# Step-by-Step Solution: Specific Volume from Specific Gravity ## 1. **Understanding the Concepts** - **Specific Gravity (SG):** \(\text{SG}\) is the ratio of the density of a substance to the density of water at \(4^\circ \text{C}\) (which is \(100~\text{kg}/\text{m}^3\)): \[ \text{SG} = \frac{\rho_{\text{liquid}}}{\rho_{\text{water}}} \] - **Specific Volume (\(v\)):** This is the volume occupied by \(1\) kg of a substance: \[ v = \frac{1}{\rho} \] where \(v\) is in \(\text{m}^3/\text{kg}\), and \(\rho\) is the density in \(\text{kg}/\text{m}^3\). ## 2. **Given Data** - Specific Gravity, \(\text{SG} = .585\) - Density of water at \(4^\circ \text{C}\), \(\rho_{\text{water}} = 100~\text{kg}/\text{m}^3\) ## 3. **Find the Density of the Liquid** \[ \rho_{\text{liquid}} = \text{SG} \times \rho_{\text{water}} \] \[ \rho_{\text{liquid}} = .585 \times 100~\text{kg}/\text{m}^3 = 585~\text{kg}/\text{m}^3 \] ## 4. **Calculate the Specific Volume** \[ v = \frac{1}{\rho_{\text{liquid}}} \] \[ v = \frac{1}{585}~\text{m}^3/\text{kg} \] \[ v \approx .001709~\text{m}^3/\text{kg} \] ## 5. **Final Answer** **The specific volume of the liquid is:** \[ \boxed{.00171~\text{m}^3/\text{kg}} \] (Rounded to 5 significant figures) --- **Summary Table:** | Property | Value | |------------------|-------------------| | Specific Gravity | .585 | | Density | \(585~\text{kg}/\text{m}^3\) | | Specific Volume | \(.00171~\text{m}^3/\text{kg}\) | --- **Key Formulae Used:** - \(\rho = \text{SG} \times 100~\text{kg}/\text{m}^3\) - \(v = \frac{1}{\rho}\)