Question

The rms velocity of a gas at a certain temperature is found to be $12,240\,cm/\sec
$. The most probable velocity of that gas in $cm/\sec $is:
A. ${10^3}$
B. ${10^4}$
C. $11280$
D. 1128

Answer 1

Hint: We know that the root mean square velocity is defined as the square root of the mean of the square of the different velocities possessed by molecules of a gas at a given temperature. While the most probable velocity is the velocity possessed by the maximum number of molecules of a gas at a given temperature.

Complete Step by step answer: To understand the concept of root means square velocity, let us state n moles of an ideal gas in a cubical box of volume V and we will keep the temperature of the box at T.
Then the question is how can we relate the pressure P of the gas that it exerts on the walls and speed of the molecules. Inside the box, we can easily see that the molecules in the box are moving randomly in all directions with varying speeds and they collide with each other and they will also bounce back off it.
For now, we ignored the fact that the molecules are colliding with each other and consider the collisions are happening only with the wall and they are elastic in nature and no loss of energy takes place.
From kinetic gas equation we can easily determine the relation between microscopic property like pressure with another microscopic property like volume in this box which is given by,

$p = \dfrac{{nM}}{{3V}}\nu _{rms}^2\,\,\,\,\,where\,{\nu _{rms\,}}is\,the\,\,root\,\,mean\,square\,velocity$
From ideal gas equation,$PV = nRT$
On substituting we get,
$\dfrac{{nRT}}{V} = \dfrac{{nM}}{{3V}}\nu _{rms}^2$
$\Rightarrow {\nu _{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
The most probable velocity is given by-${\mu _{mps}} = \sqrt {\dfrac{{2RT}}{M}} $
On dividing both speed expressions, we get,
$
\dfrac{{{\mu _{mps}}}}{{{\mu _{rms}}}} = \sqrt {\dfrac{2}{3}} \\
\Rightarrow {\mu _{mps}} = \sqrt {\dfrac{2}{3}} \times {\mu _{rms}}\\
\Rightarrow {\mu _{mps}} = \sqrt {\dfrac{2}{3}} \times 12240\,cm/\sec = 9993 \approx 10000 = {10^4}cm/\sec
$
Hence, option (B) is the correct option.

Note: Most of the students get confused between the average velocity, drift velocity and average velocity. One must go through these velocities before solving the problems. There is also another kind of velocity known as average velocity. It is defined as the arithmetic mean of a gas at a given temperature.