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If buoyancy is caused by pressure gradient and if the pressure of a gas is same everywhere in its container, which means there's no difference of pressures on surfaces of the lighter-than-air object, how come we observe such an effect?

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3 Answers 3

akhmeteli's answer is correct, and if you're going to accept an answer you should probably accept his, but I think it's worth expanding on his rather brief answer to give some physical insight as to what is going on.

Suppose your lighter than air device is a cube 1m x 1m x 1m (I'm just making it a cube for convenience). And suppose the pressure on the top surface is 1 atmosphere (101325 Pa). The cube of air has a non-zero weight. In fact a quick Google for density of air will tell you that the 1m cube of air weighs 1.225 $kg/m^3$. That means the pressure on the bottom surface of the cube is 1 atmosphere plus the weight of the air in the cube, so the pressure on the bottom is greater than the pressure on the top. In fact the difference is 1.225 $\times g$ per square metre, which is 12.02 Pa.

Because the pressure on the bottom is greater than the pressure on the top there is a net force upwards of (in this case) 12.01 Newtons, which is the same as the weight of the air. In fact this is an example of Archimedes principle, and though I've considered a cube the result is generally true i.e. the upthrust on a body in a fluid is equal to the weight of fluid displaced.

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Since you asked, you should look into the "next level" of the following concept.

the pressure of a gas is same everywhere

There can be external forces acting internally to the gas, such as gravity. By that I mean each molecule in the gas experiences the force where it is. When you have that, the above statement doesn't apply. If the gas is accelerating it's the same deal. For the case of gravity, you can write the gravitational force as a vector downward, $\vec{g} = (0,0,-9.8 m/s^2)$. The pressure then has a gradient along this vector. That statement comes from the force balance. It is required so that the system is stable, and by that I mean non-accelerating. I could write:

$$\nabla P = -\rho \vec{g}$$

And here, $\nabla$ is a scalar field to vector operator that is $(\partial/\partial x,\partial / \partial y, \partial / \partial z)$. The above statement is the fluid-mechanics analog to saying the normal force is equal to $m g$ for an object sitting on the table.

The density of the gas will also change along the the gravitational gradient. This is required from the fact that the pressure changes, and the state equation gives an equality that uses pressure. For our purposes the gas state equation may be the ideal gas law.

I think this question presents a very valid contradiction from the basic laws that students are often told early on, but aren't necessarily true in all situations. Many statements about physics, in fact, most of them, are based on a set of assumptions. It is nearly true that gases maintain a constant pressure all around us. The change in pressure is small compared to the total pressure. However, that change in pressure from elevation is the reason that a balloon floats.

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Thank you, the analogy for gravity -a gas being accelerated- was really helpful to figure how that gradient could be formed. Next question: Is the lifting force caused by this phenomenon a linear function of displaced volume and does it mean that density of a point in the gas is proportional to its depth? –  osman Sep 6 '12 at 20:19
    
@osman "does it mean that density of a point in the gas is proportional to its depth? " YES if you assume $d \rho/dP$ is constant. That's not true over the full range of pressure (even the ideal gas law contradicts this), but if the change in elevation is small compared to the total thickness of the atmosphere it is true. So on the scale of meters in air, rho is proportional to P, which is proportional to z. –  AlanSE Sep 6 '12 at 21:06
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Typically, there is pressure gradient in gas due to gravity.

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So it varies with depth, similar to hydrostatic pressure? If so, wouldn't it be controversial considering the ideal gas law? –  osman Sep 4 '12 at 20:16
    
No, as there is typically density gradient due to gravity, as well. –  akhmeteli Sep 4 '12 at 20:45
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@osman You can imagine the bulk of the gas divided into a small volumes. Ideal gas law is still valid for each small volume, where one can assume density and temperature to be constant. Though density may vary between the volumes. –  Yrogirg Sep 5 '12 at 6:35
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