Question

We have a balloon that we fill from an air tank. The empty and deflated balloon has a mass 11.625 g. After the balloon is blown up and allowed to sit in its surroundings (at T_(infinity) = 22 C and p_(infinity) = 0.99 atm) for an hour, we mass the balloon again, and find that it has a mass of 12.156 g and a diameter 8.5”. What is the gauge pressure in the balloon? What approximations and assumptions have you made, which might affect your result if not correct? NOTE: You will also need to consider Archimedes principle of buoyancy, so that if the T and p of gas were exactly the same inside and outside the balloon, the mass would still show 11.625 g (the mass of the empty balloon).

Answer 1

Connect a filled balloon to the shorter tube and see how far the water rises in the other. If it rises by 1 meter that means the pressure is 1N/cm² = 10000 N/m² = 10 kPa

Most toy balloons have around 5kPa. Modelling balloons have 10 - 20 kPa.

Buoyancy = weight of displaced fluid.

Archimedes' principle does not consider the surface tension (capillarity) acting on the body,but this additional force modifies only the amount of fluid displaced, so the principle that Buoyancy = weight of displaced fluid remains valid.

The weight of the displaced fluid is directly proportional to the volume of the displaced fluid (if the surrounding fluid is of uniform density). In simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. Thus, among completely submerged objects with equal masses, objects with greater volume have greater buoyancy. This is also known as upthrust.

Suppose a rock's weight is measured as 10 newtons when suspended by a string in a vacuum with gravity acting upon it. Suppose that when the rock is lowered into water, it displaces water of weight 3 newtons. The force it then exerts on the string from which it hangs would be 10 newtons minus the 3 newtons of buoyancy force: 10 ? 3 = 7 newtons. Buoyancy reduces the apparent weight of objects that have sunk completely to the sea floor. It is generally easier to lift an object up through the water than it is to pull it out of the water.

Assuming Archimedes' principle to be reformulated as follows,

apparent immersed weight = weight-weight of displaced fluid

then inserted into the quotient of weights, which has been expanded by the mutual volume

yields the formula below. The density of the immersed object relative to the density of the fluid can easily be calculated without measuring any volumes.:

The height to which a balloon rises tends to be stable. As a balloon rises it tends to increase in volume with reducing atmospheric pressure, but the balloon itself does not expand as much as the air on which it rides. The average density of the balloon decreases less than that of the surrounding air. The weight of the displaced air is reduced. A rising balloon stops rising when it and the displaced air are equal in weight. Similarly, a sinking balloon tends to stop sinking.

Buoyant force = weight of displaced water
= mass of water * acceleration due to gravity
= density of water * volume of cube * g
= 1x10³ kg/m ³ * 27*10^-6 m³ * 9.8 m/s² = 2.65 *10-1 N = 0.265 N