Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Text Transcription:**
The rms speed of the molecules of a gas at 133°C is 197 m/s. Calculate the mass \( m \) of a single molecule in the gas.
---
**Explanation for Educational Context:**
This question involves determining the molecular mass of a gas given its root mean square (rms) speed at a specific temperature. The rms speed is a measure of the speed of particles in a gas, and it is associated with the temperature and mass of the gas.
To solve this problem, you would typically use the formula for the rms speed:
\[
v_{\text{rms}} = \sqrt{\frac{3kT}{m}}
\]
where:
- \( v_{\text{rms}} \) is the rms speed,
- \( k \) is Boltzmann's constant (\(1.38 \times 10^{-23} \, \text{J/K}\)),
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the mass of a molecule.
First, convert the given temperature from Celsius to Kelvin:
\[
T = 133 + 273.15 = 406.15 \, \text{K}
\]
Given that \( v_{\text{rms}} \) is 197 m/s, rearrange the equation to solve for \( m \):
\[
m = \frac{3kT}{v_{\text{rms}}^2}
\]
This approach will yield the mass \( m \) of a single molecule.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F414a6cd2-98c9-4536-a651-b4f45ff08a29%2F92dc96ac-293a-4660-8196-01b4731e8b73%2F8mkjgk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Text Transcription:**
The rms speed of the molecules of a gas at 133°C is 197 m/s. Calculate the mass \( m \) of a single molecule in the gas.
---
**Explanation for Educational Context:**
This question involves determining the molecular mass of a gas given its root mean square (rms) speed at a specific temperature. The rms speed is a measure of the speed of particles in a gas, and it is associated with the temperature and mass of the gas.
To solve this problem, you would typically use the formula for the rms speed:
\[
v_{\text{rms}} = \sqrt{\frac{3kT}{m}}
\]
where:
- \( v_{\text{rms}} \) is the rms speed,
- \( k \) is Boltzmann's constant (\(1.38 \times 10^{-23} \, \text{J/K}\)),
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the mass of a molecule.
First, convert the given temperature from Celsius to Kelvin:
\[
T = 133 + 273.15 = 406.15 \, \text{K}
\]
Given that \( v_{\text{rms}} \) is 197 m/s, rearrange the equation to solve for \( m \):
\[
m = \frac{3kT}{v_{\text{rms}}^2}
\]
This approach will yield the mass \( m \) of a single molecule.
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