Question

At what temperature rms speed of \({SO}_3\) molecules is \(3.16 \times 10^2 \, {ms}^{-1}\)? (R = 8.314 J K\(^{-1}\) mol\(^{-1}\))

• A. 480 K
• B. 320 K
• C. 160 K
• D. 640 K

Hint:

Use \(v_{rms} = \sqrt{\frac{3RT}{M}}\) to find temperature when rms speed is given.

Answer 1

The Correct Option is B

Root mean square speed of gas molecules: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] Where, \(v_{rms} = 3.16 \times 10^2 \, {ms}^{-1}\), \(R = 8.314 \, {J K}^{-1} {mol}^{-1}\), \(T = ?\), \(M =\) molar mass of \({SO}_3\) in kg/mol. 
Calculate \(M\): \[ M = 32 + 3 \times 16 = 80 \, {g/mol} = 0.08 \, {kg/mol} \] Rearranging for \(T\): \[ T = \frac{M v_{rms}^2}{3R} = \frac{0.08 \times (3.16 \times 10^2)^2}{3 \times 8.314} \] Calculate numerator: \[ 0.08 \times (316)^2 = 0.08 \times 99856 = 7988.48 \] Calculate denominator: \[ 3 \times 8.314 = 24.942 \] So, \[ T = \frac{7988.48}{24.942} \approx 320.4 \, K \]