Question

At what temperature (in K) the rms velocity of SO\(_2\) molecules is equal to rms velocity of O\(_2\) molecules at \(27 \, ^\circ\text{C}\)?

• A. 300
• B. 1200
• C. 600
• D. 900

Hint:

- RMS velocity of gas molecules: \( v_{rms} = \sqrt{\frac{3RT}{M}} \). (M is molar mass in kg/mol if R is in J/mol.K) - If \( v_{rms,1} = v_{rms,2} \), then \( \frac{T_1}{M_1} = \frac{T_2}{M_2} \). - Convert Celsius to Kelvin: \( T(K) = T(^\circ C) + 273 \). (Using 273 for simplicity, 273.15 for precision). - Molar mass (M): Calculate from atomic masses. Units cancel in ratio \(M_1/M_2\).

Answer 1

The Correct Option is C

The root mean square (rms) velocity of gas molecules is given by \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where R is the ideal gas constant, T is the absolute temperature (in Kelvin), and M is the molar mass of the gas (in kg/mol).
Let \( T_{SO_2} \) be the temperature of SO\(_2\) and \( T_{O_2} \) be the temperature of O\(_2\).
Let \( M_{SO_2} \) be the molar mass of SO\(_2\) and \( M_{O_2} \) be the molar mass of O\(_2\).
Given \( v_{rms, SO_2} = v_{rms, O_2} \).
\[ \sqrt{\frac{3RT_{SO_2}}{M_{SO_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}} \] Squaring both sides and cancelling \(3R\): \[ \frac{T_{SO_2}}{M_{SO_2}} = \frac{T_{O_2}}{M_{O_2}} \] \[ T_{SO_2} = T_{O_2} \cdot \frac{M_{SO_2}}{M_{O_2}} \] Temperature of O\(_2\): \( T_{O_2} = 27 \, ^\circ\text{C} = 27 + 273 = 300 \, \text{K} \).
Molar masses: Sulfur (S) atomic mass \( \approx 32 \) g/mol.
Oxygen (O) atomic mass \( \approx 16 \) g/mol.
Molar mass of SO\(_2\) (\(M_{SO_2}\)) = \( 32 + 2 \times 16 = 32 + 32 = 64 \) g/mol.
Molar mass of O\(_2\) (\(M_{O_2}\)) = \( 2 \times 16 = 32 \) g/mol.
The ratio of molar masses \( \frac{M_{SO_2}}{M_{O_2}} = \frac{64 \text{ g/mol}}{32 \text{ g/mol}} = 2 \).
(Note: units g/mol cancel out, same as kg/mol for ratio).
Now calculate \( T_{SO_2} \): \[ T_{SO_2} = 300 \, \text{K} \times 2 = 600 \, \text{K} \] So, the temperature of SO\(_2\) should be 600 K.
This matches option (3).