What is the internal energy (total kinetic energy) of three moles of a mono-atomic gas at temperature of 27.0 °C (3.00 x 102 K)? O 9.86 x 105 J O 2.10 x 106 J ONone of the given O3.51 x 10 J O5.65 x 107 J O1.12 x 10 J
What is the internal energy (total kinetic energy) of three moles of a mono-atomic gas at temperature of 27.0 °C (3.00 x 102 K)? O 9.86 x 105 J O 2.10 x 106 J ONone of the given O3.51 x 10 J O5.65 x 107 J O1.12 x 10 J
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![### Calculation of Internal Energy of Mono-Atomic Gas
**Question:**
What is the internal energy (total kinetic energy) of three moles of a mono-atomic gas at a temperature of 27.0°C (3.00 x 10² K)?
**Options:**
1. 9.86 x 10⁵ J
2. 2.10 x 10⁶ J
3. None of the given
4. 3.51 x 10⁶ J
5. 5.65 x 10⁷ J
6. 1.12 x 10⁴ J
**Explanation:**
This question involves the calculation of the internal energy of a mono-atomic gas. The internal energy for a mono-atomic ideal gas can be found using the formula:
\[ U = \frac{3}{2} nRT \]
where:
- \( U \) is the internal energy,
- \( n \) is the number of moles of gas,
- \( R \) is the ideal gas constant (approximately \( 8.314 \, \text{J/mol·K} \)),
- \( T \) is the temperature in Kelvin.
Given:
- \( n = 3 \) moles,
- \( T = 300 \, \text{K} \) (as 27.0°C is equivalent to 300 K, since \( T(K) = T(°C) + 273.15 \)),
- \( R = 8.314 \, \text{J/mol·K} \).
Substitute these values into the formula:
\[ U = \frac{3}{2} \times 3 \times 8.314 \times 300 \]
Now, perform the calculation:
\[ U = \frac{3}{2} \times 3 \times 8.314 \times 300 \]
\[ U = \frac{3}{2} \times 3 \times 2494.2 \]
\[ U = \frac{3}{2} \times 7482.6 \]
\[ U = 1.5 \times 7482.6 \]
\[ U = 11223.9 \, \text{J} \]
Therefore, the internal energy is approximately \( 1.12 \times 10⁴ \, \text{J} \).
**Correct Answer:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18248a49-f7f9-454f-abfb-042af6624ce9%2F7cbe490a-e7f2-4576-ae27-56174aa971a4%2Fbv39zek_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculation of Internal Energy of Mono-Atomic Gas
**Question:**
What is the internal energy (total kinetic energy) of three moles of a mono-atomic gas at a temperature of 27.0°C (3.00 x 10² K)?
**Options:**
1. 9.86 x 10⁵ J
2. 2.10 x 10⁶ J
3. None of the given
4. 3.51 x 10⁶ J
5. 5.65 x 10⁷ J
6. 1.12 x 10⁴ J
**Explanation:**
This question involves the calculation of the internal energy of a mono-atomic gas. The internal energy for a mono-atomic ideal gas can be found using the formula:
\[ U = \frac{3}{2} nRT \]
where:
- \( U \) is the internal energy,
- \( n \) is the number of moles of gas,
- \( R \) is the ideal gas constant (approximately \( 8.314 \, \text{J/mol·K} \)),
- \( T \) is the temperature in Kelvin.
Given:
- \( n = 3 \) moles,
- \( T = 300 \, \text{K} \) (as 27.0°C is equivalent to 300 K, since \( T(K) = T(°C) + 273.15 \)),
- \( R = 8.314 \, \text{J/mol·K} \).
Substitute these values into the formula:
\[ U = \frac{3}{2} \times 3 \times 8.314 \times 300 \]
Now, perform the calculation:
\[ U = \frac{3}{2} \times 3 \times 8.314 \times 300 \]
\[ U = \frac{3}{2} \times 3 \times 2494.2 \]
\[ U = \frac{3}{2} \times 7482.6 \]
\[ U = 1.5 \times 7482.6 \]
\[ U = 11223.9 \, \text{J} \]
Therefore, the internal energy is approximately \( 1.12 \times 10⁴ \, \text{J} \).
**Correct Answer:**
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