Question

If the temperature of a gas is changed to \( 9 \) times the initial value, then the rms velocity of the gaseous molecule increases by

• A. \( 9 \) times
• B. \( 3 \) times
• C. \( \sqrt{3} \) times
• D. \( 18 \) times
• E. \( 12 \) times

Hint:

For gases, \[ v_{\text{rms}} \propto \sqrt{T} \] So if temperature becomes \( n \) times, rms speed becomes \( \sqrt{n} \) times.

Answer 1

The Correct Option is B

Step 1: Recall the formula for rms speed.
The root mean square speed of gas molecules is given by: \[ v_{\text{rms}}=\sqrt{\frac{3kT}{m}} \] or equivalently, \[ v_{\text{rms}} \propto \sqrt{T} \] for a given gas.

Step 2: Identify the change in temperature.
It is given that the new temperature is: \[ T_2 = 9T_1 \]

Step 3: Write the ratio of rms speeds.
Since \[ v_{\text{rms}} \propto \sqrt{T} \] we have \[ \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}} \]

Step 4: Substitute the temperature ratio.
Now, \[ \frac{v_2}{v_1}=\sqrt{\frac{9T_1}{T_1}} \] \[ =\sqrt{9} \] \[ =3 \]

Step 5: Interpret the result.
This means the new rms speed is three times the original rms speed.

Step 6: Note the important dependence.
The rms speed does not increase directly in proportion to temperature.
It increases as the square root of temperature.
That is why \( 9 \) times temperature gives only \( 3 \) times speed.

Step 7: Final conclusion.
Hence, the rms velocity increases by \[ \boxed{3 \text{ times}} \] Therefore, the correct option is \[ \boxed{(2)\ 3 \text{ times}} \]