Let's break down the problem and solve it step by step. --- ## **Step 1: Given Data** - **Water Density (\(\rho_{water}\))**: .9982 g/cm³ = 998.2 kg/m³ - **CCl₄ Density (\(\rho_{CCl4}\))**: 1.594 g/cm³ = 1594 kg/m³ - **Capillary Diameter (\(d\))**: .010 in = .010 × 2.54 cm = .0254 cm = 2.54 × 10⁻⁴ m - **Inclined Tube Length (\(L\))**: 120 in = 304.8 cm = 3.048 m - **Inclined Height Difference (\(h\))**: \(L \sin\theta = h = 1\) in = 2.54 cm = .0254 m - **Height difference in manometer (\(h_{man}\))**: 1. in = .0254 m - **Gravity (\(g\))**: 9.81 m/s² --- ## **Step 2: Find Pressure Difference (\(\Delta P\))** The manometer measures the pressure drop along the capillary length: \[ \Delta P = (\rho_{man} - \rho_{water}) g h_{man} \] Where: - \(\rho_{man} = 1594 \text{ kg/m}^3\) - \(\rho_{water} = 998.2 \text{ kg/m}^3\) - \(h_{man} = .0254 \text{ m}\) \[ \Delta P = (1594 - 998.2) \times 9.81 \times .0254 \] \[ \Delta P = 595.8 \times 9.81 \times .0254 \] \[ \Delta P = 595.8 \times .249174 \] \[ \Delta P \approx 148.5 \text{ Pa} \] --- ## **Step 3: Calculate Velocity in Capillary (\(v\))** For laminar flow in a capillary (using Hagen-Poiseuille’s law): \[ Q = \frac{\pi d^4}{128 \mu L} \Delta P \] Where: - \(Q\) is volumetric flow rate (m³/s) - \(d\) is diameter (m) - \(\mu\) is viscosity of water at 20°C (\(= 1.002 \times 10^{-3}\) Pa·s) - \(L\) is capillary length (m) - \(\Delta P\) is pressure difference (Pa) Plug in values: - \(d = 2.54 \times 10^{-4}\) m - \(\mu = 1.002 \times 10^{-3}\) Pa·s - \(L = 3.048\) m - \(\Delta P = 148.5\) Pa \[ Q = \frac{\pi (2.54 \times 10^{-4})^4}{128 \times 1.002 \times 10^{-3} \times 3.048} \cdot 148.5 \] First, calculate \(d^4\): \[ d^4 = (2.54 \times 10^{-4})^4 = (2.54^4) \times 10^{-16} = 41.62 \times 10^{-16} = 4.162 \times 10^{-15} \] Now plug values in: \[ Q = \frac{\pi \times 4.162 \times 10^{-15}}{128 \times 1.002 \times 10^{-3} \times 3.048} \cdot 148.5 \] Calculate denominator: \[ 128 \times 1.002 \times 10^{-3} \times 3.048 = 128 \times .001002 \times 3.048 = .128256 \times 3.048 = .390875 \] Now plug in: \[ Q = \frac{3.1416 \times 4.162 \times 10^{-15}}{.390875} \cdot 148.5 \] \[ = \frac{13.075 \times 10^{-15}}{.390875} \cdot 148.5 = 3.346 \times 10^{-14} \cdot 148.5 = 4.97 \times 10^{-12} \text{ m}^3/\text{s} \] --- ## **Step 4: Calculate Average Velocity (\(v\))** \[ v = \frac{Q}{A} \] Where \(A = \frac{\pi d^2}{4}\) \[ A = \frac{\pi (2.54 \times 10^{-4})^2}{4} = \frac{3.1416 \times 6.4516 \times 10^{-8}}{4} = \frac{2.026 \times 10^{-7}}{4} = 5.065 \times 10^{-8} \text{ m}^2 \] \[ v = \frac{4.97 \times 10^{-12}}{5.065 \times 10^{-8}} = 9.82 \times 10^{-5} \text{ m/s} \] --- ## **Final Answers** ### **1. Velocity of Flow (\(v\))** \[ \boxed{9.8 \times 10^{-5} \text{ m/s}} \] ### **2. Volumetric Flow Rate (\(Q\))** \[ \boxed{5. \times 10^{-12} \text{ m}^3/\text{s}} \] --- ### **Summary Table** | Quantity | Value | |---------------|----------------------------| | Velocity (\(v\)) | \(9.8 \times 10^{-5}\) m/s | | Vol. Flow (\(Q\)) | \(5. \times 10^{-12}\) m³/s | --- **Let me know if you want to see the step-by-step in a different unit (e.g., cm³/s or mL/min)!**