Question

The rms speed of a gas having diatomic molecules at temperature T (in Kelvin) is 200 m/s. If the temperature is increased to 4T and the molecules dissociate into monoatomic atoms,the rms speed will become:

• A. 400 m/s
• B. 200 m/s
• C. 800 m/s
• D. 200\(\sqrt{2}\) m/s
• E. 400\(\sqrt{2}\) m/s

Answer 1

Answer: (E)

Initial conditions: \[ v_{rms} = 200\,\text{m/s at temperature } T \text{ for diatomic gas} \]

RMS speed formula: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

After changes:

• Temperature becomes 4T
• Molecules dissociate into monoatomic atoms (mass becomes M/2)

 

New rms speed calculation: \[ v'_{rms} = \sqrt{\frac{3R(4T)}{M/2}} = \sqrt{8 \times \frac{3RT}{M}} \] \[ v'_{rms} = \sqrt{8} \times v_{rms} = 2\sqrt{2} \times 200 \] \[ v'_{rms} = 400\sqrt{2}\,\text{m/s} \]

Thus, the correct option (E) \(400\sqrt{2}\,\text{m/s}\).

Answer 2

1. Recall the formula for rms speed:

The root-mean-square (rms) speed (v_rms) of gas molecules is given by:

\[v_{rms} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}\]

where:

• k is the Boltzmann constant
• R is the universal gas constant
• T is the temperature in Kelvin
• m is the mass of one molecule
• M is the molar mass

2. Set up the initial and final conditions:

Initial conditions (diatomic, temperature T):

\[v_{rms1} = \sqrt{\frac{3kT}{m_1}} = 200 \, m/s\]

Final conditions (monoatomic, temperature 4T, mass is halved since molecules dissociate):

\[v_{rms2} = \sqrt{\frac{3k(4T)}{m_2}}\] Since the diatomic molecules dissociate into monoatomic atoms, the mass of each atom is half the mass of the diatomic molecule: \(m_2 = \frac{m_1}{2}\). \[v_{rms2} = \sqrt{\frac{3k(4T)}{m_1/2}} = \sqrt{8 \frac{3kT}{m_1}} \]

3. Relate the final rms speed to the initial rms speed:

\[v_{rms2} = \sqrt{8} \sqrt{\frac{3kT}{m_1}} = 2\sqrt{2} v_{rms1} \]

Since \(v_{rms1} = 200 \, m/s\):

\[v_{rms2} = 2\sqrt{2}(200 \, m/s) = 400\sqrt{2} \, m/s\]

Final Answer: The final answer is \(\boxed{E}\)