Question

At what temperature will the RMS velocity of$S{{O}_{2}}$be the same as that of${{O}_{2}}$at 303K?
A.850 K
B. 300 K
C. 606 K
D. 404 K

Answer 1

The individual molecules of any gas possess a velocity called as root mean square velocity or RMS velocity. This velocity depends on the molecular mass and temperature. This is inferred by the kinetic molecular theory of gases. The formula for calculating the RMS velocity is ${{V}_{RMS}}=\sqrt{\dfrac{3RT}{{{M}_{w}}}}$ , where R is gas constant, T is temperature and ${{M}_{w}}$ is molecular mass.

Complete answer:
According to the kinetic molecular theory of gases, each gaseous particle moves at a certain speed and in random motion. It is difficult to calculate the velocity of any particle or molecule at a particular time in a vast distribution of gas particles, so root mean square or RMS velocity is obtained.
The RMS velocity is calculated as ${{V}_{RMS}}=\sqrt{\dfrac{3RT}{{{M}_{w}}}}$ , where R is gas constant, T is temperature and ${{M}_{w}}$ is molecular mass.
We have been given $S{{O}_{2}}$ and ${{O}_{2}}$ which is at 303 K and the temperature of $S{{O}_{2}}$ is unknown. We have to find at what temperature of $S{{O}_{2}}$, the RMS velocity of ${{O}_{2}}$ be the same at 303K. The molecular mass of $S{{O}_{2}}$is 64 g/mol and that of ${{O}_{2}}$ is 32 g/mol. Keeping these values in the formula of RMS velocity and making them equal, we have:
$\sqrt{\dfrac{3R{{T}_{S{{O}_{2}}}}}{64}}=\sqrt{\dfrac{3R{{T}_{{{O}_{2}}}}}{32}}$
$\sqrt{\dfrac{{{T}_{S{{O}_{2}}}}}{2}}=\sqrt{303}$
${{T}_{S{{O}_{2}}}}=606\,K$
Hence, the temperature at which the RMS velocity of $S{{O}_{2}}$ be the same as that of ${{O}_{2}}$ at 303K is 606 K.

So option C is correct.

Note:
The kinetic molecular theory of gases suggests that the gas particles collide and there is transfer of kinetic energies among the particles. As kinetic energy depends on the temperature and the molecular weight of the particles, therefore the root mean square velocity is affected by the molecular weight and the temperature.