Before you read this, I suggest you read post 17.5
Liquids and gases are both examples of fluids – things that flow into the space available to them (post 16.37). In post 17.5, we saw how the pressure in a liquid increased with depth. Normally, we think of the pressure in a gas as being uniform (see post 18.25) but we know that the pressure of our atmosphere decreases with altitude – so the cabins of passenger planes must be pressurised with air. Also, the atmospheric pressure is lower than at the top of a mountain than at the base. This difference in pressure can be measured with an altimeter, to measure the height of the mountain.
Let’s now investigate why such pressure changes might occur. The picture above shows a vertical column of a fluid with a cross-sectional area A. A vertical distance above the base of the column is represented by x. We’re going to think about a horizontal slice of the column of thickness δx. Let p denote the pressure at x, and p + δp the pressure at p + δp. The vertical component (post 16.50) of the resulting force, arising from these pressures, acting on the slice is then
(p + δp)A – pA. = Aδp
This result comes from the definition of pressure (post 17.5). But there is also a gravitational force gδm acting on the slice, where δm is the mass of the slice and g is the magnitude of the gravitational field (the acceleration due to gravity, post 16.16). If the elemental slice is stationary, the sum of these forces must be zero (post 16.4) so that
Aδp + gδm =0.
δp = – gδm/A = gρAδx/A = – ρgδx. (1)
This result is obtained by noting that the volume of the slice is Aδx and that its mass is its volume multiplied by its density, ρ (post 16.44).
If ρ is a constant, we can integrate this result (post 17.19). The for a column of height h, the pressure at the base is given by
p = hρg.
This result is derived in the appendix and was derived, by a simpler method, in post 17.5. The density of a liquid is close to being constant, except under very high pressures. This result than tells us that the pressure at some depth h from the surface of a liquid is given by hρg.
But, in a gas, increasing pressure has an appreciable effect on volume and, therefore, on density. Remember that density is defined by
ρ = m/V = mp/(nRT)
for an ideal gas, where m is the mass of gas occupying a volume V, n is the number of moles of gas, R is the ideal gas constant and T is the temperature (measured on the Kelvin scale, post 16.34); see post 18.25 for more details. But the mass divided by the number of moles is the relative molecular mass, M (post 16.33), of the gas. So we can write
ρ = Mp/(RT).
Substituting this expression for ρ into equation 1 gives
δp = – [Mp/(RT)]gδx
(δp/p) = – [Mpg/(RT)/]δx.
Since M, p, g and R are all constant, we can integrate this result, at constant temperature, using a result proved in post 18.15. The result of this integration is that
logep = – Mgx/(RT) + C (2)
where C is a constant of integration (post 17.19); information on the log function is given in post and on e in post. To find the value of C, we apply the boundary condition (post 17.19) that when x = 0, p = p0, the pressure at the base of the vertical column. Then
logep0 = C.
Substituting this result into equation 2 gives
logep = – Mgx/(RT) + logep0
logep – logep0 = – Mgx/(RT).
Using information on logarithms given in post 18.3, we can write this result as
loge(p/p0) = – Mgx/(RT)
p = p0e-Mgx/(RT). (3)
The same mathematical techniques were used in post 18.15 where I explain them in more detail. From the definition of Avogadro’s number, N, given in post, M = Nm (post 17.48). Also, R = Nk (post 18.25), where k is Boltzmann’s constant (post 18.25). Putting these two results into equation 3 gives
p = p0e-mgx/(kT). (4).
Either equation 3 or equation 4 can be used to explain why the pressure in the atmosphere decreases with altitude.
I now want to use equations 3 and 4 for something different. In a fixed volume of an ideal gas, at constant temperature, p is proportional to n (post 18.25). So we can rewrite equation 3 as
n = n0e-Mgx/(RT) (5)
where n0 is the number of moles when x = 0. This is interesting because n is the number of moles at a height x above the base of our vertical column. Mgx is then the potential energy of a mole of gas at height x (post 16.21). So equation 5 tells us how the number of moles depends on potential energy. If we want to know how the number of molecules, μ, depends on potential energy, we use equation 4 (because it involves the mass of a molecule) and, noting that if the number of moles is proportional to pressure so must the number of molecules, obtain the result that
μ = μ0e-mgx/(kT) (6).
where μ0 is the number of molecules when x = 0.
Equations 5 and 6 are different ways of expressing Boltzmann’s distribution law that tells us how the distribution of particles depends on energy. Comparing equations 4 and 6 with the Gaussian distribution, in post, we can see how they resemble statistical distributions. So molecules in a gas, for example, are said to obey Boltzmann statistics. I shall return to this topic in the next post.
According to equation 1
δp = – ρgδx.
Fot a liquid, we can consider ρ to be constant. Equation 1 can then be integrated to give
p = – xρg + C’ (7)
where C’ is a constant of integration (post 17.19). To determine the value of C’, we can think about a column of liquid whose height is h. Then, when x = h, the pressure p in the liquid is zero because this value of x defines its surface. Applying this boundary condition to equation 7 gives
0 = – hρg + C’ or C’ = hρg (8)
Substituting the value for C’ from equation 8 into equation 7 gives
p = hρg – xρg
p = ρg (h – x).
Here (h – x) is the depth measured from the liquid surface. At the bottom of the column of liquid, x = 0 so that
p = hρg.
This is the result stated in the main part of the post and also derived, more simply, in post 17.5.