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At room temperature the rms speed of the...

At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1920 m/s. The gas is

A

`H_2`

B

`F_2`

C

`CI_2`

D

`O_2`

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The correct Answer is:
To solve the problem, we need to find the molecular weight of a certain diatomic gas given its root mean square (RMS) speed at room temperature. The steps are as follows: ### Step 1: Write down the formula for RMS speed The formula for the root mean square speed (v_rms) of gas molecules is given by: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where: - \( v_{rms} \) is the root mean square speed, - \( R \) is the universal gas constant (approximately \( 8.314 \, \text{J/(mol·K)} \)), - \( T \) is the absolute temperature in Kelvin, - \( M \) is the molar mass of the gas in kg/mol. ### Step 2: Substitute known values into the equation Given: - \( v_{rms} = 1920 \, \text{m/s} \) - \( T = 300 \, \text{K} \) - \( R = 8.314 \, \text{J/(mol·K)} \) We can substitute these values into the formula: \[ 1920 = \sqrt{\frac{3 \times 8.314 \times 300}{M}} \] ### Step 3: Square both sides to eliminate the square root Squaring both sides gives: \[ 1920^2 = \frac{3 \times 8.314 \times 300}{M} \] Calculating \( 1920^2 \): \[ 1920^2 = 3686400 \] So we have: \[ 3686400 = \frac{3 \times 8.314 \times 300}{M} \] ### Step 4: Calculate the right-hand side Now, calculate \( 3 \times 8.314 \times 300 \): \[ 3 \times 8.314 \times 300 = 7482.6 \] ### Step 5: Rearranging the equation to solve for M Now we can rearrange the equation to solve for \( M \): \[ M = \frac{7482.6}{3686400} \] ### Step 6: Calculate M Calculating \( M \): \[ M = \frac{7482.6}{3686400} \approx 0.0020298 \, \text{kg/mol} \] ### Step 7: Convert M to grams To convert \( M \) from kg/mol to g/mol, we multiply by 1000: \[ M \approx 0.0020298 \times 1000 \approx 2.03 \, \text{g/mol} \] ### Step 8: Identify the gas The molecular weight of approximately \( 2.03 \, \text{g/mol} \) corresponds to hydrogen (H₂), which has a molecular weight of about \( 2 \, \text{g/mol} \). ### Conclusion Thus, the gas is **hydrogen**. ---
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