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To determine the temperature of the gas, we need to use the given parameters and the provided equation for translational kinetic energy. Here is a step-by-step solution:
1. Understand the Problem:
- We have a monatomic gas.
- The mass [tex]\( m \)[/tex] of the gas is 0.008 kg.
- The average speed [tex]\( v \)[/tex] of the gas molecules is 1477 m/s.
- The amount of gas [tex]\( n \)[/tex] is 2 moles.
- The gas constant [tex]\( R \)[/tex] is 8.31 J/(mol K).
2. Translational Kinetic Energy Formula:
- The translational kinetic energy of a gas is given by the formula:
[tex]\[ KE_{\text{translational}} = \frac{1}{2} m v^2 \][/tex]
- Plug in the values:
[tex]\[ KE_{\text{translational}} = \frac{1}{2} \times 0.008 \, \text{kg} \times (1477 \, \text{m/s})^2 \][/tex]
3. Calculate the Translational Kinetic Energy:
- First, square the velocity:
[tex]\[ (1477 \, \text{m/s})^2 = 2183629 \, \text{m}^2/\text{s}^2 \][/tex]
- Then multiply by the mass and by [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ KE_{\text{translational}} = \frac{1}{2} \times 0.008 \, \text{kg} \times 2183629 \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ KE_{\text{translational}} = 0.004 \times 2183629 \][/tex]
[tex]\[ KE_{\text{translational}} \approx 8726.116 \, \text{J} \][/tex]
4. Relate Kinetic Energy to Temperature:
- Use the equation:
[tex]\[ KE_{\text{translational}} = \frac{3}{2} n R T \][/tex]
- Solve for temperature [tex]\( T \)[/tex]:
[tex]\[ T = \frac{2}{3} \frac{KE_{\text{translational}}}{n R} \][/tex]
- Plug in the values:
[tex]\[ T = \frac{2}{3} \frac{8726.116 \, \text{J}}{2 \, \text{mol} \times 8.31 \, \text{J/(mol K)}} \][/tex]
5. Calculate the Temperature:
- Multiply and divide inside the brackets:
[tex]\[ T = \frac{2}{3} \frac{8726.116}{16.62} \][/tex]
[tex]\[ \frac{8726.116}{16.62} \approx 525.037063 \][/tex]
[tex]\[ T = \frac{2}{3} \times 525.037063 \approx 350.024709 \][/tex]
Thus, the temperature of the gas is approximately [tex]\( 350 \, \text{K} \)[/tex].
Therefore, the correct answer is:
A. 350 K
1. Understand the Problem:
- We have a monatomic gas.
- The mass [tex]\( m \)[/tex] of the gas is 0.008 kg.
- The average speed [tex]\( v \)[/tex] of the gas molecules is 1477 m/s.
- The amount of gas [tex]\( n \)[/tex] is 2 moles.
- The gas constant [tex]\( R \)[/tex] is 8.31 J/(mol K).
2. Translational Kinetic Energy Formula:
- The translational kinetic energy of a gas is given by the formula:
[tex]\[ KE_{\text{translational}} = \frac{1}{2} m v^2 \][/tex]
- Plug in the values:
[tex]\[ KE_{\text{translational}} = \frac{1}{2} \times 0.008 \, \text{kg} \times (1477 \, \text{m/s})^2 \][/tex]
3. Calculate the Translational Kinetic Energy:
- First, square the velocity:
[tex]\[ (1477 \, \text{m/s})^2 = 2183629 \, \text{m}^2/\text{s}^2 \][/tex]
- Then multiply by the mass and by [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ KE_{\text{translational}} = \frac{1}{2} \times 0.008 \, \text{kg} \times 2183629 \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ KE_{\text{translational}} = 0.004 \times 2183629 \][/tex]
[tex]\[ KE_{\text{translational}} \approx 8726.116 \, \text{J} \][/tex]
4. Relate Kinetic Energy to Temperature:
- Use the equation:
[tex]\[ KE_{\text{translational}} = \frac{3}{2} n R T \][/tex]
- Solve for temperature [tex]\( T \)[/tex]:
[tex]\[ T = \frac{2}{3} \frac{KE_{\text{translational}}}{n R} \][/tex]
- Plug in the values:
[tex]\[ T = \frac{2}{3} \frac{8726.116 \, \text{J}}{2 \, \text{mol} \times 8.31 \, \text{J/(mol K)}} \][/tex]
5. Calculate the Temperature:
- Multiply and divide inside the brackets:
[tex]\[ T = \frac{2}{3} \frac{8726.116}{16.62} \][/tex]
[tex]\[ \frac{8726.116}{16.62} \approx 525.037063 \][/tex]
[tex]\[ T = \frac{2}{3} \times 525.037063 \approx 350.024709 \][/tex]
Thus, the temperature of the gas is approximately [tex]\( 350 \, \text{K} \)[/tex].
Therefore, the correct answer is:
A. 350 K
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