9.18有N个粒子,其速率分布函数为-|||-(v)=av/(v)_(0), 0 v ≤ vo-|||-(v)=a, v0 bigcirc v≤ 2v0-|||-(v)=0, v bigcirc 2v0-|||-f(v)|-|||-a-|||-O v0 2 v-|||-图9.1 习题9.18解用图-|||-(1)作速率分布曲线并求常数a;-|||-(2)分别求速率大于v0和小于v0的粒子数;-|||-(3)求粒子的平均速率。

步骤 1：作速率分布曲线并求常数a
根据题目给出的速率分布函数，我们首先需要确定常数a。速率分布函数f(v)的归一化条件是所有速率区间上的积分等于1，即：
$$
\int_{0}^{2v_0} f(v) dv = 1
$$
将速率分布函数分段代入积分，得到：
$$
\int_{0}^{v_0} \frac{av}{v_0} dv + \int_{v_0}^{2v_0} a dv = 1
$$
计算积分，得到：
$$
\frac{a}{v_0} \int_{0}^{v_0} v dv + a \int_{v_0}^{2v_0} dv = 1
$$
$$
\frac{a}{v_0} \left[\frac{v^2}{2}\right]_{0}^{v_0} + a \left[v\right]_{v_0}^{2v_0} = 1
$$
$$
\frac{a}{v_0} \cdot \frac{v_0^2}{2} + a \cdot v_0 = 1
$$
$$
\frac{a v_0}{2} + a v_0 = 1
$$
$$
\frac{3a v_0}{2} = 1
$$
$$
a = \frac{2}{3v_0}
$$
步骤 2：求速率大于v0和小于v0的粒子数
速率大于v0的粒子数为：
$$
(\Delta N)_1 = N \int_{v_0}^{2v_0} f(v) dv = N \int_{v_0}^{2v_0} a dv = N \cdot a \cdot v_0 = N \cdot \frac{2}{3v_0} \cdot v_0 = \frac{2}{3}N
$$
速率小于v0的粒子数为：
$$
(\Delta N)_2 = N - (\Delta N)_1 = N - \frac{2}{3}N = \frac{1}{3}N
$$
步骤 3：求粒子的平均速率
粒子的平均速率由下式给出：
$$
\overline{v} = \int_{0}^{2v_0} v f(v) dv
$$
将速率分布函数分段代入积分，得到：
$$
\overline{v} = \int_{0}^{v_0} v \cdot \frac{av}{v_0} dv + \int_{v_0}^{2v_0} v \cdot a dv
$$
计算积分，得到：
$$
\overline{v} = \frac{a}{v_0} \int_{0}^{v_0} v^2 dv + a \int_{v_0}^{2v_0} v dv
$$
$$
\overline{v} = \frac{a}{v_0} \left[\frac{v^3}{3}\right]_{0}^{v_0} + a \left[\frac{v^2}{2}\right]_{v_0}^{2v_0}
$$
$$
\overline{v} = \frac{a}{v_0} \cdot \frac{v_0^3}{3} + a \cdot \frac{3v_0^2}{2}
$$
$$
\overline{v} = \frac{a v_0^2}{3} + \frac{3a v_0^2}{2}
$$
$$
\overline{v} = \frac{2a v_0^2}{6} + \frac{9a v_0^2}{6}
$$
$$
\overline{v} = \frac{11a v_0^2}{6}
$$
$$
\overline{v} = \frac{11}{6} \cdot \frac{2}{3v_0} \cdot v_0^2
$$
$$
\overline{v} = \frac{11}{9} v_0
$$