Let's break down the problem and solve it **step by step**. --- ## **Step 1: List the Given Data** - **Tube diameter:** \( d = 3\, \text{mm} = 0.003\, \text{m} \) - **Initial height of toluene below top:** \( z_1 = 1.9\, \text{cm} = 0.019\, \text{m} \) - **Final height of toluene below top:** \( z_2 = 7.9\, \text{cm} = 0.079\, \text{m} \) - **Time:** \( t = 9.6 \times 10^4\, \text{s} \) - **Density of toluene:** \( \rho = 0.852\, \text{g/cm}^3 = 852\, \text{kg/m}^3 \) - **Vapor pressure of toluene at \( 39.4^\circ C \):** \( P_A^* = 57.3\, \text{torr} \) - **Barometric pressure:** \( P = 1\, \text{atm} = 760\, \text{torr} \) - **Neglect counterdiffusion of air** --- ## **Step 2: Sketch and Model** The toluene is evaporating from a vertical tube, diffusing into stagnant air (since counterdiffusion is neglected). The setup is similar to the **Stefan tube experiment**. - **Top of tube (z = 0):** Toluene vapor partial pressure is 0 (since it's open to pure air). - **Liquid surface (z = L):** Toluene vapor partial pressure is at equilibrium (\( P_A^* \)). --- ## **Step 3: The Stefan Tube Equation (Unidirectional Diffusion)** The rate of evaporation (moles per unit time) is: \[ N_A = - D_{AB} \frac{C}{L} \ln \left(\frac{1 - y_{A,2}}{1 - y_{A,1}}\right) \cdot A \] Where: - \( D_{AB} \) = diffusivity of toluene vapor in air (unknown) - \( C \) = total molar concentration of gas (\( \text{mol/m}^3 \)) - \( L \) = length of diffusion path (\( z_2 - z_1 \)), but since the liquid level drops, integration is required - \( y_{A,1}, y_{A,2} \) = mole fraction of toluene vapor at z=0 (top) and z=L (liquid surface) - \( A \) = cross-sectional area of tube But since the **length changes as the liquid drops** (from \( z_1 \) to \( z_2 \)), we need to integrate. --- ## **Step 4: Express the Rate of Drop of Liquid Level** The *rate of change of the liquid level* (\( dz/dt \)) is related to the molar flux (\( N_A \)) by: \[ N_A = \frac{1}{A} \frac{dn_A}{dt} \] But: - The rate at which the liquid level drops (\( dz/dt \)) relates to the volume change: \( dV = -A dz \) - The moles vaporized: \( dn_A = \frac{\rho}{M} (-A dz) \), where \( M \) is the molar mass of toluene (\( M = 92.13\, \text{g/mol} \)) So: \[ N_A = \frac{\rho}{M} \frac{dz}{dt} \] \[ \Rightarrow \frac{dz}{dt} = \frac{M}{\rho} N_A \] --- ## **Step 5: Stefan's Law for Unidirectional Diffusion** For *steady-state, unidirectional diffusion*, with the tube open to air at the top (\( y_{A,1} = 0 \)) and saturated at the liquid (\( y_{A,2} = y_{A}^* \)), and length \( z \): \[ N_A = \frac{D_{AB} C}{z} \ln \left( \frac{1}{1 - y_{A}^*} \right) \] --- ## **Step 6: Set Up and Integrate for Changing Length** Since z increases as the liquid level drops, we need to integrate: \[ dz = \frac{M}{\rho} N_A dt \] But \( N_A \) depends on \( z \): \[ N_A = \frac{D_{AB} C}{z} \ln \left( \frac{1}{1 - y_{A}^*} \right) \] So: \[ dz = \frac{M}{\rho} \cdot \frac{D_{AB} C}{z} \ln \left( \frac{1}{1 - y_{A}^*} \right) dt \] Separate variables: \[ z\, dz = \frac{M}{\rho} D_{AB} C \ln \left( \frac{1}{1 - y_{A}^*} \right) dt \] Integrate both sides: \[ \int_{z_1}^{z_2} z\,dz = \frac{M}{\rho} D_{AB} C \ln \left( \frac{1}{1 - y_{A}^*} \right) \int_0^{t} dt \] \[ \left. \frac{z^2}{2} \right|_{z_1}^{z_2} = \frac{M}{\rho} D_{AB} C \ln \left( \frac{1}{1 - y_{A}^*} \right) t \] \[ \frac{z_2^2 - z_1^2}{2} = \frac{M}{\rho} D_{AB} C \ln \left( \frac{1}{1 - y_{A}^*} \right) t \] Solve for \( D_{AB} \): \[ D_{AB} = \frac{z_2^2 - z_1^2}{2 t} \cdot \frac{\rho}{M} \cdot \frac{1}{C \ln \left( \frac{1}{1 - y_{A}^*} \right)} \] --- ## **Step 7: Calculate All Quantities** ### **a) \( z_1 \) and \( z_2 \) in meters** Already converted: - \( z_1 = 0.019\, \text{m} \) - \( z_2 = 0.079\, \text{m} \) \[ z_2^2 - z_1^2 = (0.079)^2 - (0.019)^2 = 0.006241 - 0.000361 = 0.00588\, \text{m}^2 \] --- ### **b) \( t \) in seconds** Given: \( t = 9.6 \times 10^4\, \text{s} \) --- ### **c) \( \rho \) and \( M \) (density and molar mass)** - \( \rho = 0.852\, \text{g/cm}^3 = 852\, \text{kg/m}^3 \) - \( M = 92.13\, \text{g/mol} = 0.09213\, \text{kg/mol} \) --- ### **d) Total molar concentration \( C \) at \( 39.4^\circ C \) and 1 atm** \[ T = 39.4^\circ C = 312.55~\text{K} \] \[ P = 1~\text{atm} = 101325~\text{Pa} \] \[ C = \frac{P}{RT} \] Where \( R = 8.205 \times 10^{-2}~\text{L atm K}^{-1} \text{mol}^{-1} = 0.08206~\text{L atm K}^{-1} \text{mol}^{-1} \) But in SI units: - \( R = 8.314~\text{J/mol K} \) So, \[ C = \frac{P}{RT} = \frac{101325}{8.314 \times 312.55} \] \[ = \frac{101325}{2597.41} \] \[ = 39.0~\text{mol/m}^3 \] --- ### **e) \( y_A^* \) (mole fraction of saturated toluene vapor)** \[ y_A^* = \frac{P_A^*}{P} \] - \( P_A^* = 57.3~\text{torr} \) - \( P = 760~\text{torr} \) \[ y_A^* = \frac{57.3}{760} = 0.0754 \] --- ### **f) \( \ln \left(\frac{1}{1 - y_A^*}\right) \)** \[ 1 - y_A^* = 1 - 0.0754 = 0.9246 \] \[ \ln \left(\frac{1}{0.9246}\right) = \ln(1.0815) = 0.0783 \] --- ## **Step 8: Plug In All Values** \[ D_{AB} = \frac{0.00588}{2 \times 9.6 \times 10^4} \cdot \frac{852}{0.09213} \cdot \frac{1}{39.0 \times 0.0783} \] Let's compute each term step by step: ### **a) \( \frac{0.00588}{2 \times 9.6 \times 10^4} \)** \[ 2 \times 9.6 \times 10^4 = 192,000 \] \[ \frac{0.00588}{192,000} = 3.0625 \times 10^{-8} \] ### **b) \( \frac{852}{0.09213} \)** \[ \frac{852}{0.09213} = 9247.3 \] ### **c) \( 39.0 \times 0.0783 = 3.0537 \)** \[ \frac{1}{3.0537} = 0.3276 \] --- Now multiply all: \[ D_{AB} = (3.0625 \times 10^{-8}) \cdot 9247.3 \cdot 0.3276 \] First, \( 3.0625 \times 10^{-8} \times 9247.3 = 2.833 \times 10^{-4} \) Then, \( 2.833 \times 10^{-4} \times 0.3276 = 9.278 \times 10^{-5} \) So, \[ \boxed{D_{AB} = 9.3 \times 10^{-5}~\text{m}^2/\text{s}} \] --- ## **Step 9: **Final Answer (with Explanation)** ### **Final Answer:** \[ \boxed{ D_{AB} = 9.3 \times 10^{-5}~\text{m}^2/\text{s} } \] ### **Explanation:** The diffusivity of toluene vapor in air at \( 39.4^\circ C \) and 1 atm, as determined from the experimental data using Stefan's Law for unidirectional diffusion, is **\( 9.3 \times 10^{-5}~\text{m}^2/\text{s} \)**. This calculation assumes negligible counterdiffusion of air and uses the observed drop in liquid level, vapor pressure, and tube dimensions to solve for the diffusivity. --- **If you need the answer in cm²/s:** \( 1~\text{m}^2/\text{s} = 10^4~\text{cm}^2/\text{s} \) So \( D_{AB} = 0.93~\text{cm}^2/\text{s} \). --- **Boxed Final Answer:** \[ \boxed{ D_{\text{toluene, air}} = 9.3 \times 10^{-5}~\text{m}^2/\text{s} = 0.93~\text{cm}^2/\text{s} } \]

Let's carefully work through the problem step by step to find the diffusivity of toluene in air. --- ### **Given Data:** | Parameter | Value | |------------------------------|------------------------------| | Tube diameter, \(d\) | 3 mm = 0.003 m | | Cross-sectional area, \(A\) | \(\pi d^2/4 = \pi \times (0.003)^2/4 \approx 7.07 \times 10^{-6}\, m^2\) | | Initial level below top, \(z_1\) | 1.9 cm = 0.019 m | | Final level below top, \(z_2\) | 7.9 cm = 0.079 m | | Time for level drop, \(t\) | 96 × 10^4 s = 96,000 s | | Density of toluene, \(\rho\) | 0.852 g/cm³ = 852 kg/m³ | | Vapor pressure of toluene, \(P_A^*\) | 57.3 torr | | Barometric pressure, \(P_{total}\) | 1 atm = 760 torr | | Temperature, \(T\) | 39.4°C = 312.55 K | | Gas constant, \(R\) | 8.314 J/(mol·K) | --- ### **Step 1: Determine the vapor mole fraction at saturation** \[ y_A^* = \frac{P_A^*}{P_{total}} = \frac{57.3}{760} \approx 0.0754 \] --- ### **Step 2: Calculate total molar concentration of air** \[ C_{air} = \frac{P_{total}}{RT} = \frac{101325\, Pa}{8.314\, J/(mol·K) \times 312.55\, K} \approx 39.1\, mol/m^3 \] --- ### **Step 3: Relate the change in liquid level to molar flux** The amount of toluene vaporized per unit time relates to the drop in liquid level: \[ \frac{dn_A}{dt} = - \frac{\rho}{M} \times A \times \frac{dz}{dt} \] where \(M = 92.13\, g/mol = 0.09213\, kg/mol\). Rearranged: \[ \frac{dz}{dt} = - \frac{M}{\rho} \times \frac{1}{A} \times \frac{dn_A}{dt} \] --- ### **Step 4: Use Stefan's law for unidirectional diffusion** The molar flux of vapor at steady state: \[ N_A = - D_{AB} \times \frac{C_{total}}{z} \times \ln\left(\frac{1}{1 - y_A^*}\right) \] Because the vapor diffuses from saturated at the surface to zero at the top (open to air): \[ N_A = \frac{1}{A} \times \frac{dn_A}{dt} \] Combine: \[ \frac{dn_A}{dt} = - D_{AB} \times C_{total} \times \frac{A}{z} \times \ln\left(\frac{1}{1 - y_A^*}\right) \] --- ### **Step 5: Derive an integral expression** As the liquid level drops from \(z_1\) to \(z_2\), integrate over time: \[ \int_{z_1}^{z_2} z\, dz = \frac{M}{\rho} D_{AB} C_{total} \times \ln\left(\frac{1}{1 - y_A^*}\right) \times t \] which simplifies to: \[ \frac{z_2^2 - z_1^2}{2} = \frac{M}{\rho} D_{AB} C_{total} \times \ln\left(\frac{1}{1 - y_A^*}\right) \times t \] Solve for \(D_{AB}\): \[ D_{AB} = \frac{z_2^2 - z_1^2}{2 t} \times \frac{\rho}{M} \times \frac{1}{C_{total} \times \ln\left(\frac{1}{1 - y_A^*}\right)} \] --- ### **Step 6: Plug in the values** - \(z_1 = 0.019\, m\) - \(z_2 = 0.079\, m\) Calculate numerator: \[ z_2^2 - z_1^2 = (0.079)^2 - (0.019)^2 = 0.006241 - 0.000361 = 0.00588\, m^2 \] - \(t = 96,000\, s\) - \(\frac{\rho}{M} = \frac{852\, kg/m^3}{0.09213\, kg/mol} \approx 9247\, mol/m^3\) - \(C_{total} = 39.1\, mol/m^3\) - \(\ln\left(\frac{1}{1 - y_A^*}\right) = \ln\left(\frac{1}{0.9246}\right) \approx 0.0783\) --- ### **Step 7: Calculate \(D_{AB}\)** \[ D_{AB} = \frac{0.00588}{2 \times 96,000} \times 9247 \times \frac{1}{39.1 \times 0.0783} \] Calculate step-by-step: - \(\frac{0.00588}{192,000} = 3.0625 \times 10^{-8}\) - \(9247 / 39.1 \approx 236.4\) - \(39.1 \times 0.0783 \approx 3.062\) Now: \[ D_{AB} = 3.0625 \times 10^{-8} \times 236.4 \times \frac{1}{3.062} \] \[ D_{AB} = 3.0625 \times 10^{-8} \times 77.2 \approx 2.36 \times 10^{-6}\, \text{m}^2/\text{s} \] --- ### **### **Final Result:** \[ \boxed{ \text{Diffusivity } D_{AB} \approx 2.36 \times 10^{-6}\, \text{m}^2/\text{s} } \] ### **Note:** This value is within the typical range for vapor diffusivities in air, confirming the reasonableness. --- ## **Final answer:** **The diffusivity of toluene in air is approximately \(\boxed{2.4 \times 10^{-6}\, \text{m}^2/\text{s}}\).**