Questions: Compute the root-mean-square speed of Ar molecules in a sample of argon gas at a temperature of 49°C.

Solution Steps

The root-mean-square speed of gas molecules is calculated using the temperature in Kelvin. First, convert the given temperature from Celsius to Kelvin:

\[ T = 49^\circ \text{C} + 273.15 = 322.15 \, \text{K} \]

The formula for the root-mean-square speed \( v_{\text{rms}} \) of gas molecules is:

\[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \]

where:

• \( k \) is the Boltzmann constant, \( k = 1.3807 \times 10^{-23} \, \text{J/K} \),
• \( T \) is the temperature in Kelvin,
• \( m \) is the mass of one molecule of the gas.

The molar mass of argon is approximately \( 39.95 \, \text{g/mol} \). Convert this to kilograms per molecule:

\[ m = \frac{39.95 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}} = \frac{39.95 \times 10^{-3} \, \text{kg/mol}}{6.022 \times 10^{23}} \approx 6.6335 \times 10^{-26} \, \text{kg} \]

Substitute the values into the root-mean-square speed formula:

\[ v_{\text{rms}} = \sqrt{\frac{3 \times 1.3807 \times 10^{-23} \, \text{J/K} \times 322.15 \, \text{K}}{6.6335 \times 10^{-26} \, \text{kg}}} \]

\[ v_{\text{rms}} = \sqrt{\frac{1.3331 \times 10^{-20} \, \text{J}}{6.6335 \times 10^{-26} \, \text{kg}}} \]

\[ v_{\text{rms}} = \sqrt{2.0103 \times 10^{5} \, \text{m}^2/\text{s}^2} \]

\[ v_{\text{rms}} \approx 448.4 \, \text{m/s} \]

Final Answer