Question

The ratio of the respective rms velocities of the gas molecules of an ideal gas at 327°C and at 627°C is:

• A. \( \sqrt{2} : 3 \)
• B. \( \sqrt{2} : 2 \sqrt{3} \)
• C. \( \sqrt{2} : \sqrt{3} \)
• D. \( \sqrt{3} : 2 \)
• E. \( \sqrt{3} : 3 \)

Hint:

The root mean square velocity of gas molecules is directly related to the square root of the temperature. To find the ratio of rms velocities at two different temperatures, take the square root of the ratio of the temperatures in Kelvin.

Answer 1

The Correct Option is C

The root mean square (rms) velocity \( v_{{rms}} \) of gas molecules is given by the formula:
\[ v_{{rms}} = \sqrt{\frac{3kT}{m}} \] where: - \( k \) is the Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the mass of the gas molecule.
The ratio of the rms velocities at two different temperatures \( T_1 \) and \( T_2 \) can be written as: \[ \frac{v_{{rms1}}}{v_{{rms2}}} = \sqrt{\frac{T_1}{T_2}} \] Given: - \( T_1 = 327^\circ C = 327 + 273 = 600 \, {K} \),
- \( T_2 = 627^\circ C = 627 + 273 = 900 \, {K} \).
Substitute these values into the formula for the ratio of rms velocities: \[ \frac{v_{{rms1}}}{v_{{rms2}}} = \sqrt{\frac{600}{900}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} \] Thus, the ratio of the rms velocities is: \[ \boxed{\sqrt{2} : \sqrt{3}} \]