Soil Mechanics – Moisture Content from Pycnometer with Wet Soil and Known Specific Gravity A 260 g sample of wet soil is placed in a pycnometer jar (empty pycnometer weight = 400 g) to determine the water (moisture) content. The specific gravity of soil solids is 2.75. The combined weight of the pycnometer with the wet soil and water added to remove all air is 1415 g. The weight of the pycnometer filled with water only is 1275 g. Calculate the moisture content of the soil sample (in %).

Difficulty: Medium

Correct Answer: 18.2%

Explanation:


Introduction / Context:
Determining water content from pycnometer readings requires careful volume–mass bookkeeping. When a wet soil is used in the pycnometer and the specific gravity of solids (Gs) is known, we can split the wet mass into dry soil and soil water by using displacement concepts and the known capacity of the pycnometer.


Given Data / Assumptions:

  • Empty pycnometer (M1) = 400 g.
  • Wet soil mass (W) = 260 g, composed of dry mass Wd and water Ww.
  • Pycnometer + wet soil + added water (no air) (M3) = 1415 g.
  • Pycnometer + water only (M4) = 1275 g.
  • Specific gravity of solids Gs = 2.75; density of water taken as 1 g/cm^3.


Concept / Approach:

When the wet soil is placed in the pycnometer and then topped up with water, the added water occupies the remaining volume. The difference between the water-only fill and the added water gives the volume of the wet soil (solids plus soil water). Using Gs, we relate dry mass to volume of solids. Two equations in unknowns Wd and Ww then yield the moisture content w = Ww / Wd.


Step-by-Step Solution:

Capacity (water-only) = M4 − M1 = 1275 − 400 = 875 g ≡ 875 cm^3.Added water with wet soil = M3 − M1 − W = 1415 − 400 − 260 = 755 g ≡ 755 cm^3.Volume of wet soil in pycnometer = 875 − 755 = 120 cm^3.Let W = Wd + Ww = 260 g and V_wet = V_s + V_w = (Wd / Gs) + Ww = 120 (since rho_w = 1).Solve: Ww = 260 − Wd. So Wd/2.75 + (260 − Wd) = 120 → 260 − 0.63636*Wd = 120 → Wd ≈ 220 g.Therefore Ww = 260 − 220 = 40 g; moisture content w = Ww / Wd = 40 / 220 = 0.1818 ≈ 18.2%.


Verification / Alternative check:

Check volumes: V_s = 220 / 2.75 = 80 cm^3; V_w (soil) = 40 cm^3; total 80 + 40 = 120 cm^3, consistent with the capacity difference method. Hence the calculation is self-consistent.


Why Other Options Are Wrong:

24.2% and 53.8% result from incorrect separation of wet mass or misuse of Gs. 'None of these' is incorrect because a consistent solution gives 18.2%.


Common Pitfalls:

Forgetting that water density equals 1 g/cm^3 (so grams convert to cm^3); mixing up added water with total water; using formulas meant for dry-soil pycnometer readings directly on wet-soil data without adjusting volumes.


Final Answer:

18.2%

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