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].
