Question

Let the r.m.s. velocity of molecules of a given mass of gas be \( C_1 \) at temperature \( 27^\circ\text{C} \). When the temperature is increased to \( 327^\circ\text{C} \), the r.m.s. velocity is \( C_2 \). Then the ratio \( \dfrac{C_2}{C_1} \) is

• A. \( \sqrt{2} \)
• B. \( 2 \)
• C. \( 4 \)
• D. \( 2\sqrt{2} \)

Hint:

R.m.s. velocity of a gas varies as the square root of absolute temperature.

Answer 1

The Correct Option is A

Step 1: Write the formula for r.m.s. velocity.
The r.m.s. velocity of gas molecules is given by \[ C = \sqrt{\frac{3RT}{M}} \] where \( T \) is absolute temperature.
Step 2: Convert temperatures into Kelvin scale.
\[ T_1 = 27 + 273 = 300 \, \text{K} \] \[ T_2 = 327 + 273 = 600 \, \text{K} \]
Step 3: Take the ratio of velocities.
\[ \frac{C_2}{C_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{600}{300}} = \sqrt{2} \]
Step 4: Conclusion.
The required ratio is \( \sqrt{2} \).