Question

The rms velocity of the gas molecule at \(327^\circ C\) is same as the rms velocity of the oxygen molecules at \(27^\circ C\). If the molecular weight of oxygen is 32 then the molecular weight of the given gas molecule is:

• A. \(32\)
• B. \(64\)
• C. \(96\)
• D. \(128\)

Hint:

For equal rms velocities, use \( \frac{T}{M} = \text{constant} \). Always convert temperature to Kelvin.

Answer 1

The Correct Option is B

Step 1: Use rms velocity formula.
The rms velocity of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

Step 2: Apply given condition.
Since rms velocities are equal:
\[ \sqrt{\frac{T_1}{M_1}} = \sqrt{\frac{T_2}{M_2}} \]

Step 3: Remove square root.
\[ \frac{T_1}{M_1} = \frac{T_2}{M_2} \]

Step 4: Convert temperatures to Kelvin.
\[ T_1 = 327 + 273 = 600\,K \]
\[ T_2 = 27 + 273 = 300\,K \]

Step 5: Substitute values.
\[ \frac{600}{M} = \frac{300}{32} \]

Step 6: Solve for molecular weight.
\[ M = \frac{600 \times 32}{300} \]
\[ M = 64 \]

Step 7: Final answer.
\[ \boxed{64} \]