# Archimedes' Principle for gases - derivation?

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• Amaterasu21
In summary, Archimedes' Principle states that the buoyant force on an object in a fluid is equal to the weight of the displaced fluid. This is true for both liquids and gases, with the pressure difference between the top and bottom of the object being dependent on the density of the fluid. This can be seen through integration and the Riemann sum, resulting in the integral outlined in #2.
Amaterasu21
Hi all,

I understand where Archimedes' Principle comes from in liquids:

If we imagine a cylinder immersed in a liquid of density ρ whose cross-sectional area is A and whose top is at depth h1 and whose bottom is at depth h2:

Force(top of cylinder) FT = ρgh1A
Force(bottom of cylinder) FB = ρgh2A
Buoyant force = ρgA(h2-h1)
= ρA(h2-h1)g = ρVcylinderg = mdisplaced liquidg = weight of displaced liquid
therefore buoyant force = weight of displaced liquid.

I also understand that this works even if the object immersed in liquid isn't a cylinder because pressure at any depth is constant, so the horizontal components of pressure on any surface of the object cancel out, and the vertical components produce the same buoyant force as the cylinder.

But what I don't understand is why this is true for gases as well. I've been told that the buoyant force in gases is equal to the weight of displaced gas. However, gases unlike liquids are compressible and a large mass of gas in a gravitational field (e.g. the atmosphere) will be far denser at the bottom than at the top. Therefore while ρ is constant during the derivation for liquids, it would certainly NOT be constant across the height of the cylinder in a gas!

Force(top of cylinder) FT = ρ1gh1A
Force(bottom of cylinder) FB = ρ2gh2A
Buoyant force = gA(ρ2h2 - ρ1h1)
...and I don't see how we get from here to "= weight of displaced gas!"

Any help would be appreciated!

Amaterasu21 said:
Force(top of cylinder) FT = ρ1gh1A
Force(bottom of cylinder) FB = ρ2gh2A
This is not correct, the pressure a priori not dependent on the density in the way you describe. What is true is that the pressure differential is given by ##dp = \rho g \,dh##. In the case where ##\rho## is constant, this implies that ##p = p_0 + \rho g h##, where ##h## is the vertical distance to some reference level where the pressure is ##p_0##.

The pressure force acting on your cylinder is given by the pressure difference between the top and bottom. This pressure difference is given by
$$\Delta p = \int_{h_1}^{h_2} dp = \int_{h_1}^{h_2} \rho g \, dh = g \int_{h_1}^{h_2} \rho\, dh.$$
If you have a cylinder with a cross-sectional area ##A##, this implies
$$\Delta p = g \int_{h_1} ^{h_2} A \rho \, dh = g M,$$
where ##M## is the mass of the displaced fluid (since it is the volume integral of the corresponding density).

Amaterasu21
Thank you, this explains it! I should have realized we'd need to integrate since we're dealing with a constantly changing density with height.

Orodruin said:
This is not correct, the pressure a priori not dependent on the density in the way you describe. What is true is that the pressure differential is given by ##dp = \rho g \,dh##. In the case where ##\rho## is constant, this implies that ##p = p_0 + \rho g h##, where ##h## is the vertical distance to some reference level where the pressure is ##p_0##.

The pressure force acting on your cylinder is given by the pressure difference between the top and bottom. This pressure difference is given by
$$\Delta p = \int_{h_1}^{h_2} dp = \int_{h_1}^{h_2} \rho g \, dh = g \int_{h_1}^{h_2} \rho\, dh.$$
If you have a cylinder with a cross-sectional area ##A##, this implies
$$\Delta p = g \int_{h_1} ^{h_2} A \rho \, dh = g M,$$
where ##M## is the mass of the displaced fluid (since it is the volume integral of the corresponding density).

Amaterasu21 said:
I've been told
Were you given any reference for this information? I appreciate that you may have reservations because it looks 'too simple'.
If you stratify your volume into sufficiently shallow layers to consider the density to be constant over the thickness and equal to the mean value then the buoyant force for that slice would in fact be the weight of the slice displaced. IMO, that should be sufficient.

sophiecentaur said:
If you stratify your volume into sufficiently shallow layers to consider the density to be constant over the thickness and equal to the mean value then the buoyant force for that slice would in fact be the weight of the slice displaced. IMO, that should be sufficient.
It should be pointed out that, in essence, this is nothing but the typical Riemann sum, which in the end results in the integral outlined in #2.

sophiecentaur

## 1. What is Archimedes' Principle for gases?

Archimedes' Principle for gases states that when an object is submerged in a gas, it experiences an upward buoyant force equal to the weight of the gas it displaces.

## 2. What is the derivation of Archimedes' Principle for gases?

The derivation of Archimedes' Principle for gases involves using the Ideal Gas Law (PV = nRT) to calculate the density of the gas at different depths, and then integrating over the volume of the object to find the buoyant force.

## 3. How does Archimedes' Principle for gases differ from the principle for liquids?

Archimedes' Principle for gases is similar to the principle for liquids, but it takes into account the compressibility of gases. This means that the density of the gas changes with depth, whereas the density of a liquid remains constant.

## 4. What are some real-world applications of Archimedes' Principle for gases?

Archimedes' Principle for gases is used in various industries, such as in the design of hot air balloons and airships, as well as in the measurement of atmospheric pressure and weather forecasting.

## 5. Are there any limitations to Archimedes' Principle for gases?

One limitation of Archimedes' Principle for gases is that it assumes the gas is in a state of hydrostatic equilibrium, meaning that there are no air currents or other disturbances. In real-world situations, these factors can affect the accuracy of the principle.

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