Let's break down the problem and solve step by step. --- ## **Step 1: Identify Key Data** - Maximum dimension, \( L = 1.7\, \text{m} \) - Particle diameter, \( d = 88\, \text{} = \times ^{-12}\ \text{m} \) - Temperature, \( T =12^\circtext{C \) - Knudsen, \( \{Kn} = 62. \) - Boltzmann constant, \( k_B = 1.3806 \times 10^{-23}\, \text{J K}^{-1} \) - **Answer in mPa (millipascal)** --- ## **Step 2: Convert Temperature to Kelvin** \[ T = 12 + 273.15 = 285.15\, \text{K} \] --- ## **Step 3: Knudsen Number Definition** \[ \text{Kn} = \frac{\lambda}{L} \] where \(\lambda\) is the mean free path. \[ \lambda = \text{Kn} \times L = 62. \times 1.7 = 105.4\, \text{m} \] --- ## **Step 4: Mean Free Path Formula** \[ \lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} \] Solve for pressure \(P\): \[ P = \frac{k_B T}{\sqrt{2} \pi d^2 \lambda} \] --- ## **Step 5: Substitute Known Values** - \(k_B = 1.3806 \times 10^{-23}\, \text{J K}^{-1}\) - \(T = 285.15\, \text{K}\) - \(d = 88 \times 10^{-12}\, \text{m}\) - \(\lambda = 105.4\, \text{m}\) - \(\sqrt{2} \approx 1.4142\) - \(\pi \approx 3.1416\) Calculate denominator: \[ \sqrt{2} \pi d^2 \lambda = 1.4142 \times 3.1416 \times (88 \times 10^{-12})^2 \times 105.4 \] First, \(d^2 = (88 \times 10^{-12})^2 = 7744 \times 10^{-24} = 7.744 \times 10^{-21}\) Now, substitute and multiply: \[ 1.4142 \times 3.1416 = 4.4429 \] \[ 4.4429 \times 7.744 \times 10^{-21} = 34.408 \times 10^{-21} \] \[ 34.408 \times 10^{-21} \times 105.4 = 3626.61 \times 10^{-21} \] \[ 3626.61 \times 10^{-21} = 3.62661 \times 10^{-18} \] --- ## **Step 6: Numerator** \[ k_B T = 1.3806 \times 10^{-23} \times 285.15 = 3.937 \times 10^{-21} \] --- ## **Step 7: Calculate Pressure \(P\)** \[ P = \frac{3.937 \times 10^{-21}}{3.62661 \times 10^{-18}} = 1.085 \times 10^{-3}\, \text{Pa} \] --- ## **Step 8: Convert to mPa (1 Pa = 100 mPa)** \[ P = 1.085 \times 10^{-3} \text{ Pa} \times 100 = 1.09\, \text{mPa} \] (Rounded to 3 significant figures.) --- ## **Final Answer** \[ \boxed{1.09\, \text{mPa}} \]