Question

Consider a mixture of \[{\rm{S}}{{\rm{O}}_{\rm{2}}}\] and \[{{\rm{O}}_{\rm{2}}}\] kept at room temperature. Compared to the oxygen molecule, the \[{\rm{S}}{{\rm{O}}_{\rm{2}}}\]molecule will hit the wall with
A. Smaller average speed
B. Greater average speed
C. Greater kinetic energy
D. Greater mass

Answer 1

Hint: Here, we have to use the formula of root mean square (rms) velocity to compare the speeds of \[{\rm{S}}{{\rm{O}}_{\rm{2}}}\]and \[{{\rm{O}}_{\rm{2}}}\]. The rms velocity defines the square root of average velocity. The formula used for calculation of this velocity is\[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \].

Complete Step by Step Solution:
The rms velocity formula is,
\[{v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} \]; where, R stands for gas constant, T stands for temperature, M stands for molar mass of the gas.
As R is gas constant and given that all gases are at same temperature, molar masses of each gas decide its rms speed.
So, \[{v_{rms}} = \sqrt {\dfrac{1}{M}} \],
That means, rms velocity is indirectly proportional to molar mass of gases. So, the gas that possesses the highest molar mass has the lowest rms velocity.
Therefore, we have to calculate the molar masses of oxygen and sulphur dioxide.

Molar mass of \[{\rm{S}}{{\rm{O}}_{\rm{2}}}\]=\[32 + 16 \times 2 = 32 + 32 = 64\,{\rm{u}}\]
Molar mass of \[{{\rm{O}}_{\rm{2}}} = 16 \times 2 = 32\,{\rm{u}}\]
So, we find that molar mass of \[{\rm{S}}{{\rm{O}}_{\rm{2}}}\] molecule is higher than \[{{\rm{O}}_{\rm{2}}}\].
So, the speed of \[{\rm{S}}{{\rm{O}}_{\rm{2}}}\]is less than the \[{{\rm{O}}_{\rm{2}}}\]molecule.
Hence, option B is right.

Note: It is to be noted that average velocity is different from rms velocity. Average velocity defines the arithmetic mean calculation of velocities of different gaseous molecules at a particular temperature. The calculation of average velocity is done by the formula of \[{v_{av}} = \sqrt {\dfrac{{8RT}}{{\pi M}}} \] . The rms velocity is always greater than average velocity.