravitational force exerted on the mass M and the restraining force exerted by the gas goes into increasing the kinetic energy of the mass with its associated piston. At the end of the fall, this kinetic energy will be dissipated by collision with the lower set of stops. If we retain the original definition of the system as the gas only, then the collision with the lower stops is irrelevant and the total work done is JPdV. This pressure will be the pressure of the gas if, and only if, the gas has a well-defined, uniform pressure a t each instant, which will be true for small piston speeds, and which in turn requires a short fall of the piston in the vertical cylinder (or eounterwcights) or a large mass for the piston in a horizontal cylinder. If the pressure within the gas is not uniform the problem cannot be solved by ordinary thermodynamics; one can then only say that the work will be somewhat greater than for a reversible compression hut certainly less than P.,tAV. Apart from trivial or non-trivial variations in the experimental arrangement, the important point is quite general and straightforward. If the pressure of a gas is
uniform during any expansion or compression, the process will be thermodynamically reversible with respect to the gas (although there may be frictional forces or heat transfers across temperature differentials that can make the total process irreversible). The requirement for a uniform pressure is normally that any motions be slow compared to the relaxation time of the gas and that there should be no internal churning of the gas, as with a rotating paddle. Experimentally, the requirement of constant temperature also requires that the process be slow when work is done. When the pressure is non-uniform it may still be possible to calculate the work done if the pressure is known at each point in the gas at all times. If this information is not available, the work cannot be calculated. It is extremely important, therefore, that examples be chosen with care. I hope we may all agree not to give such poorly stated problems as the one I had selected as a bad example. ROBERTBAUMAN POLYTECHNIC INSTITUTE OF BROOKLYN BROOKLYN, NEWYORK
To ths Editor: A recent article in THIS JOURXAL~ reaches the couclusion that when an ideal gas is compressed by an external pressure greater than the internal pressure, the work done is independent of the external pressure. I n this treatment, work and heat are assessed by effects in the system. Historically, however, heat and work are assessed by events outside the system; i.e., in the surroundings.2 The power of thermodynamics stems largely from these traditional ground rules, which make the over-all effects independent of transient events within the system. Nevertheless, the peek inside the system does not automatically invalidate the analysis; it just makes erroneous conclusions more likely. Thus, it is not surprising to find that the analysis in ref. 1 is meretricious. Let us consider a specific case of an irreversible compression of an ideal gas. Assume that initially the ideal gas is confined in an upright cylinder with a frictionless piston of mass m uppermost. At equilibrium, the pressure of the gas is equal to the downward pressure of the mass nz. Mechanistically, the gas pressure arises from elastic collisions with the piston. Before and after a collision the momentum of a gas molecule is mu, and -mucx, respectively; hence, each collision involves a momentum change of -2mu, and this is the origin of the gas pressure. I n this static situation 3, = 3" = Dz = 0. Now let us add a mass M to the piston. The piston moves and hence acquires kinetic energy. Consider, however, the effect of the moving piston on a colliding 'Textbook Error, 49, BAUMAN, R. P., J. CHEM.ED., 41, 102 (1964). 1 PLANCK, M., "Treatise on Thermodynamics," Longmans, Green and Co., Ltd., London, 1927, p. 42.
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gas molecule. The collidmg molecule approaches with a momentum mu, and departs with a momentum -m(v, 2V,) where V, is the instantaneous velocity of the piston. This means, of course, that the dynamic pressure is greater than the static pressure. If there were no equipartition of energy, we would have G , = e,= 0 and it, # 0. But there is equipartition of energy in an ideal gas. (If we exclude collisions in an ideal gas it will arise from rough walls.) Equipartition of energy will funnel this extra z-momentum of the gas into x and y directions and part of the excess work over the reversible work will appear as an increase in molecular speed, that is, as thermal energy. I n most instances the piston overshoots the equilibrium point hut the return is not reversible. We have already seen that when the piston moves toward the gas an individual collision exerts a greater pressure than the static pressure; it is correspondingly true that when the piston moves away from the gas, an individual collision exerts a pressure less than the static pressure. [In detail the momentum change for the approaching piston is IV,I); the corresponding change for the -2m(v,, retreating piston is -2m(vt, - l a ) . ] Thus, even though oscillations may occur, the statement1 that: " . . . the piston will overshoot the equilibrium point and oscillate indefinitely. The oscillations will be reversible, with work done in compressing causing a temperature rise followed by an equal and opposite change in temperature during the expansion half of the cycle. . . . " is incorrect. The oscillations die down and the excess work over the revenible work appears as thermal energy. Equipartition of energy and momentum always ultimately result in the conversion of the irreversible part of the work into thermal energy; it is this that makes paddle wheel experiments B la Joule practical in any fluid. For an isothermal process this thermal energy appears as heat. I n view of the above, it is clear that for the isothermal irreversible compression discussed in the first paragraph of Bauman's article' the overall work done is equal to P.., AV not RT In Vz/Vl as he maintained. Introduction of stops does not change the answer provided they are inside the system. If the stops are placed outside the system we are dealing with a new problem, which is far more complex than the initial problen~. Even in this case, however, the work done by these complex surroundings on the gas is not the reversible work. Finally, it should be noted that the assumption of Bauman that
+
+
is not always true. This point is mentioned by Kirkwood and Oppenheim3 and discussed at length by Bridgman.' The equality breaks down when there is a discontinuity a t the boundary of the system.
'KIR~WOOD, J. G., AND OPPENKEIM,I., 'Chemical Thermodynamics," McGraw-Hill Book Company, Inc., 1961, p 16. BRIDQWAN, R. W., "The Nature of Themodynamm," Harvard University Press, Cambridge, Mass., 1941, pp. 47-56. 676
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To the Editor: Professor Kokes has pointed toward a significant clarification of the problem. An extension of his comments will show the correlation between his molecular approach and the thermodynamic calculation. A molecule approaching a piston with speed v, will leave with speed u, 2u after an elastic collision, if u is the speed of the piston toward the molecule. The 2u) mu, = momentum transferred, Ap, is m(v, 2m(v, u). If the piston is standing still the momentum transferred is 2mu, and no work is done. Thus the excess momentum transfer is clearly the key to the energy transfer we call work. The energy transferred to a molecule in one collision, AE,, is
+
+
+
+
Letting Z, he the number of collisions with the piston per area per time, the work done on the piston of area A is
But u = &/dt and therefore uAdt = Adz = dV. Also, ApZ, is the momentum transferred to the piston per area per time, which is the pressure on the piston. Therefore
The last equation would seem to say that the work done is exactly the pressure of the gas times the change in volume of the gas, but this is not quite the whole story. The pressure in equation (3) is larger than the static pressure of the gas because Ap is larger by the factor (1 u/v,) and the number of collisions, Z, is larger by a factor of similar magnitude. For u
