Question

The mean free path of the molecules of a gas at 25°C is \(2.63 \times 10^{-5}\) meter. If the radius of the molecule is \(2.56 \times 10^{-10}\) meter, find the pressure of the gas. [\(k = 1.38 \times 10^{-23}\) Joule/K]

• A. 1 mm of mercury
• B. 10 mm of mercury
• C. 20 mm of mercury
• D. 50 mm of mercury

Hint:

Always ensure all your units are in the standard SI system (meters, Kelvin, Pascals, etc.) before plugging them into physics formulas. After calculating the result, convert it to the units required by the options. Remember the conversion: 760 mmHg \(\approx 10^5\) Pa.

Answer 1

The Correct Option is B

Step 1: Define the mean free path formula (\(\lambda\)). \[ \lambda = \frac{kT}{\sqrt{2}\pi d^2 P} \]where \(k\) is Boltzmann's constant, \(T\) is temperature in Kelvin, \(d\) is molecular diameter, and \(P\) is pressure in Pascals.

Step 2: Isolate pressure (P) in the formula. \[ P = \frac{kT}{\sqrt{2}\pi d^2 \lambda} \]

Step 3: Identify and convert given values to SI units. - \( T = 25^\circ\text{C} = 25 + 273.15 = 298.15 \) K - \( \lambda = 2.63 \times 10^{-5} \) m - Radius \( r = 2.56 \times 10^{-10} \) m, thus diameter \( d = 2r = 5.12 \times 10^{-10} \) m - \( k = 1.38 \times 10^{-23} \) J/K

Step 4: Substitute values and compute pressure in Pascals. \[ P = \frac{(1.38 \times 10^{-23})(298.15)}{\sqrt{2}\pi (5.12 \times 10^{-10})^2 (2.63 \times 10^{-5})} \]\[ P = \frac{4.114 \times 10^{-21}}{1.414 \times 3.1416 \times (2.621 \times 10^{-19}) \times (2.63 \times 10^{-5})} \]\[ P = \frac{4.114 \times 10^{-21}}{3.06 \times 10^{-23}} \approx 1344.4 \text{ Pa} \]

Step 5: Convert pressure from Pascals to mm of mercury. Given 1 atm = 760 mmHg = 101325 Pa, then \( 1 \text{ Pa} = \frac{760}{101325} \) mmHg. \[ P_{\text{mmHg}} = 1344.4 \times \frac{760}{101325} \approx 1344.4 \times 0.0075 = 10.08 \text{ mmHg} \]This is approximately 10 mm of mercury.