E1 3m15 2M1+2Mxr,又由式得E EF 5.mRT 2272 又E=E1 前 故由(2)(3)知T= )f(v) ≤2 (2)由归一化条件[(p=1得 dv N (4)从图中可看出最可几速率为1-2w各速率 (5) wf(v)dv adv 1 f(v)dv 7-5氧气未用时,氧气瓶中V=H1=32L,P1=130atm,71=T 勿1,聊1 氧气输出压强降到P2=l0atm时 氧气每天用的质量66 ( ) RT M m T T RT EF M m M m E 2 1 2 1 0 2 2 1 1 2 5 2 3 , 1 2 5 2 3 = + 后 = + 又由 式得 ③ 又 E前 = E后 ,故由(2)(3)知 (3 5 / ) 8 1 2 1 T T T T + = 7-4 (1) = 0 0 0 0 0 0 2 2 0 ( ) v v a v v v v v v v a f v (2)由归一化条件 = 0 f (v)dv 1 得 0 0 2 0 0 3 2 1 2 3 d d 0 0 0 v v v a v av a v a v v v + = = = (3) 4 ( )d d 0 0 0 0 / 2 0 / 2 N v v v a N Nf v v N v v v v = = = (4)从图中可看出最可几速率为 v0~2v0 各速率. (5) + = = 0 0 0 0 / 2 0 0 ( )d d d v v v v v v a v v a v vf v v v 0 2 0 9 11 6 11 = av = v (6) 0 / 2 0 / 2 0 9 7 d d ( )d ( )d 0 0 0 0 2 1 2 1 v v v v a v v av v f v v vf v v v v v v v v v v v = = = 7-5 氧气未用时,氧气瓶中 V =V1 = 32L, p1 =130atm,T1 =T V RT Mp V RT Mp m 1 1 1 1 1 = = ① 氧气输出压强降到 p2 =10atm 时 V RT Mp V RT Mp m 2 2 2 2 2 = = ② 氧气每天用的质量 0 0 0 V RT MP m = ③