QuestionApril 24, 2025

At what temperature (in degrees Celsius) will xenon atoms have the same average speed that Br_(2) molecules have at 17^circ C Express the temperature in degrees Celsius to two significant digits. T_(Xe)=square ^circ C

At what temperature (in degrees Celsius) will xenon atoms have the same average speed that Br_(2) molecules have at 17^circ C Express the temperature in degrees Celsius to two significant digits. T_(Xe)=square ^circ C
At what temperature (in degrees Celsius) will xenon atoms have the same average speed that Br_(2) molecules have at 17^circ C Express the temperature in degrees Celsius to two significant digits. T_(Xe)=square ^circ C

Solution
4.5(420 votes)

Answer

-154^\circ \text{C} Explanation 1. Write the formula for average speed The average speed of gas molecules is proportional to \sqrt{\frac{T}{M}}, where T is temperature in Kelvin and M is molar mass. For two gases, equating their speeds gives: \sqrt{\frac{T_{Xe}}{M_{Xe}}} = \sqrt{\frac{T_{Br_2}}{M_{Br_2}}}. 2. Solve for T_{Xe} Squaring both sides and solving for T_{Xe}: T_{Xe} = T_{Br_2} \cdot \frac{M_{Xe}}{M_{Br_2}}. 3. Convert T_{Br_2} to Kelvin T_{Br_2} = 17 + 273.15 = 290.15 \, \text{K}. 4. Substitute values Molar masses: M_{Xe} = 131.29 \, \text{g/mol}, M_{Br_2} = 159.808 \times 2 = 319.616 \, \text{g/mol}. Substituting: T_{Xe} = 290.15 \cdot \frac{131.29}{319.616} = 119.18 \, \text{K}. 5. Convert T_{Xe} back to Celsius T_{Xe} = 119.18 - 273.15 = -153.97 \, ^\circ \text{C}.

Explanation

1. Write the formula for average speed<br /> The average speed of gas molecules is proportional to $\sqrt{\frac{T}{M}}$, where $T$ is temperature in Kelvin and $M$ is molar mass. For two gases, equating their speeds gives:<br />$$\sqrt{\frac{T_{Xe}}{M_{Xe}}} = \sqrt{\frac{T_{Br_2}}{M_{Br_2}}}.$$<br /><br />2. Solve for $T_{Xe}$<br /> Squaring both sides and solving for $T_{Xe}$:<br />$$T_{Xe} = T_{Br_2} \cdot \frac{M_{Xe}}{M_{Br_2}}.$$<br /><br />3. Convert $T_{Br_2}$ to Kelvin<br /> $T_{Br_2} = 17 + 273.15 = 290.15 \, \text{K}$.<br /><br />4. Substitute values<br /> Molar masses: $M_{Xe} = 131.29 \, \text{g/mol}$, $M_{Br_2} = 159.808 \times 2 = 319.616 \, \text{g/mol}$. Substituting:<br />$$T_{Xe} = 290.15 \cdot \frac{131.29}{319.616} = 119.18 \, \text{K}.$$<br /><br />5. Convert $T_{Xe}$ back to Celsius<br /> $T_{Xe} = 119.18 - 273.15 = -153.97 \, ^\circ \text{C}$.
Click to rate:

Similar Questions