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 \( 2v_0 \).

Answer 2

Step 1: Use the formula for the root mean square (RMS) speed of an ideal gas:
\( v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \), where:
- \( v_{\text{rms}} \) is the RMS speed
- \( k \) is Boltzmann’s constant
- \( T \) is the absolute temperature
- \( m \) is the mass of a single gas molecule

Step 2: Since \( k \) and \( m \) are constants for a given gas, the RMS speed depends on the square root of the temperature:
\( v_{\text{rms}} \propto \sqrt{T} \)

Step 3: Let the initial RMS speed at \( T_1 = 200\,K \) be \( v_0 \).
Now, the temperature is increased to \( T_2 = 800\,K \).
Using the proportionality:
\( \frac{v_{\text{rms, new}}}{v_0} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2 \)

Step 4: Therefore, the new RMS speed is:
\( v_{\text{rms, new}} = 2v_0 \)

Final Answer: \( 2v_0 \)