infight/ "blleighing" Gases Is a Heavy Subject! Lols A. Nicholson West Springfield High Schwl Fairfax County. VA 22150
Students sometimes think that if a "light" gas, such as helium, is put into a rigid, empty container, the measured mass will be less than that of the empty container. They think that somehow the helium molecules inside the container "lift" the container to make i t lighter. while in realitv the helium molecules push slightly harder against the hot"tom of the container than aeainst the ton if the samnle is in a gravitational field. heir oGy experience with heliim comes from helium balloons. and thev. neelect - to consider that the buoyant effect of theair changes as the balloon is inflated. The balance reading will always increase when any gas is placed in a rigid, emity container, but the reason for this is often lamely given as "there's just more mass there now." The mechanism for how this mass makes its presence known to the balance is avoided. Sometimes a student brings up the "jar of bees" problem when discussion of "weighing gases" arises. "Does a jar of bees register a lesser mass when placed on a halance if the bees are in flight than when the bees are a t rest on the bottom of the iar?" The answer to that auestion is that the measured mass is the same whether the bees are in flight or a t rest, for in order to fly, the bees must exert a downward force on the air in the jar equal to their own weight, which is then transmitted to the bottom of the container. By the same line of reasoning, the measured mass of a sample of molecules should be the same whether the molecules are "resting" on the bottom of the container (as if condensed to liquid or solid) or "flying around"in the container in the gas nhase. .~~ Still, a container of gas is different from the jar of bees. The eas molecules are in flieht. not because thevare exerting a downward force on the helow them like thk wings of th;! bees. but because thev nossess velocitv characteristic of their mass and temperature: Textbooks emphasize that the collisions of ideal gas molecules are perfectly elastic and the pressure against the cop and sides of a container is the same as the pressure against the hottom of the container. The students can readily accept the fact that if thegascondensed to a liquid or solid that "rested" against the botcom of the containw. then the balance would "know" that these molecules are iresent and register their mass. But how can the balance "know" that there are eas molecules flvine around in a container and measure the; mass when apparently the collisions the molecules make aeainst the top and bottom of the container have the same impact? Many students confuse the pressure a gas exerts against the walls of the container with the weight of the gas. For example, when a bicycle tire is inflated to a pressure of 40 ~~~~
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Journal of Chemical Education
DONNA BOGNER Wtch~taState Unwerslty Wlchnta. KS 67208
~ounds/in.~. students are sumrised to find that there is iust a ;mall fractibn of that actual'weight of gas in the tire. kressure of a sample of gas will be exerted against all walls of the container independently of whether or not the sample is in a eravitational field. but in order to measure the mass of a sample of gas on typical laboratory balance, the sample must be in a gravitation field. What, then, is the relationship between the pressure of a gas and the mass measured on a halance, and how can the balance "know" gas is in the container if the molecules are flying randomly in all directions? A possible answer to these questions can he found by a ~ ~ l v i nsimple e laws of eravitv to the kinetic molecular model o'Egasei. If a container of gas is in a gravitational field, the nressure aeainst the bottom of the container must be sligdtly greate; than the pressure against the top of the container, although the difference in pressure is usually negligibly small. Molecules moving downward are accelerated by gravity while those moving upward are decelerated, resulting in the molecular collisions against the bottom of the container producing slightly greater force than those against the top of the container. This slight difference in force could be what "sends" the message to the balance pan that there is mass in the container of gas, and allows the balance to measure the same mass for these molecules "in flight" as would he measured if thev were "at rest" on the bottom of the container. If amolecule of mass m rests on the bottom of a container, the force imparted downward to the container is mg, whereg equals the acceleration due to gravity. This force, mg, is also called the weight of the molecule. If the balance pan "feels" on it, then the balance the force or weieht.. mz, - mshinedown . registers the mass m. The following argument shows that a freely bouncing molecule in a gravitation field making elastic collisions against the top and bottom of a container also imparts the same force, mg, to the bottom of the container. consider one gas molecule moving up and down in a container with height h. Let ut = velocity when molecule strikes the top of container and ub = velocity when molecule strikes the hottomof container. Assuming that the only force on this molecule is the constant force of gravity, then the velocity of the molecule will linearly increase from top to bottom of the container as the molecule falls. and the averaee . velocitv is given by average velocity = (u, ub)/2 (1)
a
-
-
+
The time required for a molecule to make a round trip from the bottom to the top of the container and back (or from top to bottom and backjis distance traveled secondslcollision = -~ average velocity
What if a bouncingmolecule doesnot have sufficient enerto reach the ton of the container? Surnrisinelv. this does kt affect the resilt! A molecule bouncing elast%ly off the bottom of the container hut falling hack before reaching the top of the container will still impart a net force, mg, to the bottom of the container. This can be shown as follows. If a molecule has bounced up and fallen hack, its velocity just before striking the bottom of the container will be equal to and opposite indirection to thevelocity with which it left the bottom of the container after the previous hounce. Therefore -u=u-gt (10) pv
The reciprocal of the expression in eq 2 is the number of collisions per second that tbis molecule makes against the top (and also against the bottom) of the container.
The force imparted to thesurfacedue toall thecollisionsper second by this molecule equals the change in momentum per second and is given by force = change in momentum per collision X collisionsls
The time between collisions is the time required for the molecule to make a round trip and is given by rearranging the expression in eq 10 to
where m = mass of molecule. Substitution of eq 3 into eq 4 gives the forces against the top and bottom, respectively: The reciprocal of the expression in eq 11gives the collisions per second this molecule makes against the bottom of the container. collisionsls = gI(2u) (12) The difference in force between the bottom and top is therefore
The force imparted to the bottom of the container due to all the collisions per second by this molecule equals the change in momentum per collision times the number of collisions per second. force = change in momentum per collision X collisionsls
Recall from elementary physics that ub2= ut2
+ 2gh
whereg = acceleration due to gravity. Substituting eq 8 into eq 7 gives Thus the net excess force imparted to the bottom of the container by this "bouncing" molecule is mg, the same force that the molecule would impart if it were "resting" on the bottom of the container! Therefore, if tbis container were placed on a balance to measure the mass of the contents, the balance could not distinguish between a molecule lying on the bottom of the container, and a molecule in flight and would measure the proper mass of this molecule whether it was part of a solid, liquid, or gas phase. Complications arise when one attempts to extend the preceding argument t o a sample of gas containing a large number of molecules. Molecules have a distrihution of velocities, move randomly in all directions, and make collisions with each other. However, simplifying assumptions commonly seen in textbook discussions of the kinetic molecular theory can he made here also. Collisions between molecules can be neglected since, in elastic collisions between identical particles, there is simply an exchange of velocities. We can assume all molecules have the same "average" velocity, and resolve the velocity into its x , y, and z components. We need only concern ourselves with the z or up-down component of the motion-the x and y components will not he affected by gravity. It is often stated in textbooks that, on the average one-third of the molecules are moving in the z direction, but the actual fraction does not matter. In our previous discussion of one bouncing molecule, actual velocity canceled. Therefore, as long as there are the same number of collisions in the upward direction as in the downward direction, the net excess force on the bottom of the container for N bouncing molecules is Nmg, where N is the number of molecules in the sample. The balance will feel the downward force, Mg, where M is the total mass of the sample, and measure this total mass M.
I t is also easy to show that this force difference against the top and bottom of the container causes a negligible pressure difference under typical conditions. Consider a gas with molar mass of 0.0200 kglmol in a cubical hox 1.00 m on an edge at 298 K and a t typical (nominal) atmospheric pressure of 1.01 X 105 newtons per square meter, N/m2. The ideal gas law, P V = nRT = (mass/molar mass) RT, may be used to calculate the mass of the gas sample: mass = (PVIRT) (molar mass)
The net excess force against the bottom of the box due to molecular collisions equals mg, (0.816 kg) (9.80 m/s2) or 8.00 N. Therefore the additional pressure on the bottom of the container is 8.00 N/m2, out of a total of 1.01 X lo5 N/m2, which is indeed negligible. Further questions are raised by considering the interaction of a gravitational field with gas molecules. For example, is it a co&adiction to have an i.&fhermol sample of gas in a gravitation field? Since mdecules are accelerared downward Ly gravity, the average molecular velocity of molecules close to the bottom of a sample will be slightly greater than those at the top of a sample, implying a slightly higher temperature at the bottom of a gas sample. How significant would this gravitationally increased velocity he? The average rnolecular velocity (or root-mean-square velocity) is related to absolute temperature hy where M = molar mass of the eas. D is averaee molecular velocity (or thermal velocity), R ~ the S gas c o n s k , and T is the kelvin temoerature. Ajsume a tvoical Mof 0.0200 krlmol and T of 298 K. Then the average &lecular velocity &I be
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Sunerimnosed on this thermal velocity is the additional velocity imparted to a molecule falling from the top to bottom of the container due to the acceleration of gravity. Assume the container is 1.00 m high. Then t h e additional gravitational velocity present when the molecule strikes the bottom of the container is u' = (2gh)'" = (2(9.80 m/s2)(1.00m))'"
= 4.43 m/s
This shows that the additional gravitational velocity a t the bottom of the container is less than 1%of the average velocity. Temperature depends on vZ,so the increase in temperature at the bottom of the container will be less than 0.01% of the average temperature, according to this model. Further questions and complications involving real gases could be mentioned. The preceding discussion assumed uniform density of the gas sample, but in reality the density decreases exnonentiallv with increasing height in the container. I t ha&een assumed that all mol&ulei have the same average thermal velocity, while a Boltzman distribution of velocities actually exists. What happens in mixtures of gases-do those molecules with greater mass tend to concentrate at the bottom of the container while those with less mass concentrate a t the top, a situation with lower potential energy (but also lower entropy) than a uniformly mixed samde? How could the previous discussion he modified to coniider gas samples in irregular-shaped containers in which the surface areas of the top and bottom of the container are not equal? These topics require more complex treatment beyond the scope of this paper and could be pursued by interested students.'s2 By combining gravitation laws with the kinetic molecular theory, the preceding discussion has shown that when a sample of gas is in a gravitational field, the molecular collisions aeainst the bottom of the container could be considered tohave greater impact than those against the top of the container. This difference in impact results in a slightly greater pressure on the bottom than on the top of the container and allows the balance to "know" there is gas in the container and measure its mass. If the kinetic molecular theory assumes molecules obey the laws of classical physics, then in a eravitational field the molecules, althouah flying around rakdomly, must also respond to "the gravity of the situation" and can be weighed! Addltlonal Questlons To Think About 1. If an open beaker is "weighed" on a balance, are you measuring the mass of the sir in the beaker, or just the glass that makes up the beaker? 2. If you "pour" a gas denser than air into the beaker (say carbon dioxide from a haggie), the balance reading becomes greater, implying that indeed you are "weighing" the contents of the open beaker. If so, how much above the heaker will the mass be measured? If there is a "mound" of COzabove the rim of the beaker, will that mass also he measured? If so, would you measure the mass of a plane or bird that flew over the beaker? 3. [Vhat happens when you try to measure the mass of a sample of gas by inertial methtds" .Are the gas muleculrs "piled" against the side of the container as it is being accelerated? Ix there a time lag between acceleratingthe container and acceleratingthe molecules in flight within the container? Also, if one accelerates a sample of gas and increases the velocity ofthe molecules, does the temperature of the gas increase?
'
Fleagle. R. G.; Businger, J. A. An Introduction to Atmospheric Physics: Academic: New York, 1963. Dunon. J. A. The Ceaseless Wind: An htrcduction to the Theory of Atmospheric Motio~Dover: New York, 1986. Melvin E. Zandlet, Department of Chemistry. Wichita State University. Wichita. KS 67208.
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Journal of Chemical Education
An Addendum One of the reviewers3 suggested an alternative approach that explains how a gas flying around in a container can be weiehed. This exnlanation involves the "barometric eauation;'. According to this model, the potential energy losi by falline eas molecules in a mavitational field does not eo into increased kinetic energy b;t instead into doing work comnressine the successive lavers of eas below. Thus, a aas sample is isothermal from top to b&m, but the po&ation density of molecules is neater at the hottom of the container than at the top. ~ h e i e f o r e ,the bottom of the container experiences more frequent collisions than the top, although the individual collisions against the bottom and top have the same impact. This difference in number of collisions allows the balance to "know" that there is gas in the container. Pressure and gas density decrease exponentially with heieht in a eas sample. as can be shown from the followine reasoning. conside;a cklumn of gas with molecular mass, M; constant temperature, T, and cross-sectional area, A. Now imagine dividing the column into horizontal layers or slices of height, dh. Assume each slice is thin enough so tbat the densit; and pressure of the gas are constant an> the ideal gas law holds within the slice. Consider one particular slice. The massof the gas in this slice, dm, is the number of moles in the slice times the molecular mass, nM.From the ideal gas law, n = PVIRT, and the volume of the slice is A dh, so the mass of gas in the slice is:
Now, if we can believe from the previous discussion that the molecules in this slice exert an excess force equal to their weight on the bottom of the slice even though they are in flight, we can say tbat the difference in pressure of the gas from the top to the hottom of the slice is dp=--
force at top of slice - force at bottom of-slice area of slice
Rearranging eq 2 gives
When eq 3 is integrated from height = 0 to height = H, the result is
P = P, exp (-MgHIRTJ
(4)
where Po is the pressure a t height = 0. Equation 4 is often called the "barometric equation" and shows that the pressure in an isothermal gas sample decreases exponentially with height. Since molecular density is directly proportional to pressure a t a given temperature, molecular density also decreases exponentially with height. Moreover, note how the exponent involves the molecular mass of the gas. The greater the molecular mass of a gas, the more rapid the decrease in partial pressure of tbat gas with height, implying that heavier gas molecules tend to concentrate at the bottom of a mixture of gases. In a sample of gas, according to this model, there must he more frequent molecular collisions against the bottom of the container than against the top, and this difference in number of collisions is what sends the "messaee" to the balance to "weigh" the gas. I t can be shown that tge net excess force due to the greater number of collisions aeainst the bottom of the container is just mg, where m is thetotal mass of gas in the sample, as follows.
Consider a box with cross-sectional area, A, and height, H, containing a gas with molecular mass, M. The pressure at the bottom is Pb and the pressure a t the top is Pb exp (-MgHI RT). Since force equals pressure times area, the total force against the bottom, Fb, is PbA and the total force against the top, Ft, is PbA exp (-MgHIRT). The force difference between the top and bottom is
Fb- F,= P&1 - exp (-MgHIRT)) (5) Now divide the box into horizontal layers or slices of height dh. Assume each slice is thin enough so that density and pressure of the gas are constant and the ideal gas law holds within the slice. Consider one particular slice a t height h. The mass of the gas in this slice is
The total mass of gas in the entire box, m, is the integral of dmfromh=Otoh=H.
Rearranging eq 8 gives
(6)
mg = Fb(l- exp (MgHIRT)) = PbA(l- exp (MgHIRT)) (9)
But since P = Pb exp (-MghIRT), and V = A dh, eq 6 becomes
Comparing eq 5 with eq 9 shows that the difference in force, Fb - Ft, just equ& mg!
dm = nM = (PVIRDM
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