The temperature of the gas is raised from $27^\circ C\, to\, 927^\circ C$ the root mean square velocity is

Question

What is the oxidation state of Fe in $ \mathrm{Fe_3O_4} $?

Answer 1

The root mean square velocity ($v_{\text{rms}}$) of a gas is given by the formula:

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 gas molecule.

Since the mass of the gas molecules and $k$ is constant for the given problem, the relationship can be rewritten as:

v_{\text{rms}} \propto \sqrt{T}

Given that the initial temperature ($T_1$) is $27^\circ C$ and the final temperature ($T_2$) is $927^\circ C$, we first need to convert these temperatures to Kelvin:

Step 1: Convert temperatures from Celsius to Kelvin:

$T_1 = 27^\circ C = 27 + 273.15 = 300.15\,K$
Approximate for convenience: $T_1 \approx 300\,K$
$T_2 = 927^\circ C = 927 + 273.15 = 1200.15\,K$
Approximate for convenience: $T_2 \approx 1200\,K$

Step 2: Calculate the new root mean square velocity:

The change in root mean square velocity can be expressed as the ratio of the square roots of the temperatures:

\frac{v_{\text{rms},2}}{v_{\text{rms},1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2

This means the root mean square velocity is doubled.

Conclusion: The root mean square velocity is doubled when the temperature is raised from $27^\circ C$ to $927^\circ C$. Thus, the correct answer is "doubled".

Answer 2