A soil with a dry bulk density of 1.27 g/cm3 has, at the time of sampling, 15% water by weight.

Question

Assuming a particle density of 2.65 g/cm3, what is the percentage of pore space in the soil that is NOT occupied by water?

Answer 1

You're given:

1. Dry bulk density: ρb = 1.27 g/cm³
2. Water content by weight: θw = 15%=0.15θ
3. Particle density: ρs = 2.65 g/cm³

Step 1: Calculate porosity (ϕ)

ϕ = 1− (ρb/ρs)

ϕ = 1−(1.27/2.65)

ϕ = 1−0.479

ϕ = 0.521

So, total pore space = 52.1%.

Step 2: Calculate volumetric water content (θv)

θv = θw × (ρb/ρw)

The density of water (pw) is approximately 1 g/cm³ (or 1000 kg/m³) at standard temperature and

pressure, which is usually taken as around 4°C, where water is densest.

1. This is a well-known physical constant used in soil science, hydrology, and chemistry.
2. It allows you to convert between mass and volume because 1 gram of water occupies 1 cubic centimeter (cm³).
3. In soil calculations like volumetric water content, we assume ρw=1 g/cm³ρw​=1g/cm³ unless specified otherwise.

θv = 0.15 × (1.27/1)

θv = 0.1905 = 19.1%

So, total soil volume occupied by water is 19.1%

Step 3: Calculate pore space NOT occupied by water

air-filled porosity = ϕ−θv = 0.521− 0.191= 0.33

Final answer: About 33% of the soil’s pore space is NOT occupied by water.

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