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A sample of 2 mol of diatomic gas is measured to have a temperature of 323 K. If the mass of the gas is 0.032 kg, what is the approximate average speed of the molecules in the gas?

(Recall that the equation for kinetic energy due to translation in a gas is [tex]KE_{\text{translational}} = \frac{1}{2} m v^2 = \frac{3}{2} n R T[/tex], and [tex]R = 8.31 \, \text{J} / (\text{mol} \cdot \text{K})[/tex].)

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]

Sagot :

GinnyAnswer
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].
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