Question

The r.m.s. speed of gas molecules at 800 K will be

• A. same as at 200 K
• B. twice the value at 200 K
• C. four times the value at 200 K
• D. half the value at 200 K

Hint:

The r.m.s. speed of gas molecules is directly proportional to the square root of the temperature. If the temperature doubles, the r.m.s. speed increases by a factor of \( \sqrt{2} \).

Answer 1

The Correct Option is C

Step 1: The relationship between temperature and molecular speed.
The r.m.s. speed \( v_{\text{rms}} \) of gas molecules is related to the temperature of the gas by the equation:
\[ v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}, \]
where:
- \( k_B \) is Boltzmann’s constant,
- \( T \) is the temperature in Kelvin,
- \( m \) is the mass of a gas molecule.

Step 2: Comparing the r.m.s. speeds at two different temperatures.
Let the r.m.s. speed of gas molecules at temperature \( T_1 = 200 \, \text{K} \) be \( v_1 \), and at \( T_2 = 800 \, \text{K} \), the r.m.s. speed is \( v_2 \). From the above equation:
\[ v_1 = \sqrt{\frac{3 k_B T_1}{m}}, \quad v_2 = \sqrt{\frac{3 k_B T_2}{m}}. \]
Thus, the ratio of the r.m.s. speeds is:
\[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2. \]

Step 3: Conclusion.
The r.m.s. speed of gas molecules at 800 K is twice the value at 200 K.
Final Answer:
Thus, the correct answer is:
\[ \boxed{2 \times \text{r.m.s. speed at 200 K}}. \]