why is it that if we put an inflated balloon inside th ref. it becomes smaller?
116245
I am making the assumption that the balloon is inflated with helium.  The molecules of helium contract when they are cold and expand when they are hot.  So if you put the balloon in the refrigerator or outside on a winter day, it will become smaller.  Once you put the balloon back into a warm environment, it will expand back to it's original size.
898627
The idea is that the increase in hydrostatic pressure of the water will cause a decrease in volume of the balloon. You are told that the temperature of the ballon is constant so that you can make the assumption PV = constant = Pa Va, where Pa Va are the pressure and volume at atmospheric conditions.\n\nThe bouyant force is simply the weight of water displaced, and therefore is it proportional to the volume of the balloon. You can assume that the water is incompressible and therefore constant density.\n\nSo, saying that the bouyant force is 1/2.4 of the surface value means that the volume has decreased to Va/2.4.\n\nPa Va = Pb Vb; Vb = Va/2.4 --> Pb = 2.4 Pa.\nRemember that this means absolute pressure.\nThe absolute hydrostatic pressure in the water is:\nPw = Pa + (rho) g h\nwhere h is the depth below the surface.\nSo: 2.4 Pa = Pa + (rho) g h\nh = (1.4 Pa)/((rho) g)\nPa = 101 000 N/m^2\nrho = 1000 kg/m^3\ng = 9.8 m/s^2\nh = 14.4 m
310208
The balloon rises in atmosphere and remains intact till the point when its internal pressure is lower than the atmospheric pressure. At higher altitude where the atmospheric pressure is lower and the pressure inside the balloon is higher, it bursts as the rubber cannot withstand the internal pressure.
61711
Hot air is less dense than cold air so it rises 9that's why upstairs in a house is usually warmer than downstairs).  By filling a bag with hot air you can make the bag (balloon) rise up.
881994
An balloon has positive buoyancy (floats) if the volume of air it displaces weighs more than the balloon. This is the same as saying the balloon is less dense than the air it is in.\n\nIn air, which is mostly nitrogen and only about twenty percent oxygen, a balloon will not float if its overall density is greater than air. That is, if the balloon and the gas it contains weighs more than the volume of air that it occupies the balloon will sink. So you can't fill the ball on with air and expect it to float because the balloon fabric is denser than air.\n\nHowever, there are three (and only three) gases whose densities are less than air: hydrogen, helium, and hot air. If the volume of the balloon is large enough, the weight of the hydrogen or helium or hot air when added to the weight of the fabric will be less than the weight of the air the balloon displaces. The balloon rises.\n\nAll of this is commonly referred to as manifestations of Archimedes' Principle.
30902
That last answer is slightly confused. By Charle's law (called Gay-Lussac's Law in France), V1/T1=V2/T2 at constant pressure, where Vs are volume and Ts are temperature. Therefore, if one increases the temperature T1 >T2, then one would expect the volume to increase as well, so V1>V2. \n\nHowever, this is assuming equal pressure. Obviously, as the temperature of the balloon increases, one might expect the pressure to increase as well. This situation is governed by the universal gas law, PV=nRT, and as temperature T increases, one would expect the product PV to increase as well, probably some increase in pressure P and some increase in volume V, depending on how stretchy the balloon is. Another factor that comes into play is that balloons are permeable, so as the pressure increases, more gas inside the balloon will diffuse out of the balloon, decreasing the volume of the balloon. \n\nI would expect that over a short period, the balloon would expand from the heating of the gas inside, but over a longer period, the heat and pressure would encourage enhanced diffusion, causing the ballooon to shrink. \n\nIn cold water, obviously diffusion will not operate as efficiently, and one will have a more straightforward application of the universal gas law, so that the product PV will decrease with decreasing temperature over both the short and long terms. \n\nThe rate at which the heated balloon contracts due to loss of air inside by diffusion depends on the permeability of the balloon surface.
379193
it has nothing to do with the air in the atmosphere its the air that fills the balloon up. when its hot the ballon goes up as it cools it goes down
404977
Hahahahahahaha, how amusing the previous anwers are.\n\nThe balloon would continue to rise in the atmosphere until it reached a height where the pressure inside the balloon was equal to the pressure outside the balloon...then it would remain at the same height, moving in the wind, maybe slightly up and down due to thermals...until it burst !
508366
It's expanding but it's also self-contained--much like the surface of a baloon as it expands. To creatures on the surface of a balloon it would be meaningless to ask what the balloon is "expanding into" because the surface of a balloon is contiguos in 3 (2 space 1 time) dimensions much like the universe in contiguous in 4 dimensions (3 space 1 time)
147395
A gas burner heats air inside the balloon and as the air becomes warmer it becomes less dense or lighter per unit of volume, compared to the surrounding air. The differance in the two air densities causes the balloon to float like a cork in water.  This is the reason balloonists launch their ballonons in the early morning when the air tempature is cooler.  The cooler the  outside air the less heat is needed in the balloon to get it off the ground saving fuel for a longer flight.
823437
If you take a balloon under water, and go further and further, the air becomes more dense...due to PRESSURE.   The way a submarine dives is completely different then denser/less dense ....it's that they are mostly filled with air...so they should float...but they have enough weight that they sink too,  it's almost a balance, but they have tanks that they let fill with water to sink, or push air in (and water out) to get lighter and float.
686605
Helium is lighter than air and cause the ball to float away.\n\nCarbon Dioxide is heavier than air so the ball would not bounce so well and it would be harder to kick.\n\nOxygen is lighter than air, but the difference is so small you wouldn't notice, no more that a ball being slightly over/under inflated.\n\nSince air is the cheapest, at a cost of nil, I suggest you put that in you balls.
979445
IF this is true (you need to verify this) then I have two hypotheses:\n\n1) Darker colors absorb more light as heat, and hotter air will escape faster (the extra motion of the molecules makes them pass through the baloon wall faster).\n\n2) The dyes they use for different colors, change the properties of the rubber.  E.g. the yellow dye mixed in with the rubber used in yellow balloons is more permeable than, say the red dye ... in that case, yellows baloons will lose air faster than red.\n\nIt would be a great science-fair experiment to eliminate hypothesis (1).  Two experiments you could do:\n(a) Verify whether those colors that lose air faster are always the darker colors.\n(b) Keep the same baloons in a dark closet so that they don't have light to heat them.  If they all lose air at the same rate, then that verifies hypothesis (1), if a different rate, then that eliminates hypothesis (1).\n\nRemember to limit your variables:  \n- You need to fill the baloons to exactly the same pressure and size (e.g. fill them up using a bicycle pump, using the same number of pumps ... and/or measure the circumference of each balloon with a string ... good for measuring both the starting size, and the rate that they lose air).  \n- Don't stretch the balloons before inflating them.  While this is a common trick to make blowing them up easier, it's hard to do this in exactly the same way with every balloon, so you will be affecting the results.\n- Use lots of balloons.  One of each color isn't enough.  You need to verify that (for example) red balloons *always* lose air faster then yellow balloons.\n- Compare balloons from the same manufacturer (don't compare yellow baloons from one company with red baloons from another, as they will have different types or thicknesses of rubber).
19534
ok, so lets calculate this.  a balloon with a foot diameter has a volume of .7854 cubic feet. .5 cubic feet converts into 14.158423356 liters.  The density of helium is .1785 g/L (grams per liter) and the density of air is usually approximately 1.25 g/L.  The difference between these densities is 1.07g/L.  The formula for density is d=m/V (density equals mass over volume)  so we set d=m/V, 1.07=m/14.158423356.  the 1.07 was the difference in density(how much less dense helium is than air) and the V i said(14.158423356) is how many Liters i said was in .5 cubic feet before.  it has to be in liters since the density measurement i gave you was in grams per liters, so for the formula to work we need to work in liters and receive an answer in grams.  ok, so solving for m in that density formula, we get 15.95g.  so each balloon can hold about 16g.  but i looked up the average weight of a regular party balloon and it's 2g, so each balloon can hold 14 more grams.  converting grams to pounds, 14 grams equals .030864717 pounds.  so each balloon can hold .030864717 pounds.  now if we divide your weight by the amount of weight each balloon can hold, we will get the number of balloons you need to support your weight. 160/.030864717 = 5183.9 so 5,184 balloons. \n\nyou would need 5,184 balloons to equal your weight.  this only makes you weightless, so the force of these balloons would basically perfectly counteract the force of gravity, so you would need to jump to push off and you wouldnt be pulled back down. if you wanted the balloons to actually lift you, you would need more, and depending on how many more, you would have a greater acceleration upward.  this is also assuming that each balloon is perfectly one foot in diameter, so the number of balloons may fluctuate a bit.\n\neven so, might i suggest you use a balloon with a larger diameter?\n\nif you want to find out for sure with specific balloons, you can always get like 50 balloons and some weights.  then you can measure how much weight the 50 balloons hold. then you set up a proportion saying that 50 balloons divided by the weight you measure it can hold(in pounds) equals X balloons divided by 160 lbs. then you can solve more precisely for the specific balloons you have.
634831
because den u are decreasin d volume taken up the gas molecules and so as in boyle's law d air presure wud increase causinin d molecules 2 push on d balloon wall harder and so it bursts
768769
If r,a are the actual and initial (when it's empty) radius, then the potential energy of the balloon itself is:\n\nE = (C/2)*(r-a)^2\n\nfor some constant C. A simple way to see this is to think of the balloon as a two dimensional network of springs (the potential energy of a spring is V = (k/2)(x-x0)^2) which linearly scale in size as the balloon does.\n\nTreat the air as incompressible at least for a qualitative explanation (consider how difficult it is to blow air into a fixed container (such as a pop bottle) versus a balloon). In that case the guage or net pressure (difference from atmospheric pressure) just comes from the elastic force of the balloon. The relation between the volume of air inside and the radius is:\n\ndV/dr = 4pi*r^2\n\nSo the relation between the energy and the volume of air inside it (which is the same as the pressure (dE = PdV):\n\nP(r) = dE/dV = (C/4pi)(r-a)/r^2 \n= Constant * (1/r - a/r^2)\n\nThis function increases from 0 at r=a to a maximum at r=R=2a, and from there on decreases as the radius increases. The maximum may not occur exactly at 2a, but it probably increases up to some relatively small radius R and then decreases. \n\nIf two balloons are blown up to normal sizes (>R), air should flow from the smaller into the larger. In other words the small one gets even smaller until it goes below (the very small radius) R where the pressure begins to decrease with size.
859887
Not sure if I understand. You're going to replace the old hydrogen with some new hydrogen that's less dense? The only way for the density of hydrogen to be less is if the pressure is lower or the temperature is higher. Since this is a balloon, the pressure has to match the pressure of the atmosphere. \n\nSo I'll assume the new hydrogen is at a higher temperature. In that case, the balloon can be kept inflated to the same size with fewer molecules. Therefore the mass of the hydrogen will be less because there is less of it. So the bouyancy will increase. The balloon as a whole is less dense so it is more bouyant.