QuestionJuly 15, 2025

At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 583 mph? Express your answer using one decimal place. T=square ^circ C

At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 583 mph? Express your answer using one decimal place. T=square ^circ C
At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 583 mph? Express your answer using one decimal place. T=square ^circ C

Solution
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Answer

10726.9 \text{ °C} Explanation 1. Convert airplane speed to m/s 583 \text{ mph} = 583 \times \frac{1609.34}{3600} \approx 260.7 \text{ m/s}. 2. Use formula for average speed of gas molecules Average speed v = \sqrt{\frac{3kT}{m}}, where k = 1.38 \times 10^{-23} \text{ J/K}, m = 5.32 \times 10^{-26} \text{ kg} (mass of oxygen molecule). 3. Solve for temperature T Set v = 260.7 \text{ m/s} and solve: 260.7 = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{5.32 \times 10^{-26}}}. Square both sides: 260.7^2 = \frac{3 \times 1.38 \times 10^{-23} \times T}{5.32 \times 10^{-26}}. Rearrange: T = \frac{260.7^2 \times 5.32 \times 10^{-26}}{3 \times 1.38 \times 10^{-23}}. Calculate: T \approx 1.08 \times 10^4 \text{ K}. 4. Convert temperature to Celsius T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \approx 1.08 \times 10^4 - 273.15 \approx 10726.85 \text{ °C}.

Explanation

1. Convert airplane speed to m/s<br /> $583 \text{ mph} = 583 \times \frac{1609.34}{3600} \approx 260.7 \text{ m/s}$.<br />2. Use formula for average speed of gas molecules<br /> Average speed $v = \sqrt{\frac{3kT}{m}}$, where $k = 1.38 \times 10^{-23} \text{ J/K}$, $m = 5.32 \times 10^{-26} \text{ kg}$ (mass of oxygen molecule).<br />3. Solve for temperature $T$<br /> Set $v = 260.7 \text{ m/s}$ and solve: $260.7 = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{5.32 \times 10^{-26}}}$.<br /> Square both sides: $260.7^2 = \frac{3 \times 1.38 \times 10^{-23} \times T}{5.32 \times 10^{-26}}$.<br /> Rearrange: $T = \frac{260.7^2 \times 5.32 \times 10^{-26}}{3 \times 1.38 \times 10^{-23}}$.<br /> Calculate: $T \approx 1.08 \times 10^4 \text{ K}$.<br />4. Convert temperature to Celsius<br /> $T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \approx 1.08 \times 10^4 - 273.15 \approx 10726.85 \text{ °C}$.
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