Question

The r.m.s. speed of molecules of an ideal gas at 27°C is 200 ms^-1. When the temperature is increased to 327°C, the r.m.s. speed of the molecules is changed to

• A. 490.2ms^-1
• B. 315.2ms^-1
• C. 282.8ms^-1
• D. 425.5ms^-1
• E. 517.7ms^-1

Answer 1

The Correct Option is C

The formula for the r.m.s. speed \( v_{\text{rms}} \) is given by: \[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \] where \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the mass of the molecules. Since the gas is ideal and the mass of the molecules remains constant, the r.m.s. speed of molecules is proportional to the square root of the temperature. Therefore, we can write the relation: \[ \frac{v_{\text{rms,2}}}{v_{\text{rms,1}}} = \sqrt{\frac{T_2}{T_1}} \] Given that \( v_{\text{rms,1}} = 200 \, \text{ms}^{-1} \), \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \), and \( T_2 = 327^\circ \text{C} = 600 \, \text{K} \), we can substitute into the formula: \[ \frac{v_{\text{rms,2}}}{200} = \sqrt{\frac{600}{300}} = \sqrt{2} \] Thus, \[ v_{\text{rms,2}} = 200 \times \sqrt{2} = 200 \times 1.414 = 282.8 \, \text{ms}^{-1} \] So, the new r.m.s. speed of the molecules is \( 282.8 \, \text{ms}^{-1} \).

The correct option is (C) : \(282.8\ ms^{-1}\)

Answer 2

The root mean square (r.m.s.) speed of gas molecules is given by:  

$$ v_{\text{rms}} = \sqrt{\frac{3kT}{m}} $$ 
So, r.m.s. speed is proportional to the square root of the absolute temperature: 
$$ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} $$ 
Given: \( v_1 = 200\, \text{ms}^{-1} \), \( T_1 = 27^\circ\text{C} = 300\, \text{K} \), \( T_2 = 327^\circ\text{C} = 600\, \text{K} \) 

Now: $$ v_2 = v_1 \cdot \sqrt{\frac{T_2}{T_1}} = 200 \cdot \sqrt{\frac{600}{300}} = 200 \cdot \sqrt{2} \approx 200 \cdot 1.414 = 282.8\, \text{ms}^{-1} $$ 
Correct answer: 282.8 ms^-1