If the balloon is not encased or in restricted volume, then it changes its shape to accomodate that push but prutruding on the other side and the pressure remain same. However, if there is restriction for shape change, then when you push on the balloon, the volume decreases and therefore the pressure increases.
A practical balloon, by the way, will behave between these two extremes. Pushing on one side and causing it to change shape definitely results in an increase in internal pressure. This is because the elasticity/tension of the rubber is the encased space.
Yes, it is true.
Calucus
Calculus
That is because the Pressure inside the balloon caused by the filled pressure reached equilibrium at this point based on the height so can rise no longer.
The elastic contraction of the rubber in the balloon's membrane causes the pressure in the air inside an inflated balloon. When you inflate a balloon, you have to expand the latex of the balloon, which stretches when filled with air from a pressure of 760 mm Hg to as high as 840 (about 10% higher than standard atmospheric pressure). If you inflate a non-elastic mylar balloon, it takes no effort: the air inside is at the same pressure as the air outside.
The pressure outside the balloon doesn't change when the balloon rises. By a balloon rising, I assume that air is being placed into the balloon. As the balloon fills with air, the pressure inside the balloon will increase. Since the balloon can stretch, the increasing pressure against its inner walls will cause it to rise, or more correctly put, expand. Eventually, the balloon will be stretched to its fullest capacity if more air is placed inside it. When it pops, the bang you hear is the high pressure of the atmosphere inside the balloon equalizing with the lower pressure of the atmosphere outside the balloon.
The pressure inside the balloon has to exceed the pressure outside the balloon.
The molecules in the balloon can't take the pressure inside of it so it bursts.
Calucus
Increased pressure on the inside, or decreased pressure on the outside.
That is because the Pressure inside the balloon caused by the filled pressure reached equilibrium at this point based on the height so can rise no longer.
The elastic contraction of the rubber in the balloon's membrane causes the pressure in the air inside an inflated balloon. When you inflate a balloon, you have to expand the latex of the balloon, which stretches when filled with air from a pressure of 760 mm Hg to as high as 840 (about 10% higher than standard atmospheric pressure). If you inflate a non-elastic mylar balloon, it takes no effort: the air inside is at the same pressure as the air outside.
yes because pressure that is confined by a container such as a balloon will increase if the balloon is heated.
The pressure outside the balloon doesn't change when the balloon rises. By a balloon rising, I assume that air is being placed into the balloon. As the balloon fills with air, the pressure inside the balloon will increase. Since the balloon can stretch, the increasing pressure against its inner walls will cause it to rise, or more correctly put, expand. Eventually, the balloon will be stretched to its fullest capacity if more air is placed inside it. When it pops, the bang you hear is the high pressure of the atmosphere inside the balloon equalizing with the lower pressure of the atmosphere outside the balloon.
Because of the pressure in the higher atmosphere is much lower than the pressure where it was filled. This causes the pressure inside to expand the balloon up to and past the bursting point. Most likely the tip of the troposphere.
The pressure inside the balloon has to exceed the pressure outside the balloon.
The molecules in the balloon can't take the pressure inside of it so it bursts.
The balloon is filled with a gas.
The drop in temperature will cause the atoms (or air) inside the balloon to fall in energy levels, this will result in the pressure inside the balloon dropping, and may cause the balloon to loose its shape.
The pressure will increase