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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Journol o f Chemical Education
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
