The speed distribution for a sample of N gas | vedclass

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(B) The total area under the $P(v)$ versus $v$ curve represents the total number of particles $N$.From the graph,the area is a triangle from $0$ to $v

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$0.4$

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(B) The total area under the $P(v)$ versus $v$ curve represents the total number of particles $N$.From the graph,the area is a triangle from $0$ to $v_0$ and a rectangle from $v_0$ to $2 v_0$.Area $= \frac{1}{2} 	imes v_0 	imes a + (2 v_0 - v_0) 	imes a = \frac{1}{2} v_0 a + v_0 a = \frac{3}{2} v_0 a$.Since the total area equals $N$,we have $\frac{3}{2} v_0 a = N$,which implies $v_0 a = \frac{2}{3} N$.We need to find the number of particles with speeds between $1.2 v_0$ and $1.8 v_0$. This corresponds to the area under the curve between these limits.Since $P(v) = a$ for $v > v_0$,the area is a rectangle with width $(1.8 v_0 - 1.2 v_0) = 0.6 v_0$ and height $a$.Number of particles $= 0.6 v_0 	imes a = 0.6 (v_0 a) = 0.6 	imes (\frac{2}{3} N) = 0.4 N$.

The speed distribution for a sample of $N$ gas particles is shown below. $P(v) = 0$ for $v > 2 v_0$. How many particles have speeds between $1.2 v_0$ and $1.8 v_0$ (in $N$)?

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