Question

The temperature of an ideal gas is increased from 200 K to 800 K. If the RMS speed of gas at 200 K is \( v_0 \), then the RMS speed of the gas at 800 K will be:

• A. \( v_0 \)
• B. \( 4v_0 \)
• C. \( \frac{v_0}{4} \)
• D. \( 2v_0 \)

Hint:

The RMS speed of a gas is proportional to the square root of its temperature: \[ v_{{rms}} \propto \sqrt{T} \] If temperature increases by a factor \( n \), the RMS speed increases by a factor \( \sqrt{n} \).

Answer 1

The Correct Option is D

Step 1: {Formula for RMS speed}
The root mean square (RMS) speed of gas molecules is given by: \[ v_{{rms}} = \sqrt{\frac{3RT}{M}} \] Since \( R \) and \( M \) are constants: \[ v_{{rms}} \propto \sqrt{T} \] Step 2: {Determine new RMS speed}
Given initial and final temperatures: \[ T_{{initial}} = 200 { K}, \quad T_{{final}} = 800 { K} \] Since \( v_{{rms}} \propto \sqrt{T} \), we write: \[ \frac{v_{{rms, initial}}}{v_{{rms, final}}} = \sqrt{\frac{T_{{initial}}}{T_{{final}}}} \] Step 3: {Compute new RMS speed}
\[ \frac{v_0}{v_{{rms}}} = \sqrt{\frac{200}{800}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \] \[ v_{{rms}} = 2v_0 \] Thus, the correct answer is \( 2v_0 \).