Question

If the temperature of a gas is increased from \( 27^\circ C \) to \( 159^\circ C \), the increase in the rms speed of the gas molecules is:

• A. \( 142\% \)
• B. \( 71\% \)
• C. \( 80\% \)
• D. \( 20\% \)

Hint:

Use \( v_{\text{rms}} \propto \sqrt{T} \) to calculate changes in rms speed.
- Always convert temperature to Kelvin before applying the formula.

Answer 1

The Correct Option is D

The root mean square (rms) speed of gas molecules is given by: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where \( T \) is the absolute temperature. Since \( v_{\text{rms}} \propto \sqrt{T} \), we can write: \[ \frac{v_{\text{rms final}}}{v_{\text{rms initial}}} = \sqrt{\frac{T_2}{T_1}} \] Converting temperatures to Kelvin: \[ T_1 = 27 + 273 = 300 \, K, \quad T_2 = 159 + 273 = 432 \, K \] \[ \frac{v_{\text{rms final}}}{v_{\text{rms initial}}} = \sqrt{\frac{432}{300}} \] \[ = \sqrt{1.44} = 1.2 \] Percentage increase: \[ \left(1.2 - 1\right) \times 100 = 20\% \] Thus, the correct answer is \(\boxed{20\%}\).