Question

The root mean square (rms) speed of the molecules of an ideal gas at a given temperature \(T\) is \(v\). If the temperature is increased to \(4T\) and the gas dissociates into atoms, the new rms speed becomes:

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

Hint:

Don't forget to track the mass variable when a problem explicitly mentions "dissociation"! Dissociation always drops the particle weight by half (or more depending on the chemical complexity), which simultaneously accelerates the speed of the individual particles alongside the temperature expansion.

Answer 1

The Correct Option is D

Concept: The root mean square (rms) speed (\(v_{\text{rms}}\)) of gas particles depends directly on the absolute thermodynamic temperature (\(T\)) and inversely on the molar mass (\(M\)) or molecular mass (\(m\)) of the individual gas particles according to the kinetic theory of gases: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \quad \text{or} \quad v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}} \] where \(R\) is the universal gas constant and \(k_B\) is the Boltzmann constant. When a diatomic or polyatomic gas dissociates completely into single atoms, each original molecule breaks apart, dividing the effective particle mass cleanly in half. Step 1: Establish the expression for the initial state.
Let the initial temperature be \(T_1 = T\) and the initial molar mass of the gas molecule be \(M_1 = M\). The initial rms speed is given as: \[ v = \sqrt{\frac{3RT}{M}} \quad \cdots (1) \]

Step 2: Identify the changes for the final state.
The final state introduces two distinct structural updates:
• The final temperature is quadrupled: \[ T_2 = 4T \]
• The gas dissociates into individual atoms. Assuming a standard diatomic starting condition typical for dissociation problems unless stated otherwise, breaking the molecule into its two constituent atoms splits the mass per gaseous particle in half: \[ M_2 = \frac{M}{2} \]

Step 3: Calculate the new rms speed \(v'\).
Substitute the updated final state variables into our core kinetic formula: \[ v' = \sqrt{\frac{3R(4T)}{\left(\frac{M}{2}\right)}} \] Bring the division-by-two factor up into the numerator: \[ v' = \sqrt{\frac{3RT \cdot 4 \cdot 2}{M}} = \sqrt{8 \cdot \frac{3RT}{M}} \]

Step 4: Express \(v'\) in terms of the initial speed \(v\).
Factor out the numeric constant from the radical operator: \[ v' = \sqrt{8} \cdot \sqrt{\frac{3RT}{M}} \] Since \(\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}\), and substituting equation (1) for the remaining radical component yields: \[ v' = 2\sqrt{2}\,v \]