Question

The root mean square velocity of hydrogen molecule at $27 {}^{0}C \ is \ v_H$ and that of oxygen at $402{}^{0}C \ is \ v_O$, then:
$\text{A.}\quad v_O>v_H$
$\text{B.}\quad 4v_O=9v_H$
$\text{C.}\quad 8v_O=3v_H$
$\text{D.}\quad 9v_O=13v_H$

Answer 1

By oxygen and hydrogen molecules, we mean the stable di-atomic molecules, not the individual atoms. Hence in the formulae, mass must be taken for all the atoms and must be taken in S.I. units i.e. kg. Along with this, we must also remember that the temperature should be taken in kelvin, i.e. the absolute temperature.
Formula used: $v_{rms} = \sqrt{\dfrac{3RT}{M}}$

Complete step-by-step solution:
For hydrogen ($H_2$), given;
$M = 2 \ amu = 2 \times (1.67 \times 10^{-27}) = 3.34 \times 10^{-27} kg$
But since in this question, we just have to make a comparison, let’s take masses in AMU($\mu$) only to reduce calculation.
$T = 27 {}^0C = 27+273 = 300K$
Putting the values in $v_{rms} = \sqrt{\dfrac{3RT}{M}}$, we get;
$v_H = \sqrt{\dfrac{3R\times 300}{2\mu }}$
For oxygen ($O_2$), given;
$M = 32 amu$
$T = 402{}^0C = 402 + 273= 675 K$
Putting the values in $v_{rms} = \sqrt{\dfrac{3RT}{M}}$
$v_O = \sqrt{\dfrac{3R\times 675}{32\mu }}$
Now, dividing both the velocities;
$\dfrac{v_H}{v_O} = \sqrt{\dfrac{\dfrac{3R \times 300}{2\mu }}{\dfrac{3R\times 675}{32\mu }}}$
Hence, $\dfrac{v_H}{v_O} = \sqrt{\dfrac{16 \times 300}{675}} = \dfrac83$
Thus, $3v_H = 8v_O$
Hence option C is correct.

Additional information:
In this chapter, i.e. Kinetic theory of gases, we come across several types of velocities like average speed or mean speed, most probable velocity, and root mean square velocity. Each velocity has its own importance and thus cannot be neglected at all. Mean speed is the average speed of all the molecules, irrespective of their direction. Whereas the most probable speed is the speed whose particle in number is maximum. Read that again, it does not mean that the velocity of these particles is maximum. It means that in the category of this velocity, we will find the maximum number of particles.

Note: The root means square is a very important term in energy calculations for gases. It directly corresponds to the energy of the sample unlike mean speed and most probable speed. The chance of mistake is that one could say that the formula of Mean velocity is $\sqrt{\dfrac{8RT}{\pi M}}$. But the truth is average velocity is zero because, in velocity, we have to consider the direction as well. This was one of the postulates of KTG. The above formula is for mean speed.