| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The mean translational kinetic energy (\( \overline{E_{k}} \)) of a molecule in an ideal gas can be calculated using the formula: \[ \overline{E_{k}} = \frac{3}{2} k_B T \] where: - \( k_B \) is the Boltzmann constant, approximately equal to \( 1.380649 \times 10^{-23} \, \text{J/K} \), - \( T \) is the temperature in Kelvin. Given \( T = 293 \, \text{K} \): \[ \overline{E_{k}} = \frac{3}{2} \times 1.380649 \times 10^{-23} \, \text{J/K} \times 293 \, \text{K} \] \[ \overline{E_{k}} = 6.21 \times 10^{-21} \, \text{J} \] This is the mean translational kinetic energy of a single molecule. However, the question asks for the mean translational kinetic energy of a mole of the gas. There are \( 6.022 \times 10^{23} \) molecules in a mole (Avogadro's number). \[ \overline{E_{k,\text{mole}}} = 6.21 \times 10^{-21} \, \text{J/molecule} \times 6.022 \times 10^{23} \, \text{molecules/mole} \] \[ \overline{E_{k,\text{mole}}} = 3.74 \times 10^3 \, \text{J/mole} \] Therefore, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately \( 3.74 \times 10^3 \, \text{J/mole} \). |
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| deepseekr1 | The mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) can be calculated using the formula: \[ E_k = \frac{3}{2}RT \] where \( R \) is the gas constant (\( 8.314 \, \text{J/(mol·K)} \)) and \( T \) is the temperature in Kelvin. **Steps:** 1. Plug in the values: \[ E_k = \frac{3}{2} \times 8.314 \, \text{J/(mol·K)} \times 293 \, \text{K} \] 2. Calculate: \[ E_k \approx \frac{3}{2} \times 8.314 \times 293 = 12.471 \times 293 = 3654 \, \text{J} \] **Answer:** The mean translational kinetic energy is **3.65 × 10³ J** (or **3654 J**) per mole. |
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| gpto1 | To compute the mean translational kinetic energy of one mole of an ideal gas at room temperature (293 K), we'll use the principles of kinetic theory of gases and the ideal gas law. **Fundamental Concepts:** 1. **Average Translational Kinetic Energy per Molecule:** - For an ideal gas, the average translational kinetic energy per molecule is given by: \[ \overline{E}_{\text{per molecule}} = \frac{3}{2} k T \] where: - \( k \) is the **Boltzmann constant** (\(1.380649 \times 10^{-23} \, \text{J/K}\)) - \( T \) is the **absolute temperature** in Kelvin. 2. **Total Translational Kinetic Energy per Mole:** - Since one mole of gas contains Avogadro's number (\( N_A \)) of molecules, the total kinetic energy per mole is: \[ \overline{E}_{\text{per mole}} = N_A \times \overline{E}_{\text{per molecule}} = N_A \times \left( \frac{3}{2} k T \right) \] - Recognizing that \( N_A k = R \) (the **ideal gas constant**), we simplify: \[ \overline{E}_{\text{per mole}} = \frac{3}{2} R T \] **Calculations:** 1. **Given Values:** - Temperature, \( T = 293 \, \text{K} \) - Ideal gas constant, \( R = 8.314462618 \, \text{J/mol·K} \) 2. **Compute the Mean Translational Kinetic Energy per Mole:** \[ \overline{E}_{\text{per mole}} = \frac{3}{2} R T \] 3. **Plug in the Values:** \[ \overline{E}_{\text{per mole}} = \frac{3}{2} \times 8.314462618 \, \text{J/mol·K} \times 293 \, \text{K} \] 4. **Perform the Multiplication:** - First, compute \( R \times T \): \[ R \times T = 8.314462618 \times 293 = 2436.14552 \, \text{J/mol} \] - Then, multiply by \( \frac{3}{2} \): \[ \overline{E}_{\text{per mole}} = \frac{3}{2} \times 2436.14552 = 1.5 \times 2436.14552 = 3654.21828 \, \text{J/mol} \] **Answer:** The mean translational kinetic energy of one mole of an ideal gas at 293 K is approximately **3654 Joules**. **Final Result:** \[ \boxed{\overline{E}_{\text{per mole}} = 3654 \, \text{J}} \] |
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| GPT-4o | To compute the mean translational kinetic energy of a mole of an ideal gas at a given temperature, we can use the kinetic theory of gases. According to this theory, the mean translational kinetic energy of a single molecule of an ideal gas is given by: \[ \langle E_k \rangle = \frac{3}{2} k_B T \] where: - \( \langle E_k \rangle \) is the mean translational kinetic energy of a single molecule, - \( k_B \) is the Boltzmann constant (\( k_B \approx 1.38 \times 10^{-23} \, \text{J/K} \)), - \( T \) is the absolute temperature in kelvins (K). To find the mean translational kinetic energy of a mole of gas, we need to multiply the mean energy of a single molecule by Avogadro's number (\( N_A \approx 6.022 \times 10^{23} \, \text{mol}^{-1} \)): \[ \langle E_k \rangle_{\text{mole}} = N_A \times \langle E_k \rangle \] Substituting the expression for \( \langle E_k \rangle \): \[ \langle E_k \rangle_{\text{mole}} = N_A \times \left( \frac{3}{2} k_B T \right) \] Now, let's compute this at room temperature \( T = 293 \, \text{K} \): \[ \langle E_k \rangle_{\text{mole}} = N_A \times \left( \frac{3}{2} k_B \times 293 \, \text{K} \right) \] Substituting the values of \( N_A \) and \( k_B \): \[ \langle E_k \rangle_{\text{mole}} = 6.022 \times 10^{23} \, \text{mol}^{-1} \times \left( \frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K} \times 293 \, \text{K} \right) \] Now, calculate the numerical value inside the parentheses first: \[ \frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K} \times 293 \, \text{K} = 3.0291 \times 10^{-21} \, \text{J} \] Now multiply by Avogadro's number: \[ \langle E_k \rangle_{\text{mole}} = 6.022 \times 10^{23} \, \text{mol}^{-1} \times 3.0291 \times 10^{-21} \, \text{J} \] \[ \langle E_k \rangle_{\text{mole}} \approx 18.22 \, \text{kJ/mol} \] So, the mean translational kinetic energy of a mole of an ideal gas at 293 K is approximately 18.22 kJ. |
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| llama3.1-70B-Instruct | To compute the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K), we can use the formula derived from the kinetic theory of gases. The mean translational kinetic energy per molecule of an ideal gas is given by: \[ \text{Mean translational kinetic energy per molecule} = \frac{3}{2}kT \] where: - \(k\) is the Boltzmann constant (\(1.380650399 \times 10^{-23}\) J/K), - \(T\) is the temperature in Kelvin. For a mole of gas, we multiply the mean translational kinetic energy per molecule by Avogadro's number (\(N_A = 6.02214076 \times 10^{23}\) molecules/mol) to get the total mean translational kinetic energy per mole: \[ \text{Mean translational kinetic energy per mole} = \frac{3}{2}kT \times N_A \] Substituting the given values: \[ \text{Mean translational kinetic energy per mole} = \frac{3}{2} \times (1.380650399 \times 10^{-23} \, \text{J/K}) \times 293 \, \text{K} \times (6.02214076 \times 10^{23} \, \text{mol}^{-1}) \] \[ \text{Mean translational kinetic energy per mole} = \frac{3}{2} \times 1.380650399 \times 10^{-23} \times 293 \times 6.02214076 \times 10^{23} \] \[ \text{Mean translational kinetic energy per mole} = \frac{3}{2} \times 1.380650399 \times 293 \times 6.02214076 \] \[ \text{Mean translational kinetic energy per mole} = \frac{3}{2} \times 2419.819 \] \[ \text{Mean translational kinetic energy per mole} = 3629.229 \, \text{J} \] Therefore, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately 3629.229 J. |
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| llama3.1-405B-Instruct | To compute the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K), we can use the formula for the average kinetic energy of a molecule in an ideal gas: \[ \text{KE} = \frac{3}{2}kT \] where: - \( \text{KE} \) is the average kinetic energy of a molecule, - \( k \) is the Boltzmann constant (\( 1.380650399 \times 10^{-23} \) J/K), - \( T \) is the temperature in Kelvin. For a mole of gas, we multiply this by Avogadro's number (\( N_A = 6.02214076 \times 10^{23} \) molecules/mol) to get the total kinetic energy for a mole of gas: \[ \text{KE}_{\text{mole}} = N_A \times \frac{3}{2}kT \] Plugging in the values for room temperature (293 K): \[ \text{KE}_{\text{mole}} = 6.02214076 \times 10^{23} \times \frac{3}{2} \times 1.380650399 \times 10^{-23} \times 293 \] \[ \text{KE}_{\text{mole}} = 6.02214076 \times \frac{3}{2} \times 1.380650399 \times 293 \] \[ \text{KE}_{\text{mole}} \approx 3.011 \times 3 \times 1.380650399 \times 293 \] \[ \text{KE}_{\text{mole}} \approx 3.011 \times 4049.185 \] \[ \text{KE}_{\text{mole}} \approx 12186.31 \, \text{J} \] Therefore, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately 12186.31 Joules. |
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| DeepSeek-Coder-V2 | To compute the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K), we can use the following steps: 1. **Identify the relevant formula**: The mean translational kinetic energy (E_k) of a gas molecule is given by the equation: \[ E_k = \frac{3}{2} kT \] where \( k \) is the Boltzmann constant (1.380649 \times 10^{-23} \, \text{J/K}) and \( T \) is the temperature in Kelvin. 2. **Calculate the mean translational kinetic energy per molecule**: \[ E_k = \frac{3}{2} \times 1.380649 \times 10^{-23} \, \text{J/K} \times 293 \, \text{K} \] \[ E_k = \frac{3}{2} \times 1.380649 \times 10^{-23} \times 293 \] \[ E_k = \frac{3}{2} \times 4.045 \times 10^{-21} \, \text{J} \] \[ E_k = 6.0675 \times 10^{-21} \, \text{J} \] 3. **Convert the energy per molecule to the energy per mole**: Since there are \( N_A \) (Avogadro's number) molecules in a mole, we multiply the energy per molecule by \( N_A \) (6.022 \times 10^{23} \, \text{molecules/mol}): \[ E_{k, \text{mole}} = 6.0675 \times 10^{-21} \, \text{J} \times 6.022 \times 10^{23} \, \text{molecules/mol} \] \[ E_{k, \text{mole}} = 365.38 \, \text{J/mol} \] Therefore, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately: \[ \boxed{365.38} \] |
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| Qwen2-72B-Instruct | The mean translational kinetic energy (\( \overline{E_{k}} \)) of a molecule in an ideal gas can be calculated using the formula: \[ \overline{E_{k}} = \frac{3}{2} k_B T \] where: - \( k_B \) is the Boltzmann constant, approximately equal to \( 1.380649 \times 10^{-23} \, \text{J/K} \), - \( T \) is the temperature in Kelvin. Given \( T = 293 \, \text{K} \): \[ \overline{E_{k}} = \frac{3}{2} \times 1.380649 \times 10^{-23} \, \text{J/K} \times 293 \, \text{K} \] \[ \overline{E_{k}} = 6.21 \times 10^{-21} \, \text{J} \] This is the mean translational kinetic energy of a single molecule. However, the question asks for the mean translational kinetic energy of a mole of the gas. There are \( 6.022 \times 10^{23} \) molecules in a mole (Avogadro's number). \[ \overline{E_{k,\text{mole}}} = 6.21 \times 10^{-21} \, \text{J/molecule} \times 6.022 \times 10^{23} \, \text{molecules/mole} \] \[ \overline{E_{k,\text{mole}}} = 3.74 \times 10^3 \, \text{J/mole} \] Therefore, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately \( 3.74 \times 10^3 \, \text{J/mole} \). |
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| baichuan4 | To calculate the mean translational kinetic energy of a mole of ideal gas at a given temperature, we can use the formula derived from kinetic theory: \[ \langle E_{trans} \rangle = \frac{3}{2} R T \] where: - \(\langle E_{trans} \rangle\) is the mean translational kinetic energy per mole, - \(R\) is the universal gas constant, equal to \(8.314 \, \text{J/mol}\cdot\text{K}\), - \(T\) is the absolute temperature in Kelvin. Given the problem's conditions, we have: - Room temperature \(T = 293 \, \text{K}\). Let's plug in the values: \[ \langle E_{trans} \rangle = \frac{3}{2} \times 8.314 \, \text{J/mol}\cdot\text{K} \times 293 \, \text{K} \] \[ \langle E_{trans} \rangle = 3.5 \times 8.314 \times 293 \] \[ \langle E_{trans} \rangle = 8068.61 \, \text{J/mol} \] Therefore, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately \(8068.61 \, \text{J/mol}\). |
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| doubao | The mean translational kinetic energy of an ideal gas is given by the equation: KE_avg = (3/2) * k * T, where k is the Boltzmann constant (1.38×10^-23 J/K) and T is the temperature. At 293 K, the mean translational kinetic energy per molecule is (3/2) * 1.38×10^-23 * 293. For a mole of gas (6.022×10^23 molecules), the total mean translational kinetic energy is (3/2) * 1.38×10^-23 * 293 * 6.022×10^23 = 3739.5 J (approx). |
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| minimax | The mean translational kinetic energy of a mole of an ideal gas can be calculated using the following equation derived from the kinetic theory of gases: \[ \bar{E}_k = \frac{3}{2} k T \] where: - \( \bar{E}_k \) is the mean translational kinetic energy per molecule, - \( k \) is the Boltzmann constant (\( k = 1.380649 \times 10^{-23} \) J/K), - \( T \) is the absolute temperature in Kelvin. However, since we are interested in the mean translational kinetic energy of a mole of gas, we need to multiply this value by Avogadro's number (\( N_A = 6.*** \times 10^{23} \) mol\(^{-1}\)) to convert from per molecule to per mole: \[ \bar{E}_{k, \text{mole}} = \frac{3}{2} k T N_A \] Now, let's plug in the values for room temperature (293 K): \[ \bar{E}_{k, \text{mole}} = \frac{3}{2} \times (1.380649 \times 10^{-23} \text{ J/K}) \times 293 \text{ K} \times (6.*** \times 10^{23} \text{ mol}^{-1}) \] \[ \bar{E}_{k, \text{mole}} = \frac{3}{2} \times 1.380649 \times 10^{-23} \times 293 \times 6.*** \times 10^{23} \] \[ \bar{E}_{k, \text{mole}} = \frac{3}{2} \times 2.4414 \times 10^{-21} \times 6.*** \times 10^{23} \] \[ \bar{E}_{k, \text{mole}} = 1.5 \times 2.4414 \times 6.*** \times 10^{2} \] \[ \bar{E}_{k, \text{mole}} = 3.6621 \times 6.*** \times 10^{2} \] \[ \bar{E}_{k, \text{mole}} = 22.04 \times 10^{2} \] \[ \bar{E}_{k, \text{mole}} = 2204 \text{ J/mol} \] So, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately 2204 J/mol. |
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| kimi | To compute the mean translational kinetic energy of a mole of an ideal gas at a given temperature, we can use the following formula derived from the kinetic theory of gases: \[ E_{\text{trans}} = \frac{3}{2} k_B T \] where: - \( E_{\text{trans}} \) is the mean translational kinetic energy of the gas, - \( k_B \) is the Boltzmann constant (\(1.380649 \times 10^{-23}\) J/K), - \( T \) is the temperature in Kelvin. Given the temperature \( T = 293 \) K, we can plug in the values to find the mean translational kinetic energy: \[ E_{\text{trans}} = \frac{3}{2} \times 1.380649 \times 10^{-23} \text{ J/K} \times 293 \text{ K} \] \[ E_{\text{trans}} = 1.5 \times 1.380649 \times 10^{-23} \times 293 \] \[ E_{\text{trans}} = 6.03 \times 10^{-21} \text{ J} \] This is the mean translational kinetic energy of a single molecule of the gas. To find the energy for a mole of the gas (which contains \(6.022 \times 10^{23}\) molecules, Avogadro's number), we multiply by Avogadro's number: \[ E_{\text{mole}} = 6.03 \times 10^{-21} \text{ J/molecule} \times 6.022 \times 10^{23} \text{ molecules/mole} \] \[ E_{\text{mole}} = 3.63 \times 10^{3} \text{ J/mole} \] So, the mean translational kinetic energy of a mole of an ideal gas at room temperature (293 K) is approximately \(3.63 \times 10^{3}\) Joules. |
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| wenxinyiyan | To solve this problem, we will use the formula for the mean translational kinetic energy of a mole of ideal gas, which is given by: $$\overline{E}_k = \frac{3}{2}RT$$ where $\overline{E}_k$ is the mean translational kinetic energy, $R$ is the gas constant ($8.314 \, \text{J/(mol}\cdot\text{K)}$), and $T$ is the temperature in Kelvin. Given: * Temperature $T = 293 \, \text{K}$ * Gas constant $R = 8.314 \, \text{J/(mol}\cdot\text{K)}$ We can now compute the mean translational kinetic energy: $$\overline{E}_k = \frac{3}{2} \times 8.314 \, \text{J/(mol}\cdot\text{K)} \times 293 \, \text{K}$$ $$\overline{E}_k = \frac{3}{2} \times 2447.402 \, \text{J/mol}$$ $$\overline{E}_k = 3671.103 \, \text{J/mol}$$ Therefore, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately $3671.103 \, \text{J/mol}$. |
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| zhipu | The mean translational kinetic energy of an ideal gas can be calculated using the formula derived from the kinetic theory of gases. The formula for the average translational kinetic energy (KE_trans) of a single molecule of an ideal gas is: \[ KE_{\text{trans}} = \frac{3}{2} k_B T \] where: - \( k_B \) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\)), - \( T \) is the absolute temperature in Kelvin. To find the mean translational kinetic energy of a mole of gas, we need to multiply this by Avogadro's number (\( N_A \)), which is the number of molecules in a mole (\(6.022 \times 10^{23} \, \text{mol}^{-1}\)). So, the mean translational kinetic energy of a mole of ideal gas is: \[ KE_{\text{trans, mole}} = N_A \times KE_{\text{trans}} = N_A \times \frac{3}{2} k_B T \] Given: - \( T = 293 \, \text{K} \) (room temperature), - \( k_B = 1.38 \times 10^{-23} \, \text{J/K} \), - \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \). Now, we can plug these values into the equation: \[ KE_{\text{trans, mole}} = 6.022 \times 10^{23} \times \frac{3}{2} \times 1.38 \times 10^{-23} \times 293 \] \[ KE_{\text{trans, mole}} = 6.022 \times \frac{3}{2} \times 1.38 \times 293 \] \[ KE_{\text{trans, mole}} = 6.022 \times 1.5 \times 1.38 \times 293 \] \[ KE_{\text{trans, mole}} = 6.022 \times 2.07 \times 293 \] \[ KE_{\text{trans, mole}} = 12.37754 \times 293 \] \[ KE_{\text{trans, mole}} \approx 3611.5 \, \text{J} \] Therefore, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately 3611.5 J. |
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| gpto1 | To compute the mean translational kinetic energy of one mole of an ideal gas at room temperature (293 K), we'll use the principles of kinetic theory of gases and the ideal gas law. **Fundamental Concepts:** 1. **Average Translational Kinetic Energy per Molecule:** - For an ideal gas, the average translational kinetic energy per molecule is given by: \[ \overline{E}_{\text{per molecule}} = \frac{3}{2} k T \] where: - \( k \) is the **Boltzmann constant** (\(1.380649 \times 10^{-23} \, \text{J/K}\)) - \( T \) is the **absolute temperature** in Kelvin. 2. **Total Translational Kinetic Energy per Mole:** - Since one mole of gas contains Avogadro's number (\( N_A \)) of molecules, the total kinetic energy per mole is: \[ \overline{E}_{\text{per mole}} = N_A \times \overline{E}_{\text{per molecule}} = N_A \times \left( \frac{3}{2} k T \right) \] - Recognizing that \( N_A k = R \) (the **ideal gas constant**), we simplify: \[ \overline{E}_{\text{per mole}} = \frac{3}{2} R T \] **Calculations:** 1. **Given Values:** - Temperature, \( T = 293 \, \text{K} \) - Ideal gas constant, \( R = 8.314462618 \, \text{J/mol·K} \) 2. **Compute the Mean Translational Kinetic Energy per Mole:** \[ \overline{E}_{\text{per mole}} = \frac{3}{2} R T \] 3. **Plug in the Values:** \[ \overline{E}_{\text{per mole}} = \frac{3}{2} \times 8.314462618 \, \text{J/mol·K} \times 293 \, \text{K} \] 4. **Perform the Multiplication:** - First, compute \( R \times T \): \[ R \times T = 8.314462618 \times 293 = 2436.14552 \, \text{J/mol} \] - Then, multiply by \( \frac{3}{2} \): \[ \overline{E}_{\text{per mole}} = \frac{3}{2} \times 2436.14552 = 1.5 \times 2436.14552 = 3654.21828 \, \text{J/mol} \] **Answer:** The mean translational kinetic energy of one mole of an ideal gas at 293 K is approximately **3654 Joules**. **Final Result:** \[ \boxed{\overline{E}_{\text{per mole}} = 3654 \, \text{J}} \] |
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| deepseekv3 | To compute the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K), we can use the following formula for the mean translational kinetic energy per molecule: \[ E_{\text{kin}} = \frac{3}{2} k_B T \] Where: - \( E_{\text{kin}} \) is the mean translational kinetic energy per molecule, - \( k_B \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \, \text{J/K} \)), - \( T \) is the temperature in Kelvin. For one mole of gas, we multiply this energy by Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)) to get the total mean translational kinetic energy for the mole of gas: \[ E_{\text{kin, mole}} = \frac{3}{2} k_B T \times N_A \] Substitute the values: \[ E_{\text{kin, mole}} = \frac{3}{2} (1.38 \times 10^{-23} \, \text{J/K}) \times 293 \, \text{K} \times 6.022 \times 10^{23} \, \text{mol}^{-1} \] Simplify the expression: \[ E_{\text{kin, mole}} = \frac{3}{2} \times 1.38 \times 293 \times 6.022 \, \text{J/mol} \] \[ E_{\text{kin, mole}} = \frac{3}{2} \times 242.7 \, \text{J/mol} \] \[ E_{\text{kin, mole}} = 364.0 \, \text{J/mol} \] Thus, the mean translational kinetic energy of a mole of ideal gas at room temperature (293 K) is approximately **364 J/mol**. |
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