Question

Modern vacuum pumps can evacuate a vessel down to a pressure of \(4.0 \times 10^{-15}\,\text{atm}\). At room temperature (300 K), taking \(R = 8.3\,\text{J K}^{-1}\text{mol}^{-1}\) and \(N_{\text{Avogadro}} = 6 \times 10^{23}\,\text{mol}^{-1}\), the mean distance between molecules of gas in an evacuated vessel will be of the order of:

• A. \(0.2\,\mu\text{m}\)
• B. \(0.2\,\text{mm}\)
• C. \(0.2\,\text{cm}\)
• D. \(0.2\,\text{nm}\)

Hint:

For very low-pressure gases: \[ \text{Mean separation} \sim \left(\frac{1}{\text{number density}}\right)^{1/3} \] Extremely low pressure implies extremely large intermolecular distances.

Answer 1

The Correct Option is B

Step 1: Convert pressure into SI units. \[ P = 4.0 \times 10^{-15}\,\text{atm} \] Since, \[ 1\,\text{atm} = 1.0 \times 10^{5}\,\text{Pa} \] \[ P = 4.0 \times 10^{-15} \times 10^{5} = 4.0 \times 10^{-10}\,\text{Pa} \]
Step 2: Use the ideal gas equation to find number of moles per unit volume. \[ PV = nRT \Rightarrow \frac{n}{V} = \frac{P}{RT} \] \[ \frac{n}{V} = \frac{4.0 \times 10^{-10}}{8.3 \times 300} \approx 1.6 \times 10^{-13}\,\text{mol m}^{-3} \]
Step 3: Convert moles into number of molecules per unit volume. \[ \text{Number density} = \frac{n}{V} \times N_A \] \[ = 1.6 \times 10^{-13} \times 6 \times 10^{23} \approx 9.6 \times 10^{10}\,\text{molecules m}^{-3} \]
Step 4: Find the volume available per molecule. \[ \text{Volume per molecule} \approx \frac{1}{9.6 \times 10^{10}} \approx 1.0 \times 10^{-11}\,\text{m}^3 \]
Step 5: Estimate the mean distance between molecules. The mean separation is approximately the cube root of volume per molecule: \[ d \approx (10^{-11})^{1/3} \approx 2 \times 10^{-4}\,\text{m} \] \[ d \approx 0.2\,\text{mm} \]
Hence, the correct answer is \(\boxed{0.2\,\text{mm}}\).