The percentage air voids is given by,
\(n_a=\frac{V_a}{V}\)
V_a = V_v ? V_w = e ? e_w
And V = V_s+V_v = 1+e
? n_a=\(\frac{e-e_w}{1+e}\)
Since e_w = eS
\(n_a=e*\frac{1-S}{1+e}\).

The volume of voids, water and air are represented as V_v, V_W and V_a. The volume of solids is represented as V_d or V_S in phase diagram.

The water content is given by,
\(w=\frac{W_w}{W_d}\)
hence \(1+w=\frac{(W_w + W_d)}{W_d} = \frac{W}{W_d}\)
\(W_d=\frac{W}{(1+w)}\)
\(?_d=\frac{W_d}{V} = \frac{W}{(1+w)V}\)
? ?_d = \(\frac{?}{(1+w)}\).

The relationship between relative compaction R_c and density index I_D is given by,
R_c=80+0.2I_D
When I_D=0,
R_c=80+0.2*0
R_c=80%.

In a soil sample, some space is occupied by water and rest by air. The degree of saturation is defined as ratio of volume of water to total volume of voids.

Explanation:

Porosity (η) is defined as ratio of volume of voids to total volume of soil sample in a given soil mass.

$\eta =\frac{V_v}{V}$

The porosity of soils can vary widely.

The porosity of loose soils can be about η = 50 to 60%.

Given,
Weight of soil sample W=190kN
Weight after oven drying W_d=150kN
Weight of Water W_W=W-W_d
W_W=190-150
W_W=40kN.

Given,
Voids ratio e=0.6
Specific gravity G=2.7
Dry unit weight ?_d = (G* ?_w/1+e)
Unit weight of water ?_w=9.8 kN/m³
?_d = (2.7*9.81)/(1+0.6)
?_d = 16.5 kN/m³.

Given,
Porosity n= 22%= 0.22
Void ratio e = n/(1-n)
e= 0.22/(1-0.22)
e= 0.28

Given,
Dry unit weight ?_d = 16.25 kN/m³
Saturated unit weight ?_sat = 20 kN/m³
Degree of saturation S = 50% = 0.5
Bulk unit weight ? = ?_d + S[?_sat? ?_d] ? = 16.25 + 0.5[20-16.25] ? =18.123 kN/m³.