Question

RMS velocity of one mole of an ideal gas was measured at different temperatures. A graph of \( (u_{\text{rms}})^2 \) (on y-axis) and \( T/K \) (on x-axis) gave a straight line passing through the origin, and its slope is \( 249 \, \text{m}^2 \text{s}^{-2}\text{K}^{-1} \). What is the molar mass (in kg mol\(^{-1}\)) of the ideal gas? \(( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} )\)

• A. \( 10 \)
• B. \( 1.0 \)
• C. \( 24.9 \)
• D. \( 1 \times 10^{-1} \)

Hint:

For gases, the RMS velocity formula \( u_{\text{rms}} = \sqrt{\frac{3RT}{M}} \) is useful in solving temperature-dependent kinetic energy problems.

Answer 1

The Correct Option is C

The formula for RMS velocity of an ideal gas is: \[ u_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] Squaring both sides: \[ (u_{\text{rms}})^2 = \frac{3R}{M} T \] Comparing with the given equation \( (u_{\text{rms}})^2 = 249 T \), we get: \[ \frac{3R}{M} = 249 \] Substituting \( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \): \[ \frac{3 \times 8.3}{M} = 249 \] \[ \frac{24.9}{M} = 249 \] Solving for \( M \): \[ M = \frac{24.9}{249} = 24.9 \, \text{kg mol}^{-1} \] Thus, the correct answer is: \[ \boxed{24.9} \]