Question

Consider a chamber at room temperature (27 \(^\circ\)C) filled with a gas having a molecular diameter of 0.35 nm. The pressure (in Pascal) to which the chamber needs to be evacuated so that the molecules have a mean free path of 1 km is \rule{1cm{0.15mm} \(\times 10^{-5}\) Pa. (up to two decimal places)
(Boltzmann constant \(k_B = 1.38 \times 10^{-23}\) J/K)}

Hint:

This problem illustrates the relationship between macroscopic properties (pressure, temperature) and microscopic properties (mean free path, molecular size). Note the inverse relationship between mean free path and pressure: to get a longer mean free path, you need to lower the pressure. This is the principle behind vacuum systems.

Answer 1

Correct Answer: 0.76

Step 1: Understanding the Concept:
The mean free path (\(\lambda\)) is the average distance a gas molecule travels between successive collisions. It depends on the size of the molecules (diameter \(d\)) and their number density (\(n\)). The number density, in turn, is related to pressure and temperature by the ideal gas law. To achieve a very long mean free path, the pressure must be very low (a high vacuum).
Step 2: Key Formula or Approach:
The mean free path is given by: \[ \lambda = \frac{1}{\sqrt{2} \pi d^2 n} \] where \(d\) is the molecular diameter and \(n\) is the number density. The number density is related to pressure \(P\) and temperature \(T\) by the ideal gas law in the form \(P = n k_B T\), which gives \(n = \frac{P}{k_B T}\). Substituting \(n\) into the mean free path formula allows us to solve for the pressure \(P\).
Step 3: Detailed Explanation:
Substitute \(n\) into the \(\lambda\) formula: \[ \lambda = \frac{1}{\sqrt{2} \pi d^2 (P/k_B T)} = \frac{k_B T}{\sqrt{2} \pi d^2 P} \] Rearrange to solve for the pressure \(P\): \[ P = \frac{k_B T}{\sqrt{2} \pi d^2 \lambda} \] We are given the following values: Temperature, \(T = 27 ^\circ\text{C} = 27 + 273 = 300\) K. Molecular diameter, \(d = 0.35 \text{ nm} = 0.35 \times 10^{-9}\) m. Mean free path, \(\lambda = 1 \text{ km} = 1000\) m. Boltzmann constant, \(k_B = 1.38 \times 10^{-23}\) J/K. Substitute these values into the expression for \(P\): \[ P = \frac{(1.38 \times 10^{-23} \text{ J/K}) \times (300 \text{ K})}{\sqrt{2} \pi (0.35 \times 10^{-9} \text{ m})^2 (1000 \text{ m})} \] \[ P = \frac{4.14 \times 10^{-21}}{\sqrt{2} \pi (0.1225 \times 10^{-18}) (1000)} \text{ Pa} \] \[ P = \frac{4.14 \times 10^{-21}}{1.414 \times 3.1416 \times 0.1225 \times 10^{-15}} \text{ Pa} \] \[ P \approx \frac{4.14 \times 10^{-21}}{0.544 \times 10^{-15}} \text{ Pa} \approx 7.608 \times 10^{-6} \text{ Pa} \] The question asks for the answer in the form of \(_________ \times 10^{-5}\) Pa. \[ 7.608 \times 10^{-6} = 0.7608 \times 10^{-5} \] Step 4: Final Answer:
The pressure needs to be evacuated to \(0.76 \times 10^{-5}\) Pa.