Question

The temperature of an ideal gas is increased from 75 K to 300 K. If at 75 K the r.m.s. velocity of the gas molecules is \(v\), at 300 K it becomes:

• A. \( v = 2v \)
• B. \( v = \sqrt{2}v \)
• C. \( v = 4v \)
• D. \( v = \frac{1}{2}v \)

Hint:

Remember that the r.m.s. velocity of a gas is proportional to the square root of the temperature. So, if the temperature increases, the velocity increases in proportion to the square root of the temperature.

Answer 1

The Correct Option is B

The r.m.s. velocity \(v\) of gas molecules is related to the temperature \(T\) by the equation: \[ v \propto \sqrt{T} \] This means the r.m.s. velocity of the gas molecules is proportional to the square root of the temperature.
Now, we are given:
- Initial temperature \(T_1 = 75 \, \text{K}\),
- Final temperature \(T_2 = 300 \, \text{K}\),
- The initial velocity \(v_1 = v\).
We need to find the final velocity \(v_2\). Using the formula: \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \] Substitute the values of \(T_1\) and \(T_2\): \[ \frac{v_2}{v} = \sqrt{\frac{300}{75}} = \sqrt{4} = 2 \] Thus, \(v_2 = 2v\).
The correct answer is option (B): \( v = 2v \).