Question

A sample of moist soil with a wet mass of 1.0 kg and a volume of 0.64 liters (6.4 $\times$ 10$^{-4}$ m$^3$) was dried in the oven and found to have a dry mass of 0.8 kg. Assuming the typical value of particle density (2650 kg/m$^3$) for a mineral soil, what will be the volumetric moisture content?

• A. 37.5 %
• B. 22.5 %
• C. 31.25 %
• D. 25 %

Hint:

Using liters for both water volume and total soil volume simplifies the calculation.
Keep in mind: \( 1 \text{ kg} \) of water has a volume of exactly \( 1 \text{ liter} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Volumetric moisture content ($\theta$) is defined as the ratio of the volume of water ($V_w$) in a soil sample to the total bulk volume of the soil sample ($V_t$).
Key Formula or Approach:
1. Mass of water ($M_w$) = $\text{Wet mass } (M_t) - \text{Dry mass } (M_s)$
2. Volume of water ($V_w$) = $\frac{M_w}{\rho_w}$, where $\rho_w$ is the density of water ($1.0 \text{ kg/liter}$ or $1000 \text{ kg/m}^3$)
3. Volumetric moisture content ($\theta$) = $\frac{V_w}{V_t} \times 100\%$

Step 2: Detailed Explanation:
Given values:
- Wet mass of soil ($M_t$) = $1.0 \text{ kg}$
- Dry mass of soil ($M_s$) = $0.8 \text{ kg}$
- Total volume of soil ($V_t$) = $0.64 \text{ liters}$
- Density of water ($\rho_w$) = $1.0 \text{ kg/liter}$
Calculate Mass of water ($M_w$):
\[ M_w = 1.0 \text{ kg} - 0.8 \text{ kg} = 0.2 \text{ kg} \] Calculate Volume of water ($V_w$):
\[ V_w = \frac{0.2 \text{ kg}}{1.0 \text{ kg/liter}} = 0.2 \text{ liters} \] Calculate Volumetric Moisture Content ($\theta$):
\[ \theta = \frac{0.2 \text{ liters}}{0.64 \text{ liters}} \times 100\% = 0.3125 \times 100\% = 31.25\% \]

Step 3: Final Answer:
The volumetric moisture content of the soil sample is 31.25%, corresponding to option (C).