Question

The temperature of an ideal gas is increased from 100 K to 400 K. If \( v \) is the R.M.S. velocity of its molecules at 100 K, it becomes

• A. \( \frac{v}{\sqrt{2}} \)
• B. \( 2v \)
• C. \( 3v \)
• D. \( 4v \)

Hint:

The R.M.S. velocity of gas molecules is proportional to the square root of the temperature. If the temperature increases by a factor, the R.M.S. velocity increases by the square root of that factor.

Answer 1

The Correct Option is B

Step 1: Relationship between temperature and R.M.S. velocity.
The root-mean-square (R.M.S.) velocity \( v_{\text{rms}} \) of molecules in an ideal gas is related to the temperature \( T \) by the equation:
\[ v_{\text{rms}} = \sqrt{\frac{3kT}{m}}, \]
where:
- \( k \) is the Boltzmann constant,
- \( T \) is the temperature,
- \( m \) is the mass of a molecule.

Step 2: Applying the change in temperature.
The initial temperature is \( T_1 = 100 \, \text{K} \), and the final temperature is \( T_2 = 400 \, \text{K} \). The R.M.S. velocity at the initial temperature is \( v_1 = v \). For the final temperature, the new R.M.S. velocity \( v_2 \) can be expressed as:
\[ v_2 = \sqrt{\frac{3kT_2}{m}}. \]
Using the given values:
\[ v_2 = \sqrt{\frac{3k \cdot 400}{m}} = \sqrt{4 \cdot \frac{3k \cdot 100}{m}} = 2v_1. \]

Step 3: Conclusion.
Thus, the new R.M.S. velocity when the temperature is increased from 100 K to 400 K is \( 2v \).
Final Answer:
Thus, the new R.M.S. velocity is:
\[ \boxed{2v}. \]