Question

The temperature of an ideal gas is increased from 200 K to 800 K. If r.m.s. speed of gas at 200K is v0. Then, r.m.s. speed of the gas at 800 K will be

• A. \(\frac{v_o}{4}\)
• B. \(2v_0\)
• C. \(v_0\)
• D. \(4v_0\)

Answer 1

The Correct Option is B

To solve this problem, we need to use the concept of the root mean square (r.m.s.) speed of an ideal gas, which is given by the formula:

v_{\text{rms}} = \sqrt{\frac{3kT}{m}}

Where \(k\) is the Boltzmann constant, \(T\) is the temperature in Kelvin, and \(m\) is the mass of a gas molecule.

The r.m.s. speed of a gas is directly proportional to the square root of its absolute temperature. Thus, you can relate the r.m.s. speeds at two different temperatures as follows:

\frac{v_{\text{rms}_2}}{v_{\text{rms}_1}} = \sqrt{\frac{T_2}{T_1}}

Given in the problem:

\(v_{\text{rms}_1} = v_0\) at \(T_1 = 200 \, \text{K}\)

\(T_2 = 800 \, \text{K}\)

Substituting these into the formula, we get:

\frac{v_{\text{rms}_2}}{v_0} = \sqrt{\frac{800}{200}}

\frac{v_{\text{rms}_2}}{v_0} = \sqrt{4}

\frac{v_{\text{rms}_2}}{v_0} = 2

This implies that \(v_{\text{rms}_2} = 2v_0\).

Therefore, the r.m.s. speed of the gas at 800 K is twice that at 200 K, so the correct answer is:

2v_0.