QuestionApril 24, 2025

At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 589 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 589 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 589 mph ? Express your answer using one decimal place. T=square ^circ C

Solution
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Answer

T = 0.0^\circ C Explanation 1. Convert airplane speed to SI units Convert 589 mph to m/s using 1 \, \text{mile} = 1609.34 \, \text{m} and 1 \, \text{hour} = 3600 \, \text{s}: \[ v_{\text{plane}} = 589 \times \frac{1609.34}{3600} = 263.2 \, \text{m/s}. \] 2. Use the formula for average molecular speed The average speed of a gas molecule is given by: \[ v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}}, \] where R = 8.314 \, \text{J/mol·K}, M is the molar mass in kg/mol, and T is the temperature in Kelvin. For oxygen (O_2), M = 32 \, \text{g/mol} = 0.032 \, \text{kg/mol}. Set v_{\text{avg}} = v_{\text{plane}}: \[ 263.2 = \sqrt{\frac{8(8.314)T}{\pi (0.032)}}. \] 3. Solve for T Square both sides and solve for T: \[ 263.2^2 = \frac{8(8.314)T}{\pi (0.032)}, \] \[ T = \frac{263.2^2 \cdot \pi (0.032)}{8(8.314)} = 273.15 \, \text{K}. \] 4. Convert to Celsius Convert from Kelvin to Celsius: \[ T = 273.15 - 273.15 = 0.0^\circ C. \]

Explanation

1. Convert airplane speed to SI units<br /> Convert 589 mph to m/s using $1 \, \text{mile} = 1609.34 \, \text{m}$ and $1 \, \text{hour} = 3600 \, \text{s}$:<br />\[<br />v_{\text{plane}} = 589 \times \frac{1609.34}{3600} = 263.2 \, \text{m/s}.<br />\]<br /><br />2. Use the formula for average molecular speed<br /> The average speed of a gas molecule is given by:<br />\[<br />v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}},<br />\]<br />where $R = 8.314 \, \text{J/mol·K}$, $M$ is the molar mass in kg/mol, and $T$ is the temperature in Kelvin.<br /><br />For oxygen ($O_2$), $M = 32 \, \text{g/mol} = 0.032 \, \text{kg/mol}$.<br /><br />Set $v_{\text{avg}} = v_{\text{plane}}$:<br />\[<br />263.2 = \sqrt{\frac{8(8.314)T}{\pi (0.032)}}.<br />\]<br /><br />3. Solve for $T$<br /> Square both sides and solve for $T$:<br />\[<br />263.2^2 = \frac{8(8.314)T}{\pi (0.032)},<br />\]<br />\[<br />T = \frac{263.2^2 \cdot \pi (0.032)}{8(8.314)} = 273.15 \, \text{K}.<br />\]<br /><br />4. Convert to Celsius<br /> Convert from Kelvin to Celsius:<br />\[<br />T = 273.15 - 273.15 = 0.0^\circ C.<br />\]
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