Question

The R.M.S. speed of oxygen molecules at 27 °C is $v$. At 927 °C the rms speed will be:

• A. $v$
• B. $v/2$
• C. $2v$
• D. $4v$

Hint:

If the absolute temperature (in Kelvin) becomes 4 times, the speed doubles because of the square root relationship.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The root mean square (RMS) speed of gas molecules depends on the absolute temperature of the gas. If the gas remains the same (Oxygen), the speed is proportional to the square root of the absolute temperature.
Step 2: Key Formula or Approach:
1. RMS speed formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \).
2. Ratio: \( \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \).
Step 3: Detailed Explanation:
Initial temperature \( T_1 = 27 + 273 = 300 \text{ K} \).
Final temperature \( T_2 = 927 + 273 = 1200 \text{ K} \).
Taking the ratio of the speeds: \[ \frac{v_2}{v} = \sqrt{\frac{1200}{300}} \] \[ \frac{v_2}{v} = \sqrt{4} = 2 \] \[ v_2 = 2v \]
Step 4: Final Answer:
The rms speed at 927°C will be 2v.