The volume become 197 mL. (Avogadro law)
As you twist the balloon, the volume decreases. In accordance with Boyle's Law, volume and pressure are inversely related. As the volume decreases, the pressure increases. This is pretty intuitive: you know that if you make enough twists in the balloon, eventually the pressure will cause it to pop.
49.5
the volume of the balloon decreases because the tempature decreased
3.5 litre if pressure is kept constant.
A solid has a volume and a shape a liquid has a volume but no certain shape a gas has no certain volume and no certain shape
if vigorous enough to create heat the balloon will expand.
As you twist the balloon, the volume decreases. In accordance with Boyle's Law, volume and pressure are inversely related. As the volume decreases, the pressure increases. This is pretty intuitive: you know that if you make enough twists in the balloon, eventually the pressure will cause it to pop.
The kinetic energy of the particles inside the balloon increase. This then expands the volume of the balloon.
I interpret this question to be asking how one might measure the volume of a water balloon without breaking the balloon or emptying it of the water in order to measure its volume. One method is to fill a container with water that will be large enough to contain the water balloon, and then submerging the water balloon in the container. The volume of the balloon will be the apparent volume change of the water in the container. Any measurement will introduce some error. Since water compresses hardly at all, one would expect that submerging the balloon would not significantly change the volume of the balloon. There could be some error if one had to push down on the balloon to make it fully submerge. There will also be some measurement error in determining the volume change.
It depends on many things, but if you're going by the equatorial radius as the same size, the balloon should be about 10-15% larger. Note that you can mesure the volume of irregular shapes, thanks to my good friend Archemedes ("Eurika!"). Take a mesuring container large enough to easily hold the balloon and fill it about 2/3 full. Note the volume the water occupies by the water level. Now, immerse the balloon in the water and mesure the volume by the water level again. the difference is the volume of the balloon.
The shape of the gas is determined by the shape of the container (assuming that there is enough gas to fill the container). The volume of the gas is determined by the volume of the container (again assuming that there is enough gas present to fill the container.). When a gas is introduced into any container, it will assume the size and shape of the container if the container is filled. The exception to this rule might be if the pressure of the gas introduced is great enough to influence the shape of the container (i.e. blowing up a balloon).
The volume of the balloon decreases
The air escaping out of the nozzle has momentum and creates a force on the balloon. According to the laws of motion, every action has an equal and opposite reaction; the balloon pushes the air out bacwards and the air pushes the balloon forward.
Yes, if it were loud enough and on the same resonant frequency of the balloon rubber. The problem is that a balloon is full of gas, which has no definite volume and therefore the frequency of the balloon is constantly changing.
volume decreases considering the pressure is constant
49.5
The first part of the explanation is understanding why a balloon changes when you put air into it. Before you blow up a balloon, you can see that the volume is small and that the balloon is elastic. As you put more air into the balloon you are increasing the pressure. The air is packed in tight, so it attempts to push out and escape, so the balloon's surface stretches until a balance is reached. The tension of the balloon's surface combined with the outside atmosphere's pressure matches the internal pressure of the air. This equilibrium is always held. If you increase the pressure (putting more air into the balloon) the balloon's surface gives just enough so that you equilibrium is reached again. It is this maintaining of equilibrium that answers your question. If you try and decrease the volume in one area of the balloon, the air is going to push out another area of the balloon to make up for the lost volume. The volume is always maintained and the pressure remains constant.