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青州's Question
Chemistry
Posted 2 months ago

Calculate the rms speed of gaseous F2 molecules at 23 degrees celsius and 1.00 atm1.00 \mathrm{~atm}. What is the rms speed of gaseous F2 at 23 degrees celsius and 1.45 atm1.45 \mathrm{~atm} ?
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Answer from Sia
Posted 2 months ago
Solution
1
Identify the given data: The temperature is 23C23^\circ \text{C}, which is equivalent to 296 K296 \text{ K} (since T(K)=T(°C)+273T(K) = T(°C) + 273). The molar mass of F2F_2 is 38.00g/mol38.00 \, \text{g/mol} or 0.038kg/mol0.038 \, \text{kg/mol}
2
Use the formula for rms speed: The root mean square (rms) speed of a gas is given by the formula: vrms=3RTM v_{\text{rms}} = \sqrt{\frac{3RT}{M}} where RR is the gas constant (8.314 \, \text{J/mol·K}), TT is the temperature in Kelvin, and MM 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}}} vrms=7381.8720.038 v_{\text{rms}} = \sqrt{\frac{7381.872}{0.038}} vrms=194260.316 v_{\text{rms}} = \sqrt{194260.316} vrms440.7m/s 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: vrms440.7m/s v_{\text{rms}} \approx 440.7 \, \text{m/s}
Answer
The rms speed of gaseous F2F_2 at 23C23^\circ \text{C} and 1.00atm1.00 \, \text{atm} is approximately 440.7m/s440.7 \, \text{m/s}. The rms speed of gaseous F2F_2 at 23C23^\circ \text{C} and 1.45atm1.45 \, \text{atm} is also approximately 440.7m/s440.7 \, \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.

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