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): $v = 2v$.