How does temperature affect the size of a balloon?
605190
depenz on the blow job  lol
94588
It depends on various factors but most important would be pressure. As the balloon ascends into the upper atmosphere pressure is slowly reduced as the air gets thinner farther up the balloon goes, Pressure within the balloon where the helium is held soon expands out as the air molecules in the atmosphere reduces and at one point it will burst as the helium expands past the stretchability limit of the balloon. So it depends on how strong the balloon fiber is, how much helium is stored within the balloon, how thin is the surrounding atmosphere compared to pressure within the balloon and also if the balloon meets any obstruction on its way up.
813780
Jacques Charles investigated the relationship between the Volume of a gas and how it changes with temperature. He noted that the volume of a gas increased with the temperature. Charles's Law states that the volume of a given amount of dry ideal gas is directly proportional to the Kelvin Temperature provided the amount of gas and the pressure remain fixed. When we plot the Volume of a gas against the Kelvin temperature it forms a straight line. The mathematical statement is that the V / T = a constant. For two sets of conditions the following is a math statement of Charles's Law: \n\n\nV1 / T1 = V2 / T2\nAn example of Charles's Law would be what happens when a hot air balloon has air heated. The air expands and fills the balloon. Of course, other physical principles cause the balloon to rise against the gravitational force. As the air inside the balloon expands the balloon gets bigger and displaces more air. The displaced air produces a buoyant force that counters the gravitational force and causes the balloon to rise. \n\nSo, given a problem: \n\nA gas occupies a volume of 100 mL at 300 K (All temperatures HAVE to be converted to Kelvins ). At what temperature will the gas have a volume of 200 mL? \n\nLet's rewrite that, identifying the variables: \n\nA gas occupies a volume of 100 mL (V1) at 300 K (T1). At what temperature (T2) will the gas have a volume of 200 mL (V2)? \n\nSetting up the problem, we have: \n\nV1/ T1 = V2 / T2 \n\n100 mL / 300K = 200 ml / T2 \n\nAfter cross-multiplying to solve for T2, we get 600 K.
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.
439452
Sure it does.  If there's no mass attached, you just need enough buoyancy to lift the balloon itself.  If you have a mass attached, you have to lift them both, so you need more buoyancy.  The more buoyancy you need, the more you have to heat the balloon.
950534
oxygen atoms are larger than helium atoms.\nThe helium leaches out of the balloon quicker than the oxygen.
172574
AP Physics is phun, dont let anyone tell you otherwise.\n\nFirst, find the volume of the balloon.\nThe volume of a sphere = 4/3 pi * r^3\nyou are given the radius, so this should be no problem.\n\nOnce you know the volume of the balloon, find how much Helium and air it would take to fill it.\nYou could do this two ways, (the Physics way or the Chemistry way) either by using the molar masses of the gasses....or using the densities of the gasses, the density way is probably easier, but you either need to have those given to you or you can look them up and assume what temperature the experiement is taking place at.\nYou need to find the masses of each of the gasses.\n\nThe lifting force of the balloon will equal the difference in the upward and downward forces acting on it.  The downward force is the force of the Helium's weight, the upward for is the weight of the air that the balloon is displacing (we ignore the weight of the ballooon).\n\nIf you want to believe wikipedia,\nthe density of air at 20 degrees C = 1.204 kg/m^3,\n\none mole of Helium = 4.00 grams.\nAt STP 1 mole of Helium would have a volume of about 22.4 Liters.
292220
Surface tension. The tape holds the balloon in place.
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.
378824
Temperature affects state change. If something, for instance water, is heated, it becomes a gas. Gases take up more volume than liquids. So the hotter a gas is, the larger it will expand, and the same is true, to a lesser degree, for solids and liquids. Hope I helped,\nZ
703096
Charles' law states that, under constant pressure, the ratio between a gas' volume and absolute temperature remains constant.  So if it heats up it expands and if it cools it contracts.  For example, you fill a balloon with helium at room temperature, and then put it in a refrigerator.  The helium will contract, and the balloon will shrink.
604015
The amount of material and the prevalent temperature haven't changed much, so the pressure and volume are inversely proportional.  As the external pressure on the balloon decreases, the amount of material in the balloon will expand to equalize that pressure.
899362
Actually, none of the above.  The most significant change is in the thermosphere.  Around 120km to 130km a huge curve on the temperature occurs.  Object falling into our atmosphere do generally burn up once they reach the mesospere, but when rising in altitude, it actually gets cooler, not warmer at all, throughout the mesosphere.  Hope this helps.
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.
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
331738
A full baloon will weigh more if you are filling it with room air.  The air under pressure will mean there are more molecules in the balloon.  More molecules of 'x' weigh more than fewer molecules of 'x'.
797306
Lets look at gas in a balloon.  Lets pretend we can see the gas molecules buzzing around on a micro-scale.  At the boundary of the gas and the surface there are always some of these molecules hitting the surface.  The molecules hitting the surface, much like you pushing on a surface, creates the pressure.  So, the faster the molecules are traveling (due to a rise in temperature) the harder they will impact the surface.  Also if there are more molecules present, more will be hitting the surface and also increase the pressure.  This can be verified by analyzing the ideal gas law, PV=nRT.  Where P=Pressure, n=amount of gas molecules, and T=temperature.  We can see if we raise the n or T the pressure must rise also, assuming a constant V(volume).\n\nOn a macro-scale, such as our atmosphere, the gas has a mass just like anything else.  Although the mass is small, there are miles of gas so the total mass has a noticeable affect here at the ground.  We don't notice the pressure because we are used to it (sometimes you may notice it when your ears 'pop' driving up or down a large hill) but the atmosphere is exerting pressure on everything.