Calculate the rms speed of gaseous F2 molecules at 23 degrees celsius and $1.00 \mathrm{~atm}$. What is the rms speed of gaseous F2 at 23 degrees celsius and $1.45 \mathrm{~atm}$ ?

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∻Solution∻ ‖ 1 ⋮ **Identify the given data**: The temperature is $23^\circ \text{C}$, which is equivalent to $296 \text{ K}$ (since $T(K) = T(°C) + 273$). The molar mass of $F_2$ is $38.00 \, \text{g/mol}$ or $0.038 \, \text{kg/mol}$. ‖ ‖ 2 ⋮ **Use the formula for rms speed**: The root mean square (rms) speed of a gas is given by the formula: $$ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} $$ where $R$ is the gas constant ($8.314 \, \text{J/mol·K}$), $T$ is the temperature in Kelvin, and $M$ is the molar mass in kg/mol. ‖ ‖ 3 ⋮ **Calculate the rms speed at $1.00 \, \text{atm}$**: Since the rms speed is independent of pressure, we can directly use the formula: $$ v_{\text{rms}} = \sqrt{\frac{3 \times 8.314 \, \text{J/mol·K} \times 296 \, \text{K}}{0.038 \, \text{kg/mol}}} $$ $$ v_{\text{rms}} = \sqrt{\frac{7381.872}{0.038}} $$ $$ v_{\text{rms}} = \sqrt{194260.316} $$ $$ v_{\text{rms}} \approx 440.7 \, \text{m/s} $$ ‖ ‖ 4 ⋮ **Calculate the rms speed at $1.45 \, \text{atm}$**: The rms speed is still independent of pressure, so it remains the same: $$ v_{\text{rms}} \approx 440.7 \, \text{m/s} $$ ‖ ∻Answer∻ ⚹ The rms speed of gaseous $F_2$ at $23^\circ \text{C}$ and $1.00 \, \text{atm}$ is approximately $440.7 \, \text{m/s}$. The rms speed of gaseous $F_2$ at $23^\circ \text{C}$ and $1.45 \, \text{atm}$ is also approximately $440.7 \, \text{m/s}$. ⚹ ∻Key Concept∻ ⚹ The rms speed of a gas is independent of pressure and depends only on temperature and molar mass. ⚹ ∻Explanation∻ ⚹ The root mean square speed of gas molecules is determined by the temperature and molar mass of the gas. Since the temperature and molar mass remain constant, the rms speed does not change with pressure. ⚹◊What is the relationship between the rms speed of gaseous molecules and the pressure within a system according to the kinetic molecular theory in Chemistry?⍭ Generate me a similar question◊

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