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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.
- busterwasmycatLv 72 months agoFavorite Answer
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.
X*0.5g+(1-x)*2.65g=1.33 g; X=(2.65-1.33)/(2.65-0.5)=0.61, about.
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.
- Anonymous2 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 =>
0.31 cm^3 of actual particulate
0.5 + 0.31 = 0.81
1 - 0.81 = 0.19
0.19 cm^3 of air