| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The mean molecular speed (also known as the average speed) of a gas can be found using the formula derived from the kinetic theory of gases: \[ v_{avg} = \sqrt{\frac{8RT}{\pi M}} \] where: - \(v_{avg}\) is the average speed of the molecules, - \(R\) is the ideal 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 hydrogen (\(H_2\)), the molar mass is approximately \(0.002\) kg/mol. Let's assume the temperature is \(300\) K (a typical room temperature). Substituting these values into the formula: \[ v_{avg} = \sqrt{\frac{8 \times 8.314 \times 300}{\pi \times 0.002}} \] \[ v_{avg} = \sqrt{\frac{199536}{0.006283}} \] \[ v_{avg} = \sqrt{31767198.23} \] \[ v_{avg} \approx 5636 \, \text{m/s} \] Thus, the average speed of the molecules in hydrogen gas at 300 K is approximately 5636 m/s. This is a rough estimate and assumes ideal gas behavior. Actual values can vary slightly due to factors such as intermolecular forces and the specific conditions of the gas. |
1750.0 |
| deepseekr1 | To compute the mean molecular speed \( v \) of hydrogen gas (H₂) at standard temperature (273.15 K), we use the formula for the average speed of gas molecules: \[ v = \sqrt{\frac{8RT}{\pi M}} \] Where: - \( R \) is the ideal gas constant (\( 8.314 \, \text{J/(mol·K)} \)), - \( T \) is the temperature in Kelvin (\( 273.15 \, \text{K} \)), - \( M \) is the molar mass of hydrogen gas (\( 0.00201568 \, \text{kg/mol} \)). **Step-by-Step Calculation:** 1. **Calculate the numerator \( 8RT \):** \[ 8 \times 8.314 = 66.512 \] \[ 66.512 \times 273.15 \approx 18167.7528 \] 2. **Calculate the denominator \( \pi M \):** \[ \pi \times 0.00201568 \approx 0.00633237 \] 3. **Divide the numerator by the denominator:** \[ \frac{18167.7528}{0.00633237} \approx 2869029.3 \] 4. **Take the square root of the result:** \[ \sqrt{2869029.3} \approx 1693.82 \, \text{m/s} \] Rounding to four significant figures, the mean molecular speed of hydrogen gas is: \[ \boxed{1694} \, \text{m/s} \] |
1750.0 |
| gpto1 | To compute the mean molecular speed \( v \) of hydrogen gas (H₂), we will use the kinetic theory of gases, which relates the average speed of gas molecules to the temperature and mass of the molecules. **Assumptions:** - The gas behaves ideally. - The temperature \( T \) is taken as 300 K (approximately room temperature). **Constants:** - Boltzmann constant: \( k = 1.380649 \times 10^{-23} \, \text{J/K} \) - Avogadro's number: \( N_A = 6.02214076 \times 10^{23} \, \text{mol}^{-1} \) - Molar mass of hydrogen gas (H₂): \( M = 2.01588 \times 10^{-3} \, \text{kg/mol} \) - Pi: \( \pi = 3.1415926535 \) **Steps:** 1. **Calculate the mass \( m \) of one hydrogen molecule:** \[ m = \frac{M}{N_A} = \frac{2.01588 \times 10^{-3} \, \text{kg/mol}}{6.02214076 \times 10^{23} \, \text{mol}^{-1}} \approx 3.34626 \times 10^{-27} \, \text{kg} \] 2. **Use the formula for the mean (average) molecular speed:** \[ v_{\text{avg}} = \sqrt{\frac{8kT}{\pi m}} \] 3. **Compute the numerator:** \[ \text{Numerator} = 8kT = 8 \times (1.380649 \times 10^{-23} \, \text{J/K}) \times 300 \, \text{K} = 3,313.5576 \times 10^{-23} \, \text{J} \] 4. **Compute the denominator:** \[ \text{Denominator} = \pi m = 3.1415926535 \times (3.34626 \times 10^{-27} \, \text{kg}) = 10.5148147 \times 10^{-27} \, \text{kg} \] 5. **Divide the numerator by the denominator:** \[ \frac{\text{Numerator}}{\text{Denominator}} = \frac{3,313.5576 \times 10^{-23} \, \text{J}}{10.5148147 \times 10^{-27} \, \text{kg}} = 315.34375 \times 10^{4} \, \text{m}^2/\text{s}^2 = 3,153,437.5 \, \text{m}^2/\text{s}^2 \] 6. **Compute the mean molecular speed \( v_{\text{avg}} \):** \[ v_{\text{avg}} = \sqrt{\frac{8kT}{\pi m}} = \sqrt{3,153,437.5 \, \text{m}^2/\text{s}^2} \approx 1,775 \, \text{m/s} \] **Answer:** The mean molecular speed \( v \) of hydrogen gas (H₂) at 300 K is approximately **1,775 meters per second**. |
1750.0 |
| GPT-4o | To compute the mean molecular speed \( v \) of a gas, we can use the formula from the kinetic theory of gases. The mean molecular speed, specifically the root mean square speed \( v_{rms} \), is given by: \[ v_{rms} = \sqrt{\frac{3k_BT}{m}} \] where: - \( k_B \) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\)) - \( T \) is the absolute temperature in Kelvin (K) - \( m \) is the mass of a single molecule of the gas in kilograms (kg) For hydrogen gas (H\(_2\)): - The molar mass of hydrogen \( M \) is approximately \( 2.016 \, \text{g/mol} \). - To convert the molar mass to the mass of a single molecule, we use Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \): \[ m = \frac{M}{N_A} = \frac{2.016 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \] \[ m = \frac{2.016 \times 10^{-3} \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \] \[ m \approx 3.348 \times 10^{-27} \, \text{kg} \] Now, we need the temperature \( T \) to proceed. Let's assume we are calculating the mean molecular speed at room temperature, which is approximately \( 298 \, \text{K} \). Plugging the values into the formula: \[ v_{rms} = \sqrt{\frac{3 \cdot (1.38 \times 10^{-23} \, \text{J/K}) \cdot 298 \, \text{K}}{3.348 \times 10^{-27} \, \text{kg}}} \] \[ v_{rms} = \sqrt{\frac{3 \cdot 1.38 \times 10^{-23} \times 298}{3.348 \times 10^{-27}}} \] \[ v_{rms} = \sqrt{\frac{1.23492 \times 10^{-20}}{3.348 \times 10^{-27}}} \] \[ v_{rms} = \sqrt{3.686 \times 10^{6}} \] \[ v_{rms} \approx 1920 \, \text{m/s} \] So, the mean molecular speed \( v_{rms} \) of hydrogen (H\(_2\)) at room temperature (298 K) is approximately \( 1920 \, \text{m/s} \). |
1750.0 |
| llama3.1-70B-Instruct | To compute the mean molecular speed \(v\) of hydrogen gas (H2), we can use the formula derived from the kinetic theory of gases, which relates the mean molecular speed to the temperature and the mass of the molecules. 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. For hydrogen gas (H2), the molar mass \(M\) is approximately \(2.016 \, \text{g/mol}\) or \(2.016 \times 10^{-3} \, \text{kg/mol}\). Let's assume we want to calculate the mean molecular speed at room temperature, which is approximately \(20 \, \text{°C}\) or \(293 \, \text{K}\). Plugging the values into the formula: \[v = \sqrt{\frac{8 \times 8.3145 \, \text{J/mol·K} \times 293 \, \text{K}}{\pi \times 2.016 \times 10^{-3} \, \text{kg/mol}}}\] \[v = \sqrt{\frac{8 \times 8.3145 \times 293}{\pi \times 2.016 \times 10^{-3}}}\] \[v = \sqrt{\frac{19451.372}{6.321 \times 10^{-3}}}\] \[v = \sqrt{3075511.9}\] \[v \approx 1754.4 \, \text{m/s}\] Therefore, the mean molecular speed of hydrogen gas (H2) at room temperature (approximately \(20 \, \text{°C}\) or \(293 \, \text{K}\)) is about \(1754.4 \, \text{m/s}\). |
1750.0 |
| llama3.1-405B-Instruct | To compute the mean molecular speed v in hydrogen gas (H2), we can use the formula: v = √(8RT/πM) where: v = mean molecular speed (m/s) R = gas constant (8.3145 J/mol·K) T = temperature (K) M = molar mass of hydrogen (2.016 g/mol) First, we need to convert the molar mass from g/mol to kg/mol: M = 2.016 g/mol × (1 kg / 1000 g) = 0.002016 kg/mol Now, we can plug in the values: v = √(8 × 8.3145 J/mol·K × T / (π × 0.002016 kg/mol)) To find the mean molecular speed at a specific temperature, we need to know the temperature. Let's assume we want to find the mean molecular speed at room temperature, which is approximately 293 K (20°C). v = √(8 × 8.3145 J/mol·K × 293 K / (π × 0.002016 kg/mol)) = √(1.344 × 10^6 m^2/s^2) ≈ 1160 m/s Therefore, the mean molecular speed of hydrogen gas (H2) at room temperature (293 K) is approximately 1160 m/s. |
1750.0 |
| DeepSeek-Coder-V2 | To compute the mean molecular speed \( v \) in the light gas hydrogen (H2) in m/s, we can use the following formula for the mean speed of gas molecules: \[ v = \sqrt{\frac{8k_BT}{\pi m}} \] where: - \( k_B \) is the Boltzmann constant, \( 1.380649 \times 10^{-23} \, \text{J/K} \) - \( T \) is the temperature of the gas in Kelvin (K) - \( m \) is the mass of a single hydrogen molecule in kilograms (kg) First, let's determine the mass of a single hydrogen molecule. A hydrogen molecule (H2) consists of two hydrogen atoms. The atomic mass of hydrogen is approximately 1.008 atomic mass units (u). Therefore, the mass of a hydrogen molecule is: \[ m = 2 \times 1.008 \, \text{u} = 2.016 \, \text{u} \] To convert this mass to kilograms, we use the conversion factor \( 1 \, \text{u} = 1.660539066 \times 10^{-27} \, \text{kg} \): \[ m = 2.016 \, \text{u} \times 1.660539066 \times 10^{-27} \, \text{kg/u} = 3.346 \times 10^{-27} \, \text{kg} \] 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: \[ v = \sqrt{\frac{8 \times 1.380649 \times 10^{-23} \, \text{J/K} \times 298 \, \text{K}}{\pi \times 3.346 \times 10^{-27} \, \text{kg}}} \] First, calculate the numerator: \[ 8 \times 1.380649 \times 10^{-23} \times 298 = 3.31 \times 10^{-20} \, \text{J} \] Then, calculate the denominator: \[ \pi \times 3.346 \times 10^{-27} = 1.05 \times 10^{-26} \, \text{kg} \] Now, divide the numerator by the denominator: \[ \frac{3.31 \times 10^{-20} \, \text{J}}{1.05 \times 10^{-26} \, \text{kg}} = 3.15 \times 10^6 \, \text{m}^2/\text{s}^2 \] Finally, take the square root to find the mean molecular speed: \[ v = \sqrt{3.15 \times 10^6 \, \text{m}^2/\text{s}^2} = 1.77 \times 10^3 \, \text{m/s} \] Therefore, the mean molecular speed \( v \) in hydrogen gas at room temperature is approximately: \[ \boxed{1770} \, \text{m/s} \] |
1750.0 |
| Qwen2-72B-Instruct | The mean molecular speed (also known as the average speed) of a gas can be found using the formula derived from the kinetic theory of gases: \[ v_{avg} = \sqrt{\frac{8RT}{\pi M}} \] where: - \(v_{avg}\) is the average speed of the molecules, - \(R\) is the ideal 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 hydrogen (\(H_2\)), the molar mass is approximately \(0.002\) kg/mol. Let's assume the temperature is \(300\) K (a typical room temperature). Substituting these values into the formula: \[ v_{avg} = \sqrt{\frac{8 \times 8.314 \times 300}{\pi \times 0.002}} \] \[ v_{avg} = \sqrt{\frac{199536}{0.006283}} \] \[ v_{avg} = \sqrt{31767198.23} \] \[ v_{avg} \approx 5636 \, \text{m/s} \] Thus, the average speed of the molecules in hydrogen gas at 300 K is approximately 5636 m/s. This is a rough estimate and assumes ideal gas behavior. Actual values can vary slightly due to factors such as intermolecular forces and the specific conditions of the gas. |
1750.0 |
| baichuan4 | To compute the mean molecular speed (v) of hydrogen gas (H2) molecules at a certain temperature, we can use the kinetic theory of gases. The mean molecular speed is given by the formula: \[ v = \sqrt{\frac{8 R T}{\pi M}} \] where: - \( R \) is the ideal gas constant, which is approximately \( 8.31 \, \text{J/(mol·K)} \), - \( T \) is the absolute temperature in Kelvin, - \( M \) is the molar mass of the gas in kilograms per mole. For hydrogen gas (H2), the molar mass \( M \) is approximately \( 0.002 \, \text{kg/mol} \). Let's assume the temperature \( T \) is not provided, so we will leave it as a variable. Let's plug in the values: \[ v = \sqrt{\frac{8 \times 8.31 \, \text{J/(mol·K)} \times T}{\pi \times 0.002 \, \text{kg/mol}}} \] \[ v = \sqrt{\frac{66.48 \, \text{J/(mol·K)} \times T}{\pi \times 0.002}} \] \[ v = \sqrt{\frac{66.48 \, \text{J/(mol·K)} \times T}{0.006283185307179586}} \] \[ v = \sqrt{10580.64 \times T} \] Now, if we want the mean molecular speed at a specific temperature, say \( T = 298 \, \text{K} \) (room temperature), we would calculate: \[ v = \sqrt{10580.64 \times 298} \] \[ v = \sqrt{315064.752} \] \[ v \approx 561.1 \, \text{m/s} \] So, the mean molecular speed of hydrogen gas at room temperature (298 K) is approximately 561.1 m/s. If you want the speed at a different temperature, replace the \( T \) value in the formula with the desired temperature in Kelvin and calculate accordingly. |
1750.0 |
| doubao | The mean molecular speed (v) can be calculated using the following equation: 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. For hydrogen (H₂), the molar mass M is 2.016 g/mol = 0.002016 kg/mol. Let's assume a specific temperature T. Then we can plug in the values and calculate the mean molecular speed. For example, if T = 300 K: v = √(8 * 8.314 * 300 / (π * 0.002016)) ≈ 1934.7 m/s. The actual value of the mean molecular speed will depend on the specific temperature. |
1750.0 |
| minimax | To compute the mean molecular speed \( v \) of hydrogen gas (H2) molecules at a given temperature, we can use the following equation derived from the Maxwell-Boltzmann distribution, which is valid for an ideal gas: \[ v = \sqrt{\frac{8RT}{\pi M}} \] where: - \( v \) is the mean molecular speed (in meters per second, m/s), - \( R \) is the universal gas constant (8.314 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). For hydrogen gas (H2), the molar mass \( M \) is approximately 0.002016 kg/mol (since the atomic mass of hydrogen is approximately 1 g/mol, and H2 has two hydrogen atoms). Let's assume we want to calculate the mean molecular speed at room temperature, which is approximately 298 K (25°C). Plugging in the values: \[ v = \sqrt{\frac{8 \times 8.314 \, \text{J/(mol·K)} \times 298 \, \text{K}}{\pi \times 0.002016 \, \text{kg/mol}}} \] \[ v = \sqrt{\frac{8 \times 8.314 \times 298}{\pi \times 0.002016}} \] \[ v = \sqrt{\frac{20164.32}{0.***}} \] \[ v = \sqrt{3190.96} \] \[ v \approx 56.48 \, \text{m/s} \] So, the mean molecular speed of hydrogen gas at room temperature is approximately 56.48 meters per second. |
1750.0 |
| kimi | To compute the mean molecular speed of hydrogen (H2) gas, we can use the equation derived from the kinetic theory of gases. The mean molecular speed (v) is given by the following formula: \[ v = \sqrt{\frac{8kT}{\pi m}} \] where: - \( k \) is the Boltzmann constant (\(1.380649 \times 10^{-23}\) J/K) - \( T \) is the temperature of the gas in Kelvin (K) - \( m \) is the mass of a single molecule of the gas in kilograms (kg) For hydrogen gas (H2), the molar mass is approximately 2.016 g/mol, and since we are dealing with a diatomic molecule, the mass of a single molecule is half of that, which is approximately 1.008 g/mol. To convert this mass into kilograms, we divide by the molar mass constant (1 mol = 0.001 kg): \[ m_{H2} = \frac{2.016 \times 10^{-3} \text{ kg/mol}}{1 \text{ mol}} \] Now, let's assume the temperature \( T \) is given. For the sake of this calculation, let's assume \( T = 300 \) K, which is approximately room temperature. First, we convert the mass of a single molecule of H2 to kilograms: \[ m_{H2} = \frac{2.016 \times 10^{-3} \text{ kg/mol}}{1 \text{ mol}} \times \frac{1 \text{ mol}}{6.022 \times 10^{23} \text{ molecules}} \] \[ m_{H2} = 3.344 \times 10^{-27} \text{ kg} \] Now we can plug in the values into the formula: \[ v = \sqrt{\frac{8 \times 1.380649 \times 10^{-23} \text{ J/K} \times 300 \text{ K}}{\pi \times 3.344 \times 10^{-27} \text{ kg}}} \] \[ v = \sqrt{\frac{3.44 \times 10^{-20} \text{ J}}{\pi \times 3.344 \times 10^{-27} \text{ kg}}} \] \[ v = \sqrt{1.03 \times 10^{7} \text{ m}^2/\text{s}^2} \] \[ v \approx 3.21 \times 10^{3} \text{ m/s} \] So, the mean molecular speed of hydrogen gas at 300 K is approximately 3,210 m/s. |
1750.0 |
| wenxinyiyan | To compute the mean molecular speed (v) of hydrogen (H2) gas, we can use the following formula: v = √(8kT/πm) where: - k is the Boltzmann constant (1.38064852 × 10^-23 J/K), - T is the temperature in Kelvin (assuming room temperature, which is approximately 293.15 K), - m is the mass of the hydrogen molecule (H2). The mass of a hydrogen molecule (H2) is approximately 2.01588 g/mol, which can be converted to kilograms per mole (kg/mol) by dividing by 1000. Let's perform the calculation: v = √(8 * 1.38064852 × 10^-23 J/K * 293.15 K / (π * 2.01588 g/mol / 1000)) = √(8 * 1.38064852 × 10^-23 * 293.15 / (π * 2.01588 / 1000)) = √(3.99277176 × 10^-21 / (π * 2.01588 / 1000)) = √(3.99277176 × 10^-21 / (6.324555320336759 × 10^-3)) = √(6.31230587 × 10^-19) = 7.94502769 × 10^-10 m/s Therefore, the mean molecular speed of hydrogen (H2) gas at room temperature is approximately 7.95 × 10^-10 m/s. |
1750.0 |
| zhipu | To compute the mean molecular speed \( v \) of hydrogen gas (H\(_2\)), we can use the equation derived from the kinetic theory of gases. The mean molecular speed is given by: \[ v = \sqrt{\frac{8k_B T}{\pi m}} \] 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 hydrogen molecule. First, we need to determine the mass of a single hydrogen molecule (H\(_2\)). The molar mass of hydrogen (H\(_2\)) is approximately \(2.02 \, \text{g/mol}\). To find the mass of a single molecule, we use the following conversion: \[ m = \frac{\text{molar mass}}{N_A} \] where \( N_A \) is Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\)). \[ m = \frac{2.02 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \] \[ m = \frac{2.02 \times 10^{-3} \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \] \[ m \approx 3.34 \times 10^{-27} \, \text{kg} \] Now, we can plug the values into the mean molecular speed equation. Let's assume a standard temperature \( T = 298 \, \text{K} \) (25°C): \[ v = \sqrt{\frac{8 \times 1.38 \times 10^{-23} \, \text{J/K} \times 298 \, \text{K}}{\pi \times 3.34 \times 10^{-27} \, \text{kg}}} \] Calculate the numerator: \[ 8 \times 1.38 \times 10^{-23} \times 298 \approx 3.28 \times 10^{-20} \, \text{J} \] Calculate the denominator: \[ \pi \times 3.34 \times 10^{-27} \approx 1.05 \times 10^{-26} \, \text{kg} \] Now, divide the numerator by the denominator: \[ \frac{3.28 \times 10^{-20}}{1.05 \times 10^{-26}} \approx 3.12 \times 10^{6} \] Finally, take the square root: \[ v \approx \sqrt{3.12 \times 10^{6}} \approx 1767 \, \text{m/s} \] So, the mean molecular speed of hydrogen gas (H\(_2\)) at 298 K is approximately \( 1767 \, \text{m/s} \). Note: The actual value can vary slightly depending on the precision of the constants used and the temperature chosen. |
1750.0 |
| gpto1 | To compute the mean molecular speed \( v \) of hydrogen gas (H₂), we will use the kinetic theory of gases, which relates the average speed of gas molecules to the temperature and mass of the molecules. **Assumptions:** - The gas behaves ideally. - The temperature \( T \) is taken as 300 K (approximately room temperature). **Constants:** - Boltzmann constant: \( k = 1.380649 \times 10^{-23} \, \text{J/K} \) - Avogadro's number: \( N_A = 6.02214076 \times 10^{23} \, \text{mol}^{-1} \) - Molar mass of hydrogen gas (H₂): \( M = 2.01588 \times 10^{-3} \, \text{kg/mol} \) - Pi: \( \pi = 3.1415926535 \) **Steps:** 1. **Calculate the mass \( m \) of one hydrogen molecule:** \[ m = \frac{M}{N_A} = \frac{2.01588 \times 10^{-3} \, \text{kg/mol}}{6.02214076 \times 10^{23} \, \text{mol}^{-1}} \approx 3.34626 \times 10^{-27} \, \text{kg} \] 2. **Use the formula for the mean (average) molecular speed:** \[ v_{\text{avg}} = \sqrt{\frac{8kT}{\pi m}} \] 3. **Compute the numerator:** \[ \text{Numerator} = 8kT = 8 \times (1.380649 \times 10^{-23} \, \text{J/K}) \times 300 \, \text{K} = 3,313.5576 \times 10^{-23} \, \text{J} \] 4. **Compute the denominator:** \[ \text{Denominator} = \pi m = 3.1415926535 \times (3.34626 \times 10^{-27} \, \text{kg}) = 10.5148147 \times 10^{-27} \, \text{kg} \] 5. **Divide the numerator by the denominator:** \[ \frac{\text{Numerator}}{\text{Denominator}} = \frac{3,313.5576 \times 10^{-23} \, \text{J}}{10.5148147 \times 10^{-27} \, \text{kg}} = 315.34375 \times 10^{4} \, \text{m}^2/\text{s}^2 = 3,153,437.5 \, \text{m}^2/\text{s}^2 \] 6. **Compute the mean molecular speed \( v_{\text{avg}} \):** \[ v_{\text{avg}} = \sqrt{\frac{8kT}{\pi m}} = \sqrt{3,153,437.5 \, \text{m}^2/\text{s}^2} \approx 1,775 \, \text{m/s} \] **Answer:** The mean molecular speed \( v \) of hydrogen gas (H₂) at 300 K is approximately **1,775 meters per second**. |
1750.0 |
| deepseekv3 | To compute the mean molecular speed \( v \) of hydrogen gas (\( \text{H}_2 \)) at a given temperature, we can use the following formula derived from the kinetic theory of gases: \[ v = \sqrt{\frac{8RT}{\pi M}} \] Where: 1. \( v \) = mean molecular speed (m/s) 2. \( R \) = universal gas constant = 8.314 J/(mol·K) 3. \( T \) = temperature in Kelvin (K) 4. \( M \) = molar mass of the gas in kg/mol ### Steps to solve: 1. **Determine the molar mass of \( \text{H}_2 \):** - The molar mass of hydrogen (\( \text{H} \)) is approximately 1 g/mol. - Since \( \text{H}_2 \) consists of 2 hydrogen atoms, its molar mass is: \[ M = 2 \times 1 \, \text{g/mol} = 2 \, \text{g/mol} = 0.002 \, \text{kg/mol} \] 2. **Assume a temperature:** - Let’s assume \( T = 273 \, \text{K} \) (0°C, standard temperature). 3. **Substitute values into the formula:** \[ v = \sqrt{\frac{8 \times 8.314 \times 273}{\pi \times 0.002}} \] 4. **Calculate the result:** \[ v = \sqrt{\frac{8 \times 8.314 \times 273}{3.1416 \times 0.002}} \] \[ v = \sqrt{\frac{18183.408}{0.0062832}} \] \[ v = \sqrt{2,893,000} \approx 1700 \, \text{m/s} \] ### Final Answer: The mean molecular speed of hydrogen gas (\( \text{H}_2 \)) at 273 K is approximately **1700 m/s**. |
1750.0 |