Question

The equation for the RMS velocity is given as \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M_0}} \] where \( R \) is the gas constant, \( T \) is the temperature, and \( M_0 \) is the molecular mass. If the temperature is increased, find the new RMS velocity \( v_{\text{rms}} \) when the temperature is doubled.}

• A. \( \sqrt{3} v_{\text{rms}} \)
• B. \( 2 v_{\text{rms}} \)
• C. \( \sqrt{2} v_{\text{rms}} \)
• D. \( \frac{v_{\text{rms}}}{\sqrt{2}} \)

Hint:

When the temperature of a gas is doubled, the RMS velocity of the gas molecules increases by a factor of \( \sqrt{2} \). This is due to the direct proportionality between the RMS velocity and the square root of the temperature.

Answer 1

The Correct Option is B

The RMS velocity is given by \( v_{\text{rms}} = \sqrt{\frac{3RT}{M_0}} \), where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M_0 \) is the molecular mass. If the temperature is doubled to \( T_2 = 2T_1 \), the new RMS velocity \( v_{\text{rms}}' \) can be calculated by substituting \( 2T \) into the formula: \( v_{\text{rms}}' = \sqrt{\frac{3R(2T)}{M_0}} \). This simplifies to \( v_{\text{rms}}' = \sqrt{2} \times \sqrt{\frac{3RT}{M_0}} = \sqrt{2} \times v_{\text{rms}} \). Therefore, doubling the temperature increases the RMS velocity by a factor of \( \sqrt{2} \). This outcome does not align with the provided answer choices. Please provide the options for further analysis.