Question

At what temperature will the r.m.s. velocity of a hydrogen molecule be equal to that of an oxygen molecule at \(47^\circ\text{C}\)?

• A. \(40\,\text{K}\)
• B. \(20\,\text{K}\)
• C. \(10\,\text{K}\)
• D. \(5\,\text{K}\)

Hint:

For gases, \[ v_{\text{rms}} \propto \sqrt{\frac{T}{M}} \] If two gases have equal r.m.s velocity, their temperature-to-molar-mass ratios must be equal.

Answer 1

The Correct Option is B

Concept: The r.m.s. velocity of a gas molecule is given by: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where \(T\) = absolute temperature and \(M\) = molar mass. Thus, \[ v_{\text{rms}} \propto \sqrt{\frac{T}{M}} \]

Step 1: Equate the r.m.s velocities. \[ \sqrt{\frac{T_H}{M_H}} = \sqrt{\frac{T_O}{M_O}} \] Squaring both sides: \[ \frac{T_H}{M_H} = \frac{T_O}{M_O} \]

Step 2: Substitute molar masses. \[ M_H = 2, \quad M_O = 32 \] Temperature of oxygen: \[ 47^\circ\text{C} = 320\,\text{K} \] \[ \frac{T_H}{2} = \frac{320}{32} \]

Step 3: Solve for \(T_H\). \[ \frac{T_H}{2} = 10 \] \[ T_H = 20\,\text{K} \] Thus, the required temperature is \[ \boxed{20\,\text{K}} \]