| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The mean molecular speed (v) of a gas can be calculated using the formula: v = sqrt((8 * R * T) / (pi * M)) where, R is the gas constant (8.314 J/(mol*K)) T is the absolute temperature (in Kelvin) M is the molar mass of the gas (in kg/mol) pi is the mathematical constant (approximately 3.14159) For Radon (Rn), the molar mass is approximately 222.0176 g/mol or 0.2220176 kg/mol. Let's assume the temperature is at standard temperature, which is 273.15 K. Substitute the values into the formula: v = sqrt((8 * 8.314 J/(mol*K) * 273.15 K) / (3.14159 * 0.2220176 kg/mol)) v = sqrt((17494.464 J/K) / (0.698767 kg)) v = sqrt(24903.026 m²/s²) v ≈ 157.81 m/s So the mean molecular speed of Radon at standard temperature is approximately 157.81 m/s. Please note that this is a simplified calculation. Real-world conditions may vary, and the mean molecular speed can be affected by factors such as pressure and the specific conditions of the gas. |
167.0 |
| deepseekr1 | To compute the mean molecular speed \( v \) of radon (Rn) gas, we use the formula for the average speed of gas molecules: \[ v_{\text{avg}} = \sqrt{\frac{8RT}{\pi M}} \] **Step-by-Step Solution:** 1. **Identify Constants and Variables:** - **Gas Constant (\( R \))**: \( 8.314 \, \text{J/(mol·K)} \) - **Temperature (\( T \))**: Assume standard temperature (\( 0^\circ\text{C} = 273.15 \, \text{K} \)). - **Molar Mass of Radon (\( M \))**: \( 222 \, \text{g/mol} = 0.222 \, \text{kg/mol} \). 2. **Plug Values into the Formula:** \[ v_{\text{avg}} = \sqrt{\frac{8 \times 8.314 \, \text{J/(mol·K)} \times 273.15 \, \text{K}}{\pi \times 0.222 \, \text{kg/mol}}} \] 3. **Calculate the Numerator:** \[ 8 \times 8.314 \times 273.15 = 18,167.75 \, \text{J/mol} \] 4. **Calculate the Denominator:** \[ \pi \times 0.222 \approx 0.697 \, \text{kg/mol} \] 5. **Divide Numerator by Denominator:** \[ \frac{18,167.75}{0.697} \approx 26,052.56 \, \text{m}^2/\text{s}^2 \] 6. **Take the Square Root:** \[ v_{\text{avg}} = \sqrt{26,052.56} \approx 161.4 \, \text{m/s} \] **Final Answer:** The mean molecular speed of radon gas is approximately **161 m/s**. |
167.0 |
| gpto1 | To compute the mean molecular speed \( v \) of radon gas (Rn), we'll use the formula for the mean speed of gas molecules in an ideal gas. The mean speed of molecules \( v_{\text{mean}} \) is given by: \[ v_{\text{mean}} = \sqrt{\dfrac{8kT}{\pi m}} \] where: - \( k \) is the Boltzmann constant (\( 1.380649 \times 10^{-23} \ \text{J/K} \)), - \( T \) is the absolute temperature in Kelvin (we'll assume standard temperature \( T = 273.15 \ \text{K} \) unless specified), - \( m \) is the mass of one molecule of the gas in kilograms, - \( \pi \) is the mathematical constant Pi (~3.1416). **Step 1: Calculate the mass of a radon atom.** The molar mass of radon is approximately \( M = 222 \ \text{g/mol} \). Convert grams per mole to kilograms per molecule: First, convert the molar mass to kilograms per mole: \[ M = 222 \ \text{g/mol} = 0.222 \ \text{kg/mol} \] Next, find the mass of one molecule \( m \) by dividing the molar mass \( M \) by Avogadro's number \( N_A = 6.02214076 \times 10^{23} \ \text{mol}^{-1} \): \[ m = \dfrac{M}{N_A} = \dfrac{0.222 \ \text{kg/mol}}{6.02214076 \times 10^{23} \ \text{mol}^{-1}} = 3.6865 \times 10^{-25} \ \text{kg} \] **Step 2: Calculate the mean molecular speed \( v_{\text{mean}} \).** Plugging the values into the formula: \[ v_{\text{mean}} = \sqrt{\dfrac{8kT}{\pi m}} = \sqrt{\dfrac{8 \times (1.380649 \times 10^{-23} \ \text{J/K}) \times 273.15 \ \text{K}}{\pi \times 3.6865 \times 10^{-25} \ \text{kg}}} \] Calculate the numerator: \[ \text{Numerator} = 8 \times 1.380649 \times 10^{-23} \ \text{J/K} \times 273.15 \ \text{K} = 3.0169 \times 10^{-20} \ \text{J} \] Calculate the denominator: \[ \text{Denominator} = \pi \times 3.6865 \times 10^{-25} \ \text{kg} = 1.1571 \times 10^{-24} \ \text{kg} \] Compute the ratio: \[ \dfrac{\text{Numerator}}{\text{Denominator}} = \dfrac{3.0169 \times 10^{-20} \ \text{J}}{1.1571 \times 10^{-24} \ \text{kg}} = 26065 \ \text{J/kg} \] Now, take the square root: \[ v_{\text{mean}} = \sqrt{26065 \ \text{J/kg}} \approx 161.5 \ \text{m/s} \] **Answer:** To compute the mean molecular speed \( v \) of radon gas (Rn), we use the formula for mean speed of gas molecules in an ideal gas: \[ v = \sqrt{\dfrac{8kT}{\pi m}} \] where: - \( k = 1.380649 \times 10^{-23} \ \text{J/K} \) (Boltzmann constant), - \( T = 273.15 \ \text{K} \) (assuming standard temperature), - \( m \) is the mass of one radon atom. First, find the mass \( m \) of a radon atom: \[ m = \dfrac{M}{N_A} = \dfrac{0.222 \ \text{kg/mol}}{6.02214076 \times 10^{23} \ \text{mol}^{-1}} = 3.6865 \times 10^{-25} \ \text{kg} \] Then, calculate the mean molecular speed: \[ v = \sqrt{\dfrac{8 \times 1.380649 \times 10^{-23} \ \text{J/K} \times 273.15 \ \text{K}}{\pi \times 3.6865 \times 10^{-25} \ \text{kg}}} \approx 161.5 \ \text{m/s} \] **Therefore, the mean molecular speed \( v \) in radon gas is approximately 161.5 m/s.** |
167.0 |
| GPT-4o | To compute the mean molecular speed (\( \bar{v} \)) of radon (Rn) gas, we can use the formula derived from the kinetic theory of gases. The mean molecular speed is given by: \[ \bar{v} = \sqrt{\frac{8RT}{\pi M}} \] Where: - \( R \) is the universal gas constant, \( R = 8.314 \, \text{J/(mol·K)} \) - \( T \) is the absolute temperature in Kelvin (K) - \( M \) is the molar mass of the gas in kilograms per mole (kg/mol) The molar mass of radon (Rn) is approximately \( 222 \, \text{g/mol} \). To convert this to kilograms per mole, we divide by 1000: \[ M = \frac{222 \, \text{g/mol}}{1000} = 0.222 \, \text{kg/mol} \] Let's assume a standard room temperature of \( 298 \, \text{K} \) (approximately 25°C) for this calculation. Plugging in the values: \[ \bar{v} = \sqrt{\frac{8 \times 8.314 \, \text{J/(mol·K)} \times 298 \, \text{K}}{\pi \times 0.222 \, \text{kg/mol}}} \] Next, perform the calculations step-by-step. 1. Compute the numerator: \[ 8 \times 8.314 \times 298 = 19807.504 \, \text{J·K/mol} \] 2. Compute the denominator: \[ \pi \times 0.222 = 0.697 \, \text{kg/mol} \] 3. Divide the numerator by the denominator: \[ \frac{19807.504}{0.697} = 28419.38 \] 4. Take the square root of the result: \[ \sqrt{28419.38} \approx 168.53 \, \text{m/s} \] Therefore, the mean molecular speed (\( \bar{v} \)) of radon (Rn) at 298 K is approximately \( 168.53 \, \text{m/s} \). |
167.0 |
| llama3.1-70B-Instruct | To compute the mean molecular speed \(v\) of radon (Rn), we can use the formula derived from the kinetic theory of gases, which relates the mean molecular speed to the temperature and the molar mass of the gas. The formula is: \[v = \sqrt{\frac{8RT}{\pi M}}\] where: - \(v\) is the mean molecular speed, - \(R\) is the gas constant (\(8.3145 \, \text{J/mol·K}\)), - \(T\) is the temperature in Kelvin, - \(M\) is the molar mass of the gas in kg/mol. Radon (Rn) has a molar mass of approximately \(222 \, \text{g/mol}\) or \(0.222 \, \text{kg/mol}\). Assuming we are calculating the mean molecular speed at room temperature, which is typically taken as \(20 \, \text{°C}\) or \(293 \, \text{K}\), we can substitute these values into the formula: \[v = \sqrt{\frac{8 \times 8.3145 \, \text{J/mol·K} \times 293 \, \text{K}}{\pi \times 0.222 \, \text{kg/mol}}}\] \[v = \sqrt{\frac{8 \times 8.3145 \times 293}{\pi \times 0.222}}\] \[v = \sqrt{\frac{19451.036}{0.6984}}\] \[v = \sqrt{27855.3}\] \[v \approx 166.9 \, \text{m/s}\] Therefore, the mean molecular speed of radon (Rn) at room temperature is approximately \(166.9 \, \text{m/s}\). |
167.0 |
| llama3.1-405B-Instruct | To compute the mean molecular speed of radon (Rn) gas, we can use the formula: v = √(8RT/πM) where: v = mean molecular speed (in m/s) R = gas constant (8.3145 J/mol·K) T = temperature (in Kelvin) M = molar mass of radon (in kg/mol) The molar mass of radon is 222 g/mol, which is equivalent to 0.222 kg/mol. Assuming a temperature of 298 K (25°C), we can plug in the values: v = √(8 × 8.3145 J/mol·K × 298 K / (π × 0.222 kg/mol)) = √(1991.45 m²/s²) = 44.67 m/s Therefore, the mean molecular speed of radon gas at 25°C is approximately 44.67 m/s. |
167.0 |
| DeepSeek-Coder-V2 | To compute the mean molecular speed \( v \) in the heavy gas radon (Rn) in m/s, we can use the Maxwell-Boltzmann distribution for molecular speeds. The mean speed \( v \) can be calculated using the following formula: \[ v = \sqrt{\frac{8k_B T}{\pi m}} \] where: - \( k_B \) is the Boltzmann constant (\( 1.380649 \times 10^{-23} \) J/K), - \( T \) is the temperature in Kelvin (K), - \( m \) is the mass of a radon atom in kilograms (kg). First, we need to determine the mass of a radon atom. Radon has an atomic number of 86 and an atomic mass of approximately 222 u (unified atomic mass units). The mass of a radon atom in kilograms is: \[ m = 222 \times 1.66053906660 \times 10^{-27} \, \text{kg} \] \[ m \approx 3.6864 \times 10^{-25} \, \text{kg} \] Next, we need to choose a temperature \( T \) in Kelvin. For simplicity, let's use room temperature, which is approximately 298 K. Now, we can substitute the values into the formula for the mean speed: \[ v = \sqrt{\frac{8 \times 1.380649 \times 10^{-23} \, \text{J/K} \times 298 \, \text{K}}{\pi \times 3.6864 \times 10^{-25} \, \text{kg}}} \] First, calculate the numerator: \[ 8 \times 1.380649 \times 10^{-23} \times 298 \approx 3.307 \times 10^{-20} \, \text{J} \] Next, calculate the denominator: \[ \pi \times 3.6864 \times 10^{-25} \approx 1.158 \times 10^{-24} \, \text{kg} \] Now, divide the numerator by the denominator: \[ \frac{3.307 \times 10^{-20} \, \text{J}}{1.158 \times 10^{-24} \, \text{kg}} \approx 2.856 \times 10^4 \, \text{m/s} \] Finally, take the square root: \[ v \approx \sqrt{2.856 \times 10^4} \approx 169 \, \text{m/s} \] Therefore, the mean molecular speed \( v \) in the heavy gas radon (Rn) at room temperature is approximately: \[ \boxed{169} \] |
167.0 |
| Qwen2-72B-Instruct | The mean molecular speed (v) of a gas can be calculated using the formula: v = sqrt((8 * R * T) / (pi * M)) where, R is the gas constant (8.314 J/(mol*K)) T is the absolute temperature (in Kelvin) M is the molar mass of the gas (in kg/mol) pi is the mathematical constant (approximately 3.14159) For Radon (Rn), the molar mass is approximately 222.0176 g/mol or 0.2220176 kg/mol. Let's assume the temperature is at standard temperature, which is 273.15 K. Substitute the values into the formula: v = sqrt((8 * 8.314 J/(mol*K) * 273.15 K) / (3.14159 * 0.2220176 kg/mol)) v = sqrt((17494.464 J/K) / (0.698767 kg)) v = sqrt(24903.026 m²/s²) v ≈ 157.81 m/s So the mean molecular speed of Radon at standard temperature is approximately 157.81 m/s. Please note that this is a simplified calculation. Real-world conditions may vary, and the mean molecular speed can be affected by factors such as pressure and the specific conditions of the gas. |
167.0 |
| baichuan4 | To compute the mean molecular speed \(v\) in the heavy gas radon (Rn), we can use the formula: \[ v = \sqrt{\frac{8 R T}{\pi M}} \] where: - \(R\) is the gas constant, approximately \(8.314 \, \text{J/(mol}\cdot\text{K)}\), - \(T\) is the temperature in Kelvin, - \(M\) is the molar mass of the gas in kilograms per mole. Let's assume the temperature \(T\) is given, and we know that the molar mass of radon (Rn) is approximately \(222 \, \text{g/mol}\) or \(0.222 \, \text{kg/mol}\). Given these values, we can calculate the mean molecular speed \(v\) for radon: \[ v = \sqrt{\frac{8 \times 8.314 \, \text{J/(mol}\cdot\text{K)} \times T}{\pi \times 0.222 \, \text{kg/mol}}} \] If we plug in a specific temperature, for example, \(T = 293.15 \, \text{K}\) (which is approximately room temperature), we get: \[ v = \sqrt{\frac{8 \times 8.314 \times 293.15}{\pi \times 0.222}} \] \[ v = \sqrt{\frac{18887.414}{\pi \times 0.222}} \] \[ v = \sqrt{\frac{18887.414}{0.701}} \] \[ v = \sqrt{26943.532} \] \[ v \approx 164.1 \, \text{m/s} \] So, the mean molecular speed of radon gas at room temperature would be approximately \(164.1 \, \text{m/s}\). |
167.0 |
| doubao | The mean molecular speed (v) can be calculated using the formula: v = √(8RT/πM), where R is the ideal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and M is the molar mass of the gas. The molar mass of radon (Rn) is approximately 222 g/mol. To calculate the mean molecular speed, you need to know the specific temperature at which you want to find the speed. Once you have the temperature, substitute the values of R, T, and M into the formula and calculate the square root to obtain the mean molecular speed in meters per second. |
167.0 |
| minimax | To compute the mean molecular speed \( v \) of a gas, we can use the Maxwell-Boltzmann distribution, which gives the distribution of speeds for a gas at a certain temperature. The mean molecular speed can be calculated using the following equation: \[ v = \sqrt{\frac{8RT}{\pi M}} \] where: - \( v \) is the mean molecular speed, - \( R \) is the universal gas constant (8.314 J/(mol·K)), - \( T \) is the temperature in Kelvin, - \( M \) is the molar mass of the gas in kg/mol. For radon (Rn), we need to know its molar mass and the temperature at which we want to calculate the mean molecular speed. Radon is a heavy noble gas with an atomic mass of approximately 222 g/mol, which is its molar mass since one mole of radon atoms has a mass of 222 grams. Let's assume we want to calculate the mean molecular speed at room temperature, which is approximately 298 K (25°C). First, convert the molar mass of radon from grams per mole to kilograms per mole: \[ M = 222 \, \text{g/mol} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.222 \, \text{kg/mol} \] Now, plug in the values into the equation: \[ v = \sqrt{\frac{8 \times 8.314 \, \text{J/(mol·K)} \times 298 \, \text{K}}{\pi \times 0.222 \, \text{kg/mol}}} \] \[ v = \sqrt{\frac{8 \times 8.314 \times 298}{\pi \times 0.222}} \] \[ v = \sqrt{\frac{20167.68}{0.7002}} \] \[ v = \sqrt{28798.11} \] \[ v \approx 169.7 \, \text{m/s} \] So, the mean molecular speed of radon at room temperature (298 K) is approximately 169.7 meters per second. |
167.0 |
| kimi | To compute the mean molecular speed of a gas, such as radon (Rn), we can use the equation derived from the kinetic theory of gases. The mean molecular speed (v) of a gas is given by the following equation: \[ v = \sqrt{\frac{2kT}{m}} \] where: - \( k \) is the Boltzmann constant (\(1.380649 \times 10^{-23}\) J/K) - \( T \) is the temperature of the gas in Kelvin - \( m \) is the mass of a single molecule of the gas First, we need to find the molar mass of radon. Radon has an atomic mass of approximately 222 u (atomic mass units). Since 1 u is approximately equal to \(1.660539 \times 10^{-27}\) kg, the mass of a single molecule of radon is: \[ m = 222 \times 1.660539 \times 10^{-27} \text{ kg} \] Now, let's assume the temperature \( T \) of the gas is given. If it's not, we'll need to use a standard temperature, such as room temperature, which is approximately 298 K. Using the values above, we can calculate the mean molecular speed: \[ v = \sqrt{\frac{2 \times 1.380649 \times 10^{-23} \text{ J/K} \times T}{222 \times 1.660539 \times 10^{-27} \text{ kg}}} \] Let's plug in the values: \[ v = \sqrt{\frac{2 \times 1.380649 \times 10^{-23} \times T}{222 \times 1.660539 \times 10^{-27}}} \] \[ v = \sqrt{\frac{2.761298 \times 10^{-23} \times T}{368.53 \times 10^{-27}}} \] \[ v = \sqrt{\frac{2.761298 \times T}{368.53}} \] \[ v = \sqrt{7.5 \times 10^{-23} \times T} \] If we assume \( T = 298 \) K (room temperature), then: \[ v = \sqrt{7.5 \times 10^{-23} \times 298} \] \[ v = \sqrt{2.235 \times 10^{-20}} \] \[ v \approx 1.5 \times 10^5 \text{ m/s} \] So, the mean molecular speed of radon at room temperature is approximately \(1.5 \times 10^5\) m/s. |
167.0 |
| wenxinyiyan | 要计算重气体氡（Rn）的平均分子速度v（以m/s为单位），我们需要使用平均分子速度的公式，该公式与气体的摩尔质量M（以kg/mol为单位）和温度T（以开尔文K为单位）有关。平均分子速度v的公式为： v = √(8RT/πM) 其中R是通用气体常数，其值为8.314 J/(mol·K)。 氡（Rn）的摩尔质量M约为222 g/mol，即0.222 kg/mol。假设我们在标准温度T = 273.15 K（即0°C）下计算。 将这些值代入上述公式，我们可以计算出氡的平均分子速度。 计算结果为：v ≈ 161.40 m/s 所以，在标准温度下，氡（Rn）的平均分子速度约为161.40 m/s。 |
167.0 |
| zhipu | To compute the mean molecular speed \( v \) of radon gas (Rn), we can use the equation derived from the kinetic theory of gases for the average speed of molecules: \[ v = \sqrt{\frac{8k_B T}{\pi m}} \] where: - \( k_B \) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\)), - \( T \) is the absolute temperature in Kelvin, - \( m \) is the mass of a single molecule of radon. The mass of a single molecule of radon can be calculated from the molar mass of radon (\( M \)) and Avogadro's number (\( N_A \)): \[ m = \frac{M}{N_A} \] For radon (Rn), the molar mass \( M \) is approximately \( 222 \, \text{g/mol} \) (or \( 222 \times 10^{-3} \, \text{kg/mol} \)), and Avogadro's number \( N_A \) is \( 6.022 \times 10^{23} \, \text{mol}^{-1} \). Let's plug in the values: 1. Calculate the mass of a single radon molecule: \[ m = \frac{222 \times 10^{-3} \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 3.69 \times 10^{-25} \, \text{kg} \] 2. Assuming a standard temperature \( T = 298 \, \text{K} \) (25°C), we can now calculate the mean molecular speed: \[ v = \sqrt{\frac{8 \times 1.38 \times 10^{-23} \, \text{J/K} \times 298 \, \text{K}}{\pi \times 3.69 \times 10^{-25} \, \text{kg}}} \] 3. Simplify the expression under the square root: \[ v = \sqrt{\frac{8 \times 1.38 \times 10^{-23} \times 298}{\pi \times 3.69 \times 10^{-25}}} \] 4. Calculate the numerical value: \[ v \approx \sqrt{\frac{3.29 \times 10^{-20}}{1.15 \times 10^{-24}}} \] \[ v \approx \sqrt{2.86 \times 10^{4}} \] \[ v \approx 169 \, \text{m/s} \] Therefore, the mean molecular speed of radon gas at 298 K is approximately \( 169 \, \text{m/s} \). Note: The actual value can vary slightly depending on the precision of the constants used and the temperature chosen. Adjust the temperature \( T \) as needed for different conditions. |
167.0 |
| gpto1 | To compute the mean molecular speed \( v \) of radon gas (Rn), we'll use the formula for the mean speed of gas molecules in an ideal gas. The mean speed of molecules \( v_{\text{mean}} \) is given by: \[ v_{\text{mean}} = \sqrt{\dfrac{8kT}{\pi m}} \] where: - \( k \) is the Boltzmann constant (\( 1.380649 \times 10^{-23} \ \text{J/K} \)), - \( T \) is the absolute temperature in Kelvin (we'll assume standard temperature \( T = 273.15 \ \text{K} \) unless specified), - \( m \) is the mass of one molecule of the gas in kilograms, - \( \pi \) is the mathematical constant Pi (~3.1416). **Step 1: Calculate the mass of a radon atom.** The molar mass of radon is approximately \( M = 222 \ \text{g/mol} \). Convert grams per mole to kilograms per molecule: First, convert the molar mass to kilograms per mole: \[ M = 222 \ \text{g/mol} = 0.222 \ \text{kg/mol} \] Next, find the mass of one molecule \( m \) by dividing the molar mass \( M \) by Avogadro's number \( N_A = 6.02214076 \times 10^{23} \ \text{mol}^{-1} \): \[ m = \dfrac{M}{N_A} = \dfrac{0.222 \ \text{kg/mol}}{6.02214076 \times 10^{23} \ \text{mol}^{-1}} = 3.6865 \times 10^{-25} \ \text{kg} \] **Step 2: Calculate the mean molecular speed \( v_{\text{mean}} \).** Plugging the values into the formula: \[ v_{\text{mean}} = \sqrt{\dfrac{8kT}{\pi m}} = \sqrt{\dfrac{8 \times (1.380649 \times 10^{-23} \ \text{J/K}) \times 273.15 \ \text{K}}{\pi \times 3.6865 \times 10^{-25} \ \text{kg}}} \] Calculate the numerator: \[ \text{Numerator} = 8 \times 1.380649 \times 10^{-23} \ \text{J/K} \times 273.15 \ \text{K} = 3.0169 \times 10^{-20} \ \text{J} \] Calculate the denominator: \[ \text{Denominator} = \pi \times 3.6865 \times 10^{-25} \ \text{kg} = 1.1571 \times 10^{-24} \ \text{kg} \] Compute the ratio: \[ \dfrac{\text{Numerator}}{\text{Denominator}} = \dfrac{3.0169 \times 10^{-20} \ \text{J}}{1.1571 \times 10^{-24} \ \text{kg}} = 26065 \ \text{J/kg} \] Now, take the square root: \[ v_{\text{mean}} = \sqrt{26065 \ \text{J/kg}} \approx 161.5 \ \text{m/s} \] **Answer:** To compute the mean molecular speed \( v \) of radon gas (Rn), we use the formula for mean speed of gas molecules in an ideal gas: \[ v = \sqrt{\dfrac{8kT}{\pi m}} \] where: - \( k = 1.380649 \times 10^{-23} \ \text{J/K} \) (Boltzmann constant), - \( T = 273.15 \ \text{K} \) (assuming standard temperature), - \( m \) is the mass of one radon atom. First, find the mass \( m \) of a radon atom: \[ m = \dfrac{M}{N_A} = \dfrac{0.222 \ \text{kg/mol}}{6.02214076 \times 10^{23} \ \text{mol}^{-1}} = 3.6865 \times 10^{-25} \ \text{kg} \] Then, calculate the mean molecular speed: \[ v = \sqrt{\dfrac{8 \times 1.380649 \times 10^{-23} \ \text{J/K} \times 273.15 \ \text{K}}{\pi \times 3.6865 \times 10^{-25} \ \text{kg}}} \approx 161.5 \ \text{m/s} \] **Therefore, the mean molecular speed \( v \) in radon gas is approximately 161.5 m/s.** |
167.0 |
| deepseekv3 | To compute the mean molecular speed \( v \) of radon (Rn) gas, we use the following formula derived from the kinetic theory of gases: \[ v = \sqrt{\frac{8RT}{\pi M}} \] Where: - \( v \) = mean molecular speed (m/s) - \( R \) = universal gas constant = \( 8.314 \, \text{J/mol·K} \) - \( T \) = temperature in Kelvin (K) - \( M \) = molar mass of the gas in kg/mol ### Step 1: Determine the molar mass of radon (Rn) The molar mass of radon (Rn) is approximately \( 222 \, \text{g/mol} \). Convert this to kg/mol: \[ M = 222 \, \text{g/mol} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.222 \, \text{kg/mol} \] ### Step 2: Assume a temperature The problem does not specify a temperature, so we assume standard temperature \( T = 298 \, \text{K} \). ### Step 3: Plug values into the formula \[ v = \sqrt{\frac{8 \times 8.314 \times 298}{\pi \times 0.222}} \] ### Step 4: Calculate the result \[ v = \sqrt{\frac{8 \times 8.314 \times 298}{3.1416 \times 0.222}} \] \[ v = \sqrt{\frac{19811.792}{0.698}} \] \[ v = \sqrt{28383.6} \] \[ v \approx 168.5 \, \text{m/s} \] Thus, the mean molecular speed of radon gas at \( 298 \, \text{K} \) is approximately \( 168.5 \, \text{m/s} \). **Note:** If the temperature is different, substitute the appropriate value for \( T \) in the formula. |
167.0 |