GUEST AUTHOR Robert P. Bauman
Polytechnic Institute of Brooklyn Brooklyn, New York
Textbook Errors,
Work of Compressing an Ideal Gas
Calculation of work done in the compression or expansion of an ideal gas is sufficiently simple so that it is a favorite introductory problem for thermodynamics. Unfortunately, it also represents one of the more confusing problems of notation and definition which leads to frequent errors by the students as well as some textbooks.' A typical pitfall is a problem stated as follows: "One mole of an ideal gas a t 25'C and 1 atm is compressed isothermally by a constant external pressure of 5 atm. Find the work done by the gas." Depending upon the book or the student, the answer may be given as -960 cal or -2380 cal. If the compression is assumed reversible, the first answer follows from the equation w = J P dV = n R T In K/V1 = -960 cal, whereas if emphasis is placed on the constant pressure aspect, the calculation is w = P A V = -2380 cal. Surprisingly (from the student's standpoint) the first method is the correct one. Consider first the straightforward calculation according to the definition2 of work. Work is defined to be the integral w = f f dx where f represents the force opposing the motion. Thus, in the familiar example of an ideal gas expanding into a vacuum, the pressure of the gas is greater than zero but the opposing pressure is zero and therefore the work is zero. I n any compression a t speeds small compared to the speed of sound, the force opposing the motion is very clearly the pressure of the gas itself. Hence the work is
for any (ordinary) compression of an ideal gas. The most obvious and common cause of misunderstanding lies in the meaning to he assigned the symbol P. It is customary in nearly all thermodynamic equations to let such symbols represent the appropriate property of the system. I n writing the equation for Suggestions of material suitable for this column and guest columns suitable for publication directly are eagerly solicited. They should be sent with as many details as possible, and particularly with references to modem textbooks, to Karol J. Mysels, Department of Chemistry, University of Southern California, 120s Angeles 7, California. ' Since the purpose of this column is to prevent the spread and continuation of errom and not the evaluation of individual text% the source of errors discussed will not be cited. The error must occur in at least two independent standard books to be presented. a It iis nssumed that work done by the system must be done on the surroundings. That is, w.,,,,,
=
-waum"ndinm
and
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49
work, the symbol sometimes (without warning) takes on the significance of a property of the surroundings. The equation for work is sometimes written w = J P,, dV, but this is only correct when P,,,,represents the pressure opposing the motion. Unless the two pressures are equal one would calculate a different value for w depending on the labeling of system and surroundings. The work done by the system would then not he equal and opposite to the work done by the surroundings. A more fundamental difficulty underlying problems such as these (including that given above) is that the problem itself is not completely defined. We seldom encounter serious difficulties by assuming weightless, frictionless systems. When difficulties do arise, as in this instance, we are not expecting them. It should be intuitively obvious that it is not possible to exert a greater pressure on a gas than the pressure exerted by the gas itself. To demonstrate this, consider the forces acting on the piston illustrated in Figure 1.
The left-hand side initially holds 0.20 mole He a t 300°K and has a volume of 2 liters; the right-hand side initially holds 0.10 mole He a t 300°K and has a volume of 2 liters. The piston is released and the isolated system is allowed to reach equilibrium. Fmd the work done by the gas on the right-hand side. Let the pressure on the left be P,, the force on the left P,A = fL,the pressure on the right P, and the force exerted by the gas on the piston fa = PEA. If fR < f,, the piston will be accelerated by an amount a = (f, - f,)/m, where m is the mass of the piston. If the system is absolutely adiabatic and absolutely frictionless, but the piston has a non-zero mass, then the inertial reaction force of the piston decreases the net force applied to the gas to the value f, -ma = fL -m(fi f,)/m = fR. The energy difference, f (P,- PR) dV, is stored in the piston as kinetic energy which must cause the piston to overshoot the equilibrium point and oscillate indefinitely. The oscillations will be reversible, with the work done in compression causing a temperature rise followed by an equal and opposite change in temperature during the expansion half of the cycle. In practice the oscillations will be damped by irreversible heat transfer, through the walls and piston, of the thermal energy arising from compression and from friction. Friction will always act to decrease
the force applied to the gas during a compression step. Some examples of possible solutions under reasonable assumptions are given in the appendix. If the mass of the piston is zero, the problem is again insufficiently defined. The best example would seem to he a shock tube in which both oscillation and appreciable energy transfer to the container are observed. This problem is not very suitable for equilibrium thermodynamics since pressures are not well defmed when speeds approach the speed of sound in the medium. The classic experiment of Gay-Lussac, repeated by Joule, is quite different. I n this experiment the attainment of pressure equilibrium is delayed by the process of effusion or viscous flow through the aperture. Normally there is no piston, but a light piston placed in the aperture would be thrown against the wall of the containing vessel because the large pressure differential in this region would produce a large acceleration. A piston placed beyond the aperture would move slowly, subject to a vanishingly small pressure differential. This experiment is an example in which the gas does work on itself, but the net work done by the (total) system is zero since no work is done on the surroundings. The conclusion that all compressions with frictionless pistons must be treated as reversible has some interesting consequences. I n finding maximum work done by a gas in an expansion one can argue that w = JP,, dV 5 JP,,, dV so that w,., = JP,,, dV. By a similar argument, however, the maximum work done by a gas in compression should be given by the minimum absolute value of the integral J P dV. Since J P dV is a definite integral for a compression (equal to the negative of the change of the Helmholtz free energy if the temperature is constant), this distinction is unimportant except that for a closed cycle the total work done by the system is a maximum if the expansion step is reversible. Conclusion
Thermodynamics in its customary form deals with the flow of energy under equilibrium conditions. Non-equilibrium processes are acceptable when changes in properties, or "state functions," are desired provided that the initial and final states are independently defined. Sometimes small deviations from true equilibrium, or thermodynamic reversibility, can be tolerated, as in electrochemical processes with junction potentials or overvoltages or in the slow escape of a gas through an orifice. One must be very careful not to build into a problem a type of irreversibility that cannot be ignored. Thus in formulatmg examples of compression problems there should be an explicit statement that the process is reversible, or a t least slow (hence nearly re~ersible).~It should also be made quite clear that rapid motions are of practical interest; hut, because they involve inelastic collisions or friction, they depend on rates and are outside the domain of classical thermodynamics. a LANDAU, L. D., AND LIPSHITZ,E. M.,"Statisticd Physics," Addison-m-esley, Reading, Mass., 1958, p. 37.
Appendix
The problem depicted in Figure 1 can be solved in a variety of ways depending upon the specific assumptions that are made. Three examples are given here. A. If, for example, one assumes that temperature equilibration is fast, pressure equilibration is slow, and the friction is sufficient to prevent oscillation but is otherwise negligible, then the work done by the left side is equal and opposite to the work done by the right side and the entire process is isothermal.
Apart from the fact that pressure equilibration is usually much faster, rather than slower, than temperature equilibration, the answer must be discounted if one examines the magnitude of the work done against inertia. The actual work done by the left side against the opposing gas and the inertial force of the piston is the reversible, isothermal work,
The discrepancy between the two results is 10.3 cal, which is not negligible with respect to the total work done. B. It would he more realistic to assume that pressure equilibrium is achieved first, followed by a slower temperature equilibration. Again assuming first that inertial work and friction can be neglected, w,. = -w,. The new pressure can he calculated by adding volumes. V r = n~ R(To ATr,)/P; VB = n.e R(To+ A T B ) / P
+
4P =
+
VL
( n ~ nd RTo
+ V R= 4
+(
n R~AT&
+ nR RA TR)
But and therefore
The temperatures can now be calculated from the usual equation for adiihatic, reversible expansion of an ideal gas, Cpln T , / = R ln P2/P1. This gives
and the work is As the temperatures approach each other by thermal conduction through the piston additional work will he done. (This will be a constant pressure process, since A E, = - A E, requires that ( nA T),= - ( nA T), and therefore A (PV)&= - A (PV)n. But dVz dV, = 0 and P, = Pa requires that V L dPL V R dPR = 0. On the other hand, since pressure equilibrium is always maintained, dPL = dP,. It follows that dPL = dPR = 0.) The work in this second step is thus P A V .
+
wr. = 9.22 cal =
+
-WZ
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/
103
For the total process WL
=
23.lcal =
-WE
C. When the inertial force of the piston is taken into consideration the work terms for the two gases are no longer equal in the initial calculation. Applying the equation for reversible, adiibatic expansion to each side and requiring that the 6nal pressures be equal, it followsthat Tn/Tr. = ( 2 ) R G = 1.32
This result can be inserted into the equation obtained by adding the two volumes to yield the condition
+
TI, = 4 P / ( n ~ 1.32 nn)R
which can be combined with one of the earlier equations for reversible, adiabatic expansion to find P = 1.8046 atm TI.= 265'K, T R = 350°K WL
104
/
=
21 oal, wn = -14.9 eal
Journal of Chemical Education
and the difference in work terms is about 6.2 cal. If this extra work is assumed lost to the system through the containing walls, the second step of temperature equilibration, in which the pressures are equal and no significant momentum is given to the piston, leads to the values
and the final temperature of 293OK. If, however, the energy that went into kinetic energy of the piston in the initial step is considered to reappear as thermal energy in the gas, then the final answer will depend on the exact manner of distribution of this energy. For example, this would he sufficient thermal energy to raise the pressure of each subsystem to 1.846 atm and this larger pressure, acting on the same volume change, gives slightly larger work terms,
