A hot air balloon is filled to a volume of 44.5 L at 758 torr. What will be the volume of the?
The pressure, that the barometers measures, is the atmospheric pressure. If the balloon is in your house, the barometer is measuring the air pressure inside your house. This pressure pushes on the outside surface of the balloon. As the air pressure increases the elastic surface of the balloon moves inward, so the volume decreases. As the air pressure decreases the elastic surface of the balloon moves outward, so the volume increases The volume is inversely proportional to the pressure. P1 * V1 = P2 * V2 758 * 44.5 = 748 * V2 V2 = (758 * 44.5) ÷ 748 = 45.1 torr = 45.1 mm Hg http://en.wikipedia.org/wiki/Torr The website above describes a torr. The torr (symbol: Torr) is a non-SI unit of pressure with the ratio of 1 to 760 standard atmosphere, chosen to be roughly equal to the fluid pressure exerted by a millimetre of mercury, i.e., a pressure of 1 Torr is approximately equal to 1 mmHg A pressure of 414 mm Hg = 414 torr, = (414/766) atmospheres
A toy balloon filled with air has an internal pressure of 1.25 atm and a volume of 2.50 L....?
If the cylinder you're conversing approximately is a CL length. And that is crammed to 1850kpa... nicely with a CL cylinder crammed to 2000kpa you will desire to have the potential to fill 50 x 11 inch balloons. So in short i think of you will desire to have the potential to fill around 40 5 x 11 inch balloons.... desire this helps
Thermodynamics Help An initially flat balloon is filled?
An initially flat balloon is filled with helium by connecting it to a tank of helium at a pressure of 10 bar and temperature of 30 deg. cel. Until the balloon takes its final spherical shape, with a diameter of 6 m, the pressure within the balloon is approx. equal to the ambient pressure of 1.01 bar. During the filling process, which occurs at a constant temperature, the membrane of the balloon is not stretched. At the end of the process, the pressures in the tank and the balloon are equal to the ambient pressure. Calculate, the work done on the atmosphere during the filling process. Neglect the initial volume of the balloon (answer:11.42 MJ)
It partly depends on just how full it is to start with. I'm assuming we are talking about rubber party balloons. As the balloon goes up in the air, the helium expands. But the air it is rising in also gets less dense with altitude. As a first approximation, the net excess lift does not change as the balloon goes up. For a rubber balloon, as the helium expands, the rubber stretches. If the balloon is fairly full to start with, then the increase in volume can cause it to pop. However, if the balloon is not very full, but still with enough helium to cause it to rise slowly, then it is possible for the balloon to last much longer. Unlike most materials that have a linear relation between the stress in the material and how much it is stretched, the stress in rubber increases roughly in proportion to the cube of how much it is stretched. As a result, the pressure inside the balloon due to the stretching rubber will go up as it gets more full*. That increase in pressure can be enough to increase the density of the helium just enough to stop it rising through the air. It can come to an equilibrium altitude. So it doesn't pop. The helium is able to diffuse slowly through the rubber. This lets it deflate just a little and the balloon will descend to a lower altitude. Because of the stretch in the rubber, it will still be essentially in equilibrium with the surrounding air (meaning the net force is zero) so it floats stably at the new lower altitude. Over time, this equilibrium altitude slowly falls as the helium leaks out. Eventually, the balloon will drift back down to the ground. Sometimes it will pop on some sharp branch. Other times, its string can get caught and it will sit there bobbing in the breeze. Eventually the helium will diffuse out and some air will diffuse in and you can end up with a flaccid balloon with just a little air in it dangling from a branch. * Note that for most solid materials, if you make a balloon out of it and fill it completely and then force in a little more gas, but not enough to pop it, you get the surprising result that the pressure goes down.
Most latex/rubber party balloons may pop. Many stronger mylar balloons may not. It depends on how strong the balloon is, how much lifting gas is in the balloon, the conditions of the surrounding air and the weight of the balloon and any payload (i.e. string).It is possible for a mylar balloon to reach a height where it will not rise further. and it will float along more or less level at that altitude. When the sun sets the lifting gas will cool and decrease in volume and the balloon will probably descend to the ground.The lift of a balloon is the difference of the weight of the atmosphere that is displaced by the inflated balloon and the weight of the lifting gas contained in the balloon minus the weight of the balloon and any payload.Altitude, temperature, humidity, and the expansion and contraction of the balloon are all variable factors.So a balloon may rise to a point of equilibrium before the balloon fails. At this point it will float more or less at this altitude until atmospheric changes occur.This could be increasing ambient heat during the day. Or the sun going behind clouds or overcast. Or even a change in humidity (humid air is LESS dense than dry air.) I know it seems counter intuitive. Also loss of lifting gas by slow leaks or leaks through the membrane. Loss in a rubber balloon is reasonably rapid. A mylar balloon is considerably less porous.There is also the condition of a super pressure balloon staying at pressure height but that is a tune sung on a different day.
A balloon has a volume of 1.4 L at a pressure of 758 torr. What will the volume, in liters, of the balloon be?
By Boyle's law:- P1V1 = P2V2 =>V2 = P1V1/P2 =>V2 = [758 x 1.4]/979 =>V2 = 1.08 L