Question

The bulk density and water content of a soil are \(1800\ \text{kg m}^{-3}\) and \(18%\), respectively. Find the dry density of the soil. \([\,\text{round off to 2 decimal places}\,]\)

Hint:

Memorize: \(\rho_d=\rho_b/(1+w)\) with \(w\) as a decimal. If given as a percent, divide by 100 first.

Answer 1

Step 1: Define symbols and basic relations.
Let \\(M_s\\) = mass of dry solids, \\(M_w\\) = mass of pore water, \\(V\\) = total volume of the specimen (assumed unchanged between bulk and dry states). \\[ \\text{Bulk density: }\\; \\rho_b = \\frac{M_s + M_w}{V}, \\qquad \\text{Dry density: }\\; \\rho_d = \\frac{M_s}{V}. \\] Water content (gravimetric): \\(w = \\dfrac{M_w}{M_s} = 18\\% = 0.18 \\; \\Rightarrow \\; M_w = 0.18 M_s\\).

Step 2: Express \\(\\rho_b\\) in terms of \\(\\rho_d\\) and \\(w\\).
\\[ \\rho_b = \\frac{M_s + M_w}{V} = \\frac{M_s(1+w)}{V} = (1+w) \\frac{M_s}{V} = (1+w) \\rho_d. \\] Hence the fundamental conversion: \\[ \\boxed{\\rho_d = \\frac{\\rho_b}{1+w}} \\]

Step 3: Substitute numerical values with units.
\\[ \\rho_d = \\frac{1800\\; \\text{kg m}^{-3}}{1+0.18} = \\frac{1800}{1.18}\\; \\text{kg m}^{-3} = 1525.4237\\ldots\\; \\text{kg m}^{-3}. \\]

Step 4: Rounding and reasonableness check.
– Rounding to 2 decimals: \\(1525.42\\; \\text{kg m}^{-3}\\).
– Sanity: since \\(w>0\\), \\(\\rho_d < \\rho_b\\). Indeed, \\(1525.42 < 1800\\) ✓.
– Extreme checks: if \\(w \\to 0 \\; \\Rightarrow \\; \\rho_d \\to \\rho_b\\) (consistent); if \\(w \\to 100\\% \\; \\Rightarrow \\; \\rho_d = \\rho_b / 2\\) (trend consistent).

Final Answer: \\(\\boxed{1525.42\\; \\text{kg m}^{-3}}\\)