Chemistry for Everyone
Why Does a Helium-Filled Balloon “Rise”? Richard W. Ramette† 765 West Fountain Creek Dr., Green Valley, AZ 85614;
[email protected] I have this colleague, a physicist, who thinks too much. One day, during lunch at La Fuente restaurant, he asked with a mischievous grin, “May I try out an idea on you?” I winced, remembering past “ideas,” but said, “Only if you pick up the tab.” “Deal. Here it is: consider a sealed, perfectly insulated room containing only air at normal temperature and pressure. Do you think the air in the room is homogeneous and that the molecules are moving randomly?” “Of course,” I replied, “anyone who understands kinetic theory knows that.” Not trusting him, I added, “I assume you mean homogeneous at the macro level.” “OK,” he continued, looking a bit smug I thought, “now tell me, why is it that a helium-filled balloon in such a room would rise to the ceiling?” I said, “I know you are leading up to something, but I will play along. By Archimedes’ principle, a body (balloon) immersed in a fluid (air) is buoyed up with a force equal to the weight of the displaced fluid. The weight of the helium is less than that of the displaced air.” “Now,” he gently chided as he looked me in the eye, “that is a good statement of an observation, but is it an explanation? That is, why does Archimedes’ principle ‘work’? How can you explain, at the molecular level, why the balloon rises or, for that matter, why a canoe floats in water?” I must have looked a bit bemused, because he prodded me with “Think! What is the real source of the force that holds up the balloon?” “Oh, I see what you are getting at,” I said with relief. “It has to be the molecular collisions of air molecules with the balloon. Each collision transfers momentum, and because there are zillions of collisions per second it adds up to a force that pushes the balloon up.” “Hmm,” he gloated with his finger pointed at my chest, “but didn’t you say earlier that the air molecules in the room are moving randomly? If so, wouldn’t there be the same number of collisions per second on the top of the balloon, as well as on the bottom or on the sides? So wouldn’t the balloon just hover in a stationary way?” “In fact,” he mused, “isn’t that exactly what would happen if you released a helium balloon in the space station where everything just floats because there is no gravity?” Now it was my turn to score a point. “Well, it is not that there is no gravity, because Earth’s mass is still as attractive as ever. It is just that the orbiting space station and its contents are in free fall, so there does not seem to be any gravity!” “Good point,” he admitted with a slight blush. “So what do you say next?” “OK, I think I am getting a grip on this little exercise. I should not have said that the air is homogeneous and random. †
Laurence M. Gould Professor of Chemistry, Emeritus, Carleton College, Northfield, MN.
Because of Earth’s gravity the air establishes a density gradient: the molecules have a tendency to be attracted downward.” That is why they pressurize airplane cabins, after all. A glint appeared in his eye: “If that is so, then wouldn’t the balloon be hammered down a bit more than it is hammered up, because of more molecular collisions on the top? Why wouldn’t the balloon sink instead of rise?” I finished my burrito and excused myself, not because I really needed to visit the room labeled ‘Hombres’, but to gain a little time to think. When I returned I said, “The answer, of course, is that the density gradient caused by gravity results in more molecules per milliliter at the bottom of the balloon than at the top. So there are more collisions per second on the bottom, giving a net force upwards.” “My compliments!” he said with a genuine smile. “You are the first one I have tried this game on who finally came up with a good explanation.” I did not think the ‘finally’ was necessary. But I was not all that pleased with myself, because a nagging thought intruded. I said, “Thanks, but I still see a problem. Wouldn’t this argument about a density gradient apply just as well to the helium in the balloon? If so, wouldn’t the inside bottom be hammered more times per second than the inside top? Wouldn’t this counterbalance the density gradient effect in the surrounding air?” My colleague looked uncomfortable and, for a change, was at a loss for words. “Oh! Look at the time!” he exclaimed. “I have to rush to my class. Same time tomorrow for lunch?” “Fine with me,” I replied. I was glad for the chance to do a little research on density gradients. That evening I logged onto the Web, called up the JCE Web site (1) and used the search feature to find an article on the ‘lifting power’ of helium compared to hydrogen (2). This was a good set of calculations based on Archimedes’ principle. Another article dealt nicely with the determination of molecular weight of light gases (3), but neither addressed the buoyancy question as I wanted. I had a vague recollection of an equation that related air pressure to altitude, so I went to http://www.google.com (accessed Jul 2003), and entered the search words ‘atmospheric pressure equation’. Google returned 76,200 hits, and I especially liked the 4th one: The Barometric Formula (4). The barometric equation as presented there is, P P°
= exp
− mgh kT
where P is the pressure at altitude h, P ⬚ is the pressure at ground level, m is the average mass of the gas molecules, g is the acceleration due to gravity, k is Boltzmann’s constant, and T is the Kelvin temperature. Multiplying the exponent top and bottom by Avogadro’s number converts the equation to, P P°
= exp
− Mgh RT
JChemEd.chem.wisc.edu • Vol. 80 No. 10 October 2003 • Journal of Chemical Education
1149
Chemistry for Everyone
where M is the molar mass of the gas and R is the ideal gas law constant. For air we may use M = 0.029 kg兾mol, g = 9.8 m兾s2, R = 8.314 J兾K mol, and T = 298 K. This gives:
P P°
= exp (−0.000115 h )
If we have a typical 10-in. balloon, say 0.25 m in diameter, then the pressure ratio, top to bottom, is about 0.999971, barely different from being the same, but enough to cause ‘lift’. This is only 0.022 mm Hg difference in barometer reading, not even discernible with the usual instrument. But if I did my calculation right, this amounts to about 14 grams in lifting power, almost enough to lift 3 nickels. As for the helium, inside the balloon, M = 0.004 kg兾mol, and
P P°
= exp (−0.0000158 h) = 0.999996
so the gradient effect inside the balloon is far from enough to counterbalance the outside effect. This also explains why our atmosphere is so poor in helium: it escapes gravity easily. I had it! My colleague would be impressed. I almost couldn’t stand the wait until tomorrow’s lunch. There he was, at our usual table, already dipping chips into hot salsa. Before I could open my briefcase, he said, “You know, I have been thinking, since the molecular mass of helium is so much lower than the average value for air, its density gradient is almost negligible.” Imagine my disappointment! I showed him my stuff anyway, and he said he had done about the same thing, but had derived the barometric formula himself. I smiled, rather tightly. Oh well, at least we both had a better appreciation
1150
of why a helium balloon ‘rises’, and the enchiladas were great. He also told me that Hyperphysics was his favorite Web site. As we parted he said, “Oh, by the way, since you understand the rising helium balloon so well, this should be easy. Suppose you have two balloons in your car. One is filled with carbon dioxide and hangs from the ceiling. The other is filled with helium and is tethered to the seat. When you accelerate from a dead stop, your body presses backward into the seat owing to inertia. What do the balloons do?” He left, with a jaunty wave and an ‘adios’, as I glared at his retreating back. I went to the stockroom, prepared the two balloons, and took them to my car when I quit for the day. I picked up my grandson Joshua as an unbiased observer so I could keep my eyes on the road. What happened during acceleration, and during braking, was quite surprising: I knew I would not get much sleep that night until I had figured it out. I also vowed to ask my colleague to explain how a siphon works, and perhaps why we believe that H2O is the formula for water! I know few people who can clearly explain either. Literature Cited 1. Journal of Chemical Education Online index. http:// jchemed.chem.wisc.edu/Journal/Search/index.html (accessed Jul 2003). 2. Ball, D. W. J. Chem. Educ. 1998, 75, 726. 3. Lieu, V. T.; Kalbus, G. E. J. Chem. Educ. 2002, 79, 473. 4. The Barometric Formula. http://hyperphysics.phy-astr.gsu.edu/ hbase/kinetic/barfor.html (accessed Jul 2003). This is the work of Carl Rod Nave, Department of Physics and Astronomy, Georgia State University. His main Web page is at: http:// hyperphysics.phy-astr.gsu.edu/hphys.html (accessed Jul 2003) and provides an enormously valuable, easy-to-use resource for all sorts of physics principles and problems.
Journal of Chemical Education • Vol. 80 No. 10 October 2003 • JChemEd.chem.wisc.edu
