Question

The root mean square (R.M.S) speed $ V $ of the molecules of an ideal gas is given by the expression, $ v = sqrt(\dfrac{{3RT}}{M}) $ and $ v = sqrt(\dfrac{{3KT}}{m}) $ where R is universal gas constant, $ T $ is the absolute (Kelvin) temperature, m is the molar mass, K is Boltzmann’s constant and M is the molecular mass. The R.M.S speed of oxygen molecules $ ({O_2}) $ at temperature $ T1 $ is $ V1 $ . 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)

Answer 1

Hint :Gases consist of atoms or molecules that move at different speeds in random directions. The root mean square velocity (RMS velocity) is a way to find a single velocity value for the particles. The average velocity of a gas particle is found using the root mean square velocity formula.

Complete Step By Step Answer:
When examining the root mean square speed equation $ v = sqrt(\dfrac{{3RT}}{M}) $ , we can see that the change in temperature and molar mass affect the speed of gas molecules. The speed of the molecules in a gas is proportional to the temperature and is inversely proportional to the molar mass of the gas. In other words, as the temperature of a sample of gas is increased, the molecules speed up and the root mean square molecular speed increases as a result.
For equation $ v = sqrt(\dfrac{{3KT}}{m}) $ , it says that the speed of gas particles is related to their absolute temperature. In other words, as their temperature increases, their speed increases, and finally their total energy increases as well. However, it is impossible to define the speed of any one gas particle. As such, the speed of gases are defined in terms of their root mean square speed.
For the above question,
We have $ v = sqrt(\dfrac{{3RT}}{M}) $ or $ v = sqrt(\dfrac{{3KT}}{m}) $ and temperature is $ T1 $ , on dissociation the molar mass as well as the molecular mass get halved. Using the second equation, the R.M.S speed $ v $ after dissociation as given by
 $ v = {(\dfrac{{3K \times 2T1}}{{\dfrac{m}{2}}})^{1/2}} $ = $ 2{(\dfrac{{3KT1}}{m})^{1/2}} $ = $ 2V1 $ .

Note :
The RMS calculation gives you the root mean square speed not velocity. This is because velocity is a vector quantity that has magnitude and direction. The RMS calculation only gives the magnitude or speed. The temperature must be converted to kelvin and the molar mass must be found in $ Kg $ to complete the other problems.