# A soil with a bulk density of 1.33 g/cm3 has half of its pore space filled with water. What percentage of the bulk soil is air? ?

Assume a particle density of 2.65 g/cm3.

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• 2 months ago

this is a mass balance problem with a volume-density component.

The trick is to determine the density of the pore-space portion.  We are given the density of the solids portion and the bulk density.

50% of the pore space is filled with water, at 1 g/cm3.  Half is "empty", so 0 g/cm3.  thus, the pore-space has an effective average density of 0.5 g/cm3 (=(0.5*1+0.5*0)g/cm3).

We are given that the mass of a unit volume (mass of 1 cm3) is 1.33 g.  We are given that density2 (the solid portion of the volume) is 2.65 g/cm3 so a unit volume would have a mass of 2.65g, and we have determined that the average density of the pore space is 0.5 g/cm3 (a unit volume would have a mass of 0.5g).

Normalize the mass balance equation to unit volume and you get X*0.5g+(1-X)*2.65g=1.33 g. Solve for X.

Half of that 61% by volume pore space is filled with air, so the percentage of air by volume is 30.5%, about (you ought to be more precise; I have rounded too early; the true value is more like 30.7%). Basically, about 1/3 is water, 1/3 is air, and 1/3 is solid.

• Anonymous
2 months ago

Surprising how many people get try to get their education on Yahoo! Answers or get others to do their coursework and then pass it off as their own.

• 2 months ago

Water has a density of 1 g/cm^3

We can pretty much treat the air density as 0 g/cm^3.  It really isn't, obviously, but it's so low in comparison to the densities of soil and water, we needn't worry about it.

In a cubic cm of soil, we have 0.5 grams of water and 1.33 grams of soil

1.33 - 0.5 = 0.83

0.83 grams / (2.65 g/cm^3) =>

0.83/2.65  cm^3 =>

83/265  cm^3 =>

0.3132075471698113207547169811320... cm^3

0.31 cm^3 of actual particulate

0.5 + 0.31 = 0.81

1 - 0.81 = 0.19

0.19 cm^3 of air

19%, roughly.