Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To solve this problem, let's proceed step-by-step using the given data and the relevant equations.
1. Given Data:
- Number of moles [tex]\( n = 2 \)[/tex] mol
- Temperature [tex]\( T = 323 \)[/tex] K
- Mass of the gas [tex]\( m = 0.032 \)[/tex] kg
- Universal gas constant [tex]\( R = 8.31 \)[/tex] J/(mol·K)
2. Transitional Kinetic Energy:
The translational kinetic energy of the gas molecules can be calculated using the formula:
[tex]\[ KE_{\text{translational}} = \frac{3}{2} n R T \][/tex]
Substituting the given values:
[tex]\[ KE_{\text{translational}} = \frac{3}{2} \times 2 \times 8.31 \times 323 \][/tex]
Working this out gives:
[tex]\[ KE_{\text{translational}} \approx 8052.39 \text{ J} \][/tex]
3. Relating Kinetic Energy to Speed:
The average speed [tex]\( v \)[/tex] of the molecules can be found from the kinetic energy expression:
[tex]\[ KE_{\text{translational}} = \frac{1}{2} m v^2 \][/tex]
Solving for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{\frac{2 \times KE_{\text{translational}}}{m}} \][/tex]
Substituting the known values:
[tex]\[ v = \sqrt{\frac{2 \times 8052.39}{0.032}} \][/tex]
[tex]\[ v \approx \sqrt{503274.375} \][/tex]
[tex]\[ v \approx 709.42 \text{ m/s} \][/tex]
4. Determine the Correct Option:
Comparing the calculated average speed [tex]\( v \approx 709.42 \text{ m/s} \)[/tex] to the provided options:
- A. [tex]\( 681 \text{ m/s} \)[/tex]
- B. [tex]\( 709 \text{ m/s} \)[/tex]
- C. [tex]\( 652 \text{ m/s} \)[/tex]
- D. [tex]\( 621 \text{ m/s} \)[/tex]
The closest option to the calculated speed is option B.
Therefore, the approximate average speed of the molecules in the gas is [tex]\( \boxed{709 \text{ m/s}} \)[/tex].
1. Given Data:
- Number of moles [tex]\( n = 2 \)[/tex] mol
- Temperature [tex]\( T = 323 \)[/tex] K
- Mass of the gas [tex]\( m = 0.032 \)[/tex] kg
- Universal gas constant [tex]\( R = 8.31 \)[/tex] J/(mol·K)
2. Transitional Kinetic Energy:
The translational kinetic energy of the gas molecules can be calculated using the formula:
[tex]\[ KE_{\text{translational}} = \frac{3}{2} n R T \][/tex]
Substituting the given values:
[tex]\[ KE_{\text{translational}} = \frac{3}{2} \times 2 \times 8.31 \times 323 \][/tex]
Working this out gives:
[tex]\[ KE_{\text{translational}} \approx 8052.39 \text{ J} \][/tex]
3. Relating Kinetic Energy to Speed:
The average speed [tex]\( v \)[/tex] of the molecules can be found from the kinetic energy expression:
[tex]\[ KE_{\text{translational}} = \frac{1}{2} m v^2 \][/tex]
Solving for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{\frac{2 \times KE_{\text{translational}}}{m}} \][/tex]
Substituting the known values:
[tex]\[ v = \sqrt{\frac{2 \times 8052.39}{0.032}} \][/tex]
[tex]\[ v \approx \sqrt{503274.375} \][/tex]
[tex]\[ v \approx 709.42 \text{ m/s} \][/tex]
4. Determine the Correct Option:
Comparing the calculated average speed [tex]\( v \approx 709.42 \text{ m/s} \)[/tex] to the provided options:
- A. [tex]\( 681 \text{ m/s} \)[/tex]
- B. [tex]\( 709 \text{ m/s} \)[/tex]
- C. [tex]\( 652 \text{ m/s} \)[/tex]
- D. [tex]\( 621 \text{ m/s} \)[/tex]
The closest option to the calculated speed is option B.
Therefore, the approximate average speed of the molecules in the gas is [tex]\( \boxed{709 \text{ m/s}} \)[/tex].
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.