Question

Graph shows a hypothetical speed distribution for a sample of $N$ gas particle (for \[V > {V_0}\]​;\[\dfrac{{dN}}{{dV}} = 0\],\[\dfrac{{dN}}{{dV}}\]​ is a rate of change of number of particles with change velocity.)

A. The value of \[a{V_0}\] is \[2N\]
B. The ratio of \[\dfrac{{{V_{avg}}}}{{{V_0}}}\] is equal to \[\dfrac{2}{3}\]
C. The ratio of \[\dfrac{{{V_{rms}}}}{{{V_0}}}\] is equal to \[\dfrac{1}{{\sqrt 2 }}\]
D. Three fourth of the total particle has a speed between \[0.5{V_0}\]and \[{V_0}\]

Answer 1

Hint: To solve this problem find the value of the rate of change of number of particles with change in velocity. Then find the relation between the \[{V_0}\] and the number of particles \[N\].

Formula used:
The equation of straight line is given by,
\[y = mx + c\]
where, \[m\] is the slope of the straight line and \[c\] is the interception on the Y-axis.

Complete step by step answer:
From the graph we can see that the slope of the curve is, \[\dfrac{a}{{{V_0}}}\] since the maximum value of Y-axis is \[a\] and on the X-axis it is \[{V_0}\] minimum value is zero. The interception on the Y-axis is zero. The Y variable is here the rate of change of number of particles with change in velocity \[\dfrac{{dN}}{{dV}}\] and the X-axis plot is \[{V_0}\].

Hence, we can write the equation of the curve as, \[\dfrac{{dN}}{{dV}} = \dfrac{a}{{{V_0}}}V\]
Now, we can rewrite this as, \[dN = \dfrac{a}{{{V_0}}}VdV\]
Integrating over the limit for \[N\] \[0 \to N\] and for \[V\] \[0 \to {V_0}\] we have,
\[\int\limits_0^N {dN} = \dfrac{a}{{{V_0}}}\int\limits_0^{{V_0}} {VdV} \]
Putting the value of the limit we have,
\[N = \dfrac{a}{{{V_0}}}\dfrac{{{V_0}^2}}{2}\]
Up on simplifying we have, \[N = \dfrac{{a{V_0}}}{2}\]
So, we can see, \[2N = a{V_0}\]
Hence, the value of \[a{V_0}\] is \[2N\]

Hence, option A is the correct answer.

Note: The r.m.s value of \[V\] is \[{V_{rms}} = \dfrac{1}{N}\int\limits_0^{{V_0}} {{V^2}dN} \] which is equal to the \[\sqrt {\dfrac{{a{V_0}^3}}{{4N}}} \] and the average value of \[V\] is \[{V_{avg}} = \dfrac{1}{N}\int\limits_0^{{V_0}} {VdN} \] which is equal to the value \[\dfrac{{a{V_0}^2}}{{3N}}\]. Hence, all the options except option (A) are incorrect. You can perform the integration by yourself to check the answers for further insight.