Question

If \(m\) represents the mass of each molecule of a gas and \(T\), its absolute temperature, then the root mean square velocity of the gaseous molecule is proportional to

• A. \(m T\)
• B. \(m^{1/2} T^{1/2}\)
• C. \(m^{-1/2} T\)
• D. \(m^{-1/2} T^{1/2}\)
• E. \(m T^{-1/2}\)

Hint:

At the same temperature, hydrogen molecules move much faster than oxygen molecules because hydrogen has a much smaller molecular mass.

Answer 1

The Correct Option is D

Concept:
The root mean square (rms) speed of gas molecules is obtained from the kinetic theory of gases. It represents the effective average speed of molecules moving randomly in all directions. For an ideal gas, the rms speed is given by \[ v_{rms} = \sqrt{\frac{3k_B T}{m}} \] where,
• \(v_{rms}\) = root mean square speed of gas molecules
• \(k_B\) = Boltzmann constant
• \(T\) = absolute temperature of the gas
• \(m\) = mass of one molecule of the gas Physical Meaning:
• The rms speed increases when the temperature increases.
• The rms speed decreases when the molecular mass increases.
• Thus, lighter molecules move faster than heavier molecules at the same temperature.

Step 1: Write the rms speed formula.
\[ v_{rms} = \sqrt{\frac{3k_B T}{m}} \]

Step 2: Remove the constant terms for proportionality analysis.
Since \(3\) and \(k_B\) are constants, they do not affect proportionality. Hence, \[ v_{rms} \propto \sqrt{\frac{T}{m}} \]

Step 3: Express the square root in power form.
Using the property \[ \sqrt{x} = x^{1/2} \] we get \[ v_{rms} \propto \left(\frac{T}{m}\right)^{1/2} \]

Step 4: Separate temperature and mass terms.
\[ v_{rms} \propto T^{1/2} m^{-1/2} \] Final Result: \[ \boxed{v_{rms} \propto T^{1/2} m^{-1/2}} \] Conclusion:
• rms speed is directly proportional to the square root of temperature.
• rms speed is inversely proportional to the square root of molecular mass.