Question

A container is filled with a sample of gas having n molecules with speed $\alpha, 2\alpha, 3\alpha,…,n\alpha$. The ratio of average speed to root mean square speed is:
A. $\sqrt{\dfrac{3(n+1)}{2(2n+1)}}$
B. $\sqrt{\dfrac{(n+1)}{2(2n+1)}}$
C. $\sqrt{\dfrac{5(2n+2)}{7(3n+1)}}$
D. $\sqrt{\dfrac{3(n+2)}{5(3n+2)}}$

Answer 1

Recall that the average velocity is the arithmetic mean over all individual molecular speeds, whereas the rms speed is the square root of the sum of squares of individual molecular speeds over the total number of molecules. Use these definitions to arrive at expressions corresponding to the respective kinds of speed and dividing the two should give you the desired ratio.

Formula Used:
The root mean square speed of n individual molecular speeds $v_{rms} = \sqrt{\dfrac{1}{n}(v_1^2 + v_2^2 + ….. + v_n^2)}$
Average speed:
$v_{average} = \dfrac{1}{n}(v_1 + v_2 +….. +v_n)$

Complete step by step answer:
Let us begin by differentiating between average speed and root mean square speed.
The average speed is defined as the mean speed of all gas molecules in the sample, whereas the root mean square speed corresponds to the speed of molecules having the same kinetic energy as the average kinetic energy of the sample.
Quantitatively, they are expressed as follows:
The root mean square velocity is the square root of the average of the square of n individual velocities, i.e.,
$v_{rms} = \sqrt{\dfrac{1}{n}(v_1^2 + v_2^2 + ….. + v_n^2)}$
Whereas, $v_{average} = \dfrac{1}{n}(v_1 + v_2 +….. +v_n)$
Given that the individual molecular speeds are $\alpha, 2\alpha, 3\alpha,…,n\alpha$,
The average speed:
$v_{avg} = \dfrac{\alpha+2\alpha+3\alpha+…+n\alpha}{n} = \alpha \dfrac{(1+2+3+….+n)}{n}$
The series of numbers in the numerator are in an arithmetic progression (AP) since successive numbers have a common difference of 1 between them.
The sum of n terms for an AP is given by $S_{AP} = \dfrac{n}{2}(2a +(n-1)d)$ where a is the first term of the series and d is the difference between successive numbers in the series.
Therefore, for our series, we have $S = \dfrac{n}{2}(2 +n-1) = \dfrac{n(n+1)}{2}$
$\Rightarrow v_{avg} = \dfrac{\alpha}{n}.\dfrac{n(n+1)}{2} = \dfrac{\alpha(n+1)}{2}$
Now, the root mean square velocity is given as:
$v_{rms} = \sqrt{\dfrac{\alpha^2 + (2\alpha)^2 + (3\alpha)^2 +…..+(n\alpha)^2}{n}} = \alpha\sqrt{\dfrac{1+2^2+3^2+….+n^2}{n}}$
The series in the numerator have all numbers raised to their second power. The sum of n squares is given by the relation: $S = \dfrac{n(n+1)(2n+1)}{6}$
$\Rightarrow v_{rms} = \alpha\sqrt{\dfrac{1}{n}. \dfrac{n(n+1)(2n+1)}{6}}= \alpha\sqrt{\dfrac{(n+1)(2n+1)}{6}}$
Therefore, the ratio of average speed to rms speed is given as:
$\dfrac{v_{avg}}{v_{rms}} = \dfrac{\left(\dfrac{\alpha(n+1)}{2}\right)}{\left(\alpha\sqrt{\dfrac{(n+1)(2n+1)}{6}}\right)}$
$\dfrac{v_{avg}}{v_{rms}} = \sqrt{\dfrac{\left(\dfrac{(n+1)^2}{4}\right)}{\left(\dfrac{(n+1)(2n+1)}{6}\right)}}$
$\Rightarrow \dfrac{v_{avg}}{v_{rms}} = \sqrt{\dfrac{6(n+1)^2}{4(n+1)(2n+1)}} = \sqrt{\dfrac{3(n+1)}{2(2n+1)}}$
Therefore, the correct choice would be A. $\sqrt{\dfrac{3(n+1)}{2(2n+1)}}$.

Note:
Remember that for a typical gas sample, the net velocity is zero since the constituent particles are moving in all directions. Therefore, the average velocity would yield no quantifiable result. This is where the rms speed comes in, since it does not indicate the resultant velocity direction. In fact, the rms speed is independent of the direction of any of its individual velocities and considers only the magnitude. Thus, we can say that the average speed always gives a magnitude and direction (and is expressed as a vector) of the mean speed, whereas the rms speed gives only the magnitude (and is expressed as a scalar quantity).