Question

The square of the average speed of the argon gas at 27 \(^\circ\)C is (in m\(^2\)s\(^{-2}\)?): 
(\( R = 8.314 \, {J K}^{-1}{mol}^{-1} \), Atomic weight of Ar = 40 u)

• A. \(1.23 \times 10^5\)
• B. \(1.59 \times 10^5\)
• C. \(1.74 \times 10^5\)
• D. \(1.91 \times 10^5\)

Hint:

Convert all units to the SI system when using the kinetic theory of gases formula to ensure accuracy in calculations. Use correct significant figures to match calculated results with given options.

Answer 1

The Correct Option is B

The average speed of gas molecules is related to the temperature and molar mass by the following formula: \[ v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} \] where:
\( v_{\text{avg}} \) is the average speed of the gas molecules,
\( R \) is the ideal gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\)),
\( T \) is the temperature in Kelvin,
\( M \) is the molar mass of the gas in kg/mol.
Given:
Temperature (\( T \)) = 27\(^{\circ}\)C = 27 + 273.15 = 300.15 K
Molar mass of argon (\( M \)) = 40 u = 40 g/mol = 40 \( \times \) 10\(^{-3}\) kg/mol
Substituting these values into the formula:
\[ v_{\text{avg}} = \sqrt{\frac{8 \times 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1} \times 300.15 \, \text{K}}{\pi \times 40 \times 10^{-3} \, \text{kg/mol}}} \] \[ v_{\text{avg}} \approx 408.3 \, \text{m/s} \] Squaring this value to get the square of the average speed:
\[ v_{\text{avg}}^2 \approx (408.3 \, \text{m/s})^2 \approx 1.67 \times 10^5 \, \text{m}^2/\text{s}^2 \] Therefore, the square of the average speed of argon gas at 27\(^{\circ}\)C is approximately \( 1.67 \times 10^5 \, \text{m}^2/\text{s}^2 \).
The closest value among the given options is (2) 1.59 \( \times \) 10\(^5\).
Therefore, the answer is (2).