The average velocity of a molecule in an ideal gas is related to the temperature, which can be determined using the formula for the root mean square speed (v_rms), derived from the Maxwell-Boltzmann distribution. The formula for v_rms is:

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

where:

• k is the Boltzmann constant,
• T is the absolute temperature in Kelvin,
• m is the mass of a molecule of the gas.

To find the average velocity at two different temperatures, it is important to note that while the average velocity and the root mean square speed are not exactly the same, they are proportional to the square root of the temperature. Therefore, for the purpose of comparing speeds at two temperatures, we can use the formula:

$$ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} $$

Where:

• v₁ and v₂ are the speeds at temperatures T₁ and T₂ respectively.

Here, temperatures must be converted to Kelvin:

• T₁ = 27^o C = 27 + 273.15 = 300.15 K
• T₂ = 927^o C = 927 + 273.15 = 1200.15 K

Plugging these values into the proportion formula gives:

$$ \frac{v_2}{0.3} = \sqrt{\frac{1200.15}{300.15}} $$

Calculating the square root:

$$ \sqrt{\frac{1200.15}{300.15}} \approx \sqrt{4} = 2 $$

Therefore:

$$ v_2 = 0.3 \times 2 = 0.6 \, \text{m/sec} $$

Thus, the average velocity of the gas molecules at 927ºC is 0.6 m/sec.

Therefore, the correct answer is: Option A (0.6 m/sec).