What will be the porosity when a soil has its bulk density and particle density of \(1.50\ Mg/m^3\) and \(2.65\ Mg/m^3\), respectively?

Question

Which of the following statements is false?

Answer 1

Porosity is the empty space left in a soil sample after the solid grains are packed together. If we take a fixed total volume of soil and imagine pulling out just the solid particles and pressing them together with no gaps, the volume they would occupy comes from dividing bulk density by particle density.

Take exactly \(1\ m^3\) of soil as it sits in the field. Its bulk density is \(1.50\ Mg/m^3\), so this cubic metre of soil, solids plus pores together, weighs 1.50 Mg. Now ask: what volume would 1.50 Mg of pure solid material occupy, with no pores at all, given a particle density of \(2.65\ Mg/m^3\)?

\[ \text{Volume of solids} = \frac{\text{mass}}{\text{particle density}} = \frac{1.50}{2.65} = 0.566\ m^3 \]

So out of the 1 cubic metre we started with, only 0.566 cubic metre is actually solid soil grains. The rest of the volume has to be pore space, since the total we started with was exactly 1 cubic metre:

\[ \text{Volume of pores} = 1 - 0.566 = 0.434\ m^3 \]

Since the total volume taken was exactly 1 cubic metre, this pore volume of 0.434 cubic metre is also directly the fraction of pore space, which as a percentage is 43.4%.

Let's summarize:

Dividing bulk density by particle density gives the solid fraction of the soil volume, here 0.566 or 56.6%.
Whatever is left after removing the solid fraction from the whole is the pore space, here 43.4%.
The other listed values do not match this direct calculation and come from rounding errors.

So the porosity of this soil comes to 43.4%, matching option 2.

Answer 2

Porosity is the empty space left in a soil sample after the solid grains are packed together. If we take a fixed total volume of soil and imagine pulling out just the solid particles and pressing them together with no gaps, the volume they would occupy comes from dividing bulk density by particle density.

Take exactly \(1\ m^3\) of soil as it sits in the field. Its bulk density is \(1.50\ Mg/m^3\), so this cubic metre of soil, solids plus pores together, weighs 1.50 Mg. Now ask: what volume would 1.50 Mg of pure solid material occupy, with no pores at all, given a particle density of \(2.65\ Mg/m^3\)?

\[ \text{Volume of solids} = \frac{\text{mass}}{\text{particle density}} = \frac{1.50}{2.65} = 0.566\ m^3 \]

So out of the 1 cubic metre we started with, only 0.566 cubic metre is actually solid soil grains. The rest of the volume has to be pore space, since the total we started with was exactly 1 cubic metre:

\[ \text{Volume of pores} = 1 - 0.566 = 0.434\ m^3 \]

Since the total volume taken was exactly 1 cubic metre, this pore volume of 0.434 cubic metre is also directly the fraction of pore space, which as a percentage is 43.4%.

Let's summarize:

Dividing bulk density by particle density gives the solid fraction of the soil volume, here 0.566 or 56.6%.
Whatever is left after removing the solid fraction from the whole is the pore space, here 43.4%.
The other listed values do not match this direct calculation and come from rounding errors.

So the porosity of this soil comes to 43.4%, matching option 2.

What will be the porosity when a soil has its bulk density and particle density of \(1.50\ Mg/m^3\) and \(2.65\ Mg/m^3\), respectively?

Show Hint

The Correct Option is B

Solution and Explanation

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