I've been interested in this. I'm wondering if it's because I'm at about 5400 ft. above sea level. Less air pressure on a _fluid_ would mean less density, and therefore less weight, for a given volume. Right? I've been trying to research this to see what sort of variance I could expect. I've noticed on several recipes that I seem to have to add more water. Again, I presume this is because I actually get less water for the same volume measurement.

The difference is more likely due to lower humidity at altitude than to lower density of the water, because the difference in the volume of water owing to the difference in its density is so small as to be negligible.

Here's the math. (Can you tell I'm bored and have waaaay too much time on my hands?

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The hydrostatic equation for a fluid is defined as:

dP = - Dg dz Equation 1

where dP is the difference in pressure, D and g are, respectively, the density of air and the acceleration due to gravity at the altitude of the air layer, and the minus sign ensures that pressure decreases when the altitude increases (positive dz). This is the hydrostatic equation for a fluid.

At temperatures and pressures normally present in the troposphere, the behaviour of air closely approximates the behavior of an ideal gas, and therefore can be given as:

PV = RT Equation 2

where one mole of air occupies the volume V at pressure P and absolute temperature T, and R is the universal gas constant. The density of air is

D = M/V Equation 3

where M is the molecular weight of air. Combining Equation 1 and Equation 3 we get

dP = - (Mg/V)dz Equation 4.

Substituting Equation 2 for V in Equation 4 yields:

dP/P = -Mg/RT Equation 5.

The acceleration due to gravity ('g') is the sum of two terms: (i) the attractive force per unit mass as given by Newton's universal law of gravitation, and (ii) the "repulsive" centrifugal force caused by the earth's rotation. The latter effect is very small, contributing no more than 0.35% at sea level. The former drops off inversely as the square of the distance from the center of the earth. but even at an altitude of 10,000 meters it is only 0.3% less than its value (g0) at sea level. Therefore, the only factor on the right hand side of Equation 5 that shows a significant variation with altitude is the temperature.

The atmospheric temperature profile in the troposphere (which extends to 11Km above sea level) decreases linearly with altitude to -55° C at the tropopause, at which point it starts to increase because of the solar heating of the stratospheric ozone. The drop in air temperature per unit increase in altitude, i.e. the lapse rate, is B=6.5° K /km.

Below the tropopause, therefore:

T = T0 - Bz Equation 6

where T0 is the sea level air temperature and z is the altitude in km. Substituting this into Equation 5,

dP/P = - (Mg/R)/(T0 -Bz) Equation 7

Integrating both sides from sea level (z=0) to z, we get

P(z)/P0 = [1 - z(B/T0)]^(Mg/BR) Equation 8.

Using the definition of the standard atmosphere adopted by the International Civil Aviation Organization, we have: M=28.9644 (carbon-12 scale), T0=1° C = 288.1° K, g=g0=9.80665 m/s^2, R=8.314 Joules/gram-mole/deg K, P0=101.325 kPa (1 atmosphere). Substituting into Equation 8 we have, finally,

P(z)/P0 = (1 - 0.02255z)^5.256 Equation 9

Assuming an altitude of 1.5 km (4921'), P(z)/P0 is 0.834.

The bulk modulus of water, B, is 2 x 10^9 pascals so water with volume V will change volume by DV under an additional pressure P by DV/V = P/B.

One atmosphere is defined as 101,325 pascals, or approx. 10^5 pa, so given a pressure difference of .166 (between sea level and 5000'), the difference in the volume of water DV/V works out to .166 x 10^-5 or slightly more than one ten-thousandth of one percent (.000166%).