Question

At what temperature will the RMS velocity of $S{O_2}$ be the same as that of ${O_2}$ at $303K$ ?
A.$850K$
B.$300K$
C.$606K$
D.$404K$

Answer 1

Hint: We have to know that, in science and its applications, the root mean square is characterized as the square foundation of the mean square (the math mean of the squares of a bunch of numbers). The RMS is otherwise called the quadratic mean. RMS can likewise be characterized for a consistently differing capacity as far as a fundamental of the squares of the momentary qualities during a cycle.

Complete answer:
We have to know that the atoms are moving an alternate way with various speeds slamming into each other just as with the dividers of the compartment. Subsequently, their individual speed and thus the dynamic energies continue changing even at a similar temperature. Nonetheless, it is discovered that at a specific temperature, the normal active energy of the gas stays steady.
In the given details,
${T_{{O_2}}} = 303K$
By using the RMS velocity expression, of the given two gases,
For $S{O_2}$ ,
${U_{S{O_2}}} = \sqrt {\dfrac{{3R{T_{S{O_2}}}}}{{{M_{S{O_2}}}}}} $
For${O_2}$ ,
${U_{{O_2}}} = \sqrt {\dfrac{{3R{T_{{O_2}}}}}{{{M_{{O_2}}}}}} $ \[\]
Here, ${M_{S{O_2}}} = 64g$ and ${M_{{O_2}}} = 32g$
Now, equating the both the above expression,
$\sqrt {\dfrac{{3R{T_{S{O_2}}}}}{{{M_{S{O_2}}}}}} = \sqrt {\dfrac{{3R{T_{{O_2}}}}}{{{M_{{O_2}}}}}} $
Then, squaring on both sides,
$\dfrac{{3R{T_{S{O_2}}}}}{{{M_{S{O_2}}}}} = \dfrac{{3R{T_{{O_2}}}}}{{{M_{{O_2}}}}}$
Applying molecular mass values in the above expression,
$\dfrac{{{T_{S{O_2}}}}}{{64}} = \dfrac{{{T_{{O_2}}}}}{{32}}$
Applying given, ${T_{{O_2}}} = 303K$ value in the above expression,
$\dfrac{{{T_{S{O_2}}}}}{{64}} = \dfrac{{303}}{{32}}$
Then, to calculate the temperature of $S{O_2}$ ,
${T_{S{O_2}}} = \dfrac{{303 \times 64}}{{32}}$
Hence, the temperature of the $S{O_2}$ is $606K$ .

Therefore, option (D) is correct.

Note:
We have to know that the root mean square speed is the square base of the normal of the square of the speed. Accordingly, it has units of speed. The explanation we use for the RMS speed rather than the normal is that for a common gas test the net speed is zero since the particles are moving every which way.