The root mean square (R.M.S.) Speed V of the molecules of an ideal gas is given by the expressions, $v = \sqrt{(\frac{3 RT}{M})}$ and $v = \sqrt{(\frac{3 KT}{m})}$ where R is universal gas constant, T is the absolute (Kelvin) temperature M is the molar mass, K is Boltzman's constant and m is the molecular mass. The R.M.S. speed of oxygen molecules ( $O_{2}$ ) at temperature $T_{1} is V_{1}$ . When the temperature is doubled, if the oxygen molecules are dissociated into atomic oxygen, what will be R.M.S. speed of oxygen atoms? (Treat the gas as ideal).

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Solution

Root mean square speed equation :
Root mean square speed equation, $v = \sqrt{(\frac{3 RT}{M})}$
From the equation, it is clear that the speed of gas molecules is affected by changes in temperature and molar mass.
The speed of molecules in a gas is directly proportional to its temperature and inversely proportional to its molar mass.
In other words, as the temperature of a gas sample rises, the molecules accelerate, increasing the root mean square molecular speed.
Root mean square speed equation in terms of “K” :
Root mean square speed equation, $v = \sqrt{(\frac{3 KT}{m})}$
This equation states that the speed of gas particles is related to their absolute temperature.
In other words, as their temperature rises, their speed increases, and eventually their total energy also increases.
However, it is not possible to define the speed of a single gas particle.
Therefore, the velocity of the gas is defined by the root mean square speed.
Root mean square speed of oxygen atoms:
When the temperature is doubled, and if the oxygen molecules are dissociated into atomic oxygen, the molar mass, as well as the molecular mass, gets halved.
Hence, root mean square speed, $v = {(\frac{3 K \times 2 T_{1}}{\frac{m}{2}})}^{1 / 2} = 2 {(\frac{3 {KT}_{1}}{m})}^{1 / 2} = 2 v_{1}$

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