Question

If the temperature of a gas is increased from 127 °C to 527 °C, then the rms speed of the gas molecules

• A. increases by 4 times
• B. becomes \(\sqrt{2}\) times
• C. becomes half
• D. decreases by \(\sqrt{2}\) times

Hint:

In gas law and kinetic theory problems, always convert temperatures to the absolute scale (Kelvin) before using them in any formula. Forgetting this conversion is a very common mistake. \( T(K) = T(^\circ C) + 273.15 \).

Answer 1

The Correct Option is B

The root-mean-square (rms) speed of gas molecules is directly proportional to the square root of the absolute temperature (in Kelvin).
The formula is \( v_{rms} = \sqrt{\frac{3RT}{M}} \), so \( v_{rms} \propto \sqrt{T} \).
First, we must convert the given Celsius temperatures to Kelvin.
Initial temperature, \( T_1 = 127^\circ\text{C} + 273 = 400 \text{ K} \).
Final temperature, \( T_2 = 527^\circ\text{C} + 273 = 800 \text{ K} \).
Let \(v_1\) be the initial rms speed and \(v_2\) be the final rms speed.
The ratio of the final speed to the initial speed is:
\( \frac{v_2}{v_1} = \frac{\sqrt{T_2}}{\sqrt{T_1}} = \sqrt{\frac{T_2}{T_1}} \).
Substitute the temperatures in Kelvin.
\( \frac{v_2}{v_1} = \sqrt{\frac{800}{400}} = \sqrt{2} \).
So, the final rms speed is \( v_2 = \sqrt{2} \times v_1 \).
The rms speed becomes \(\sqrt{2}\) times the initial speed.