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At room temperature (300K), the rms spee...

At room temperature `(300K)`, the rms speed of the molecules of a certain diatomic gas is found to be `1930 m//s`. Can you gusess name of the gas? Find the temperature at which the rms speed is double of the speed in part one `(R = 25//3 J// mol -K)`

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To solve the problem, we will follow these steps: ### Step 1: Use the RMS speed formula The root mean square (RMS) speed \( v_{rms} \) of a gas is given by the formula: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where: - \( R \) is the universal gas constant, - \( T \) is the temperature in Kelvin, - \( M \) is the molar mass of the gas. ### Step 2: Rearrange the formula to find the molar mass Rearranging the formula to solve for \( M \): \[ M = \frac{3RT}{v_{rms}^2} \] ### Step 3: Substitute the known values Given: - \( R = \frac{25}{3} \, \text{J/mol-K} \) - \( T = 300 \, \text{K} \) - \( v_{rms} = 1930 \, \text{m/s} \) Substituting these values into the equation: \[ M = \frac{3 \cdot \frac{25}{3} \cdot 300}{(1930)^2} \] ### Step 4: Calculate the molar mass Calculating the numerator: \[ 3 \cdot \frac{25}{3} \cdot 300 = 25 \cdot 300 = 7500 \] Calculating the denominator: \[ (1930)^2 = 3724900 \] Now substituting back: \[ M = \frac{7500}{3724900} \approx 0.002013 \, \text{kg/mol} = 2.013 \, \text{g/mol} \] ### Step 5: Identify the gas The molar mass of approximately \( 2.013 \, \text{g/mol} \) corresponds to hydrogen gas (H₂), which has a molar mass of about \( 2 \, \text{g/mol} \). ### Step 6: Find the temperature for double the RMS speed We need to find the temperature at which the RMS speed is double: \[ v_{rms, final} = 2 \cdot v_{rms, initial} \] From the relationship: \[ v_{rms} \propto \sqrt{T} \] This implies: \[ \frac{v_{rms, final}}{v_{rms, initial}} = \sqrt{\frac{T_{final}}{T_{initial}}} \] Substituting the known values: \[ 2 = \sqrt{\frac{T_{final}}{300}} \] Squaring both sides: \[ 4 = \frac{T_{final}}{300} \] Thus: \[ T_{final} = 4 \cdot 300 = 1200 \, \text{K} \] ### Final Answer The gas is hydrogen (H₂), and the temperature at which the RMS speed is double the initial speed is \( 1200 \, \text{K} \). ---

To solve the problem, we will follow these steps: ### Step 1: Use the RMS speed formula The root mean square (RMS) speed \( v_{rms} \) of a gas is given by the formula: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where: ...
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