第14章 气体动理论_例题

14
solution:
(1)
N =
∫0
2v0
2v0 a 2N n ( v ) dv = ∫ vdv + ∫ adv ⇒ a = 0 v 0 3v0 0 v0
(2) dN (3) v =
= n ( v ) dv ⇒ ∆ N = ∫ dN =
∫
2 .0 v 0
1 .5 v 0
N adv ⇒ ∆ N = 3
2
Solution: 1.
P = nkT
5 × 10
−6
N = nV
mmHg 5 3 −6 × 1 × 10 × 10 cm × 10 760 N = × 8.31 × 300 K 23 6.02 × 10
N = 1.61×10
12
3
3 N 3 −8 2. Ek = N kT = RT = 1.00 × 10 J 2 N0 2
2 −8 3. Ek ' = N kT = 0.667 ×10 J 2
5 −8 4 . E int = N kT = 1 .667 × 10 J 2
4
Discuss:
What’s possible microscope interpretation of heat Q? (热量Q的可能微观解释是什 热量Q 么?)
vp 99 Solution: Solution: v = vp − ＝ vp 100 100 vp vp vp ∆v = vp + − vp − = 100 100 50
10
Suppose W = v/vp ，therefor , Maxwell Distribution of Speed ：
∫
2 v0
0
vn ( v ) dv N
= (∫
v0
0
a v vdv + v0
∫
2 v0
v0
vadv ) / N
1 2 11 2 2 = ( av 0 + av 0 ) / N = v0 3 3 9
(4)
v
2
=
∫
2 v0
0
v 2 n ( v ) dv N
15
Chapter 14 Kinetic Theory of Gases
第十四章
气体动理论
Example:
A chamber’s volume is 10 cm3 and it contains atmosphere. The chamber is drawn through a vacuum pump to a pressure P=5×10-6 mmHg at temperature T=300K . Find:1. Find:1. The number of molecules in the chamber at this time. 2. The total translational kinetic energy of all molecules. 3. Total rotational kinetic energy 4. Total kinetic energy.
13
Example:
A hypothetical gas of N particle has the speed distribution shown in Fig. [n(v)=0 for v> v0 ]. (a) Express a in terms of N and v0 . (b) How many of the particles have speeds n(v) between 1.5v0 and 2.0 v0 ? (c) Express the average speed a of the particles in terms of v0 . 0 v0 2v0 v (d) Find vrms .
n (v ) Note: f (v) = N
0 0 v
v0
12
Solution:
(a ) N =
∫0
2v0
3 cv 0 3N 2 n ( v ) dv = ∫ cv vdv = =N⇒c= 3 0 3 v0 v0
Answer: (a) 3 N / v 03 (b) 0 . 750 v 0 (c) 0 . 775 v 0 (d) 0 . 633 v 0
6
1.average speed
（平均速率） 平均速率）
v=
vdN ∫ N
=∫
∞
0
8kT 8RT vf (v)dv = = πm πµ
7
2.the root-mean-square speed （方均根速率） root-mean方均根速率）
v =
2
v dN ∫
2
N
=∫
2
∞
0
3kT 3RT v f (v)dv = = m µ
∆N
Hence: Hence:
Nபைடு நூலகம்
＝ (W)∆W = f
4
π
We
2 −W 2
∆W
∆v 1 ∆W = ＝ v p 50 2 2 99 ∆N 4 99 −100 1 ＝ e =1.66% N π 100 50
v 99 W= ＝ vp 100
11
Example:
For the hypothetical gas speed distribution of N particles shown in Fig. [n(v) = C v2 , 0<V< vo ; n(v)=0, v> vo ], find: (a) an expression for C in terms of N and vo , (b) the average speed of the particles, (c) the rms speed of the particles, and (d) the n(v) most probable speed vp .
2
3kT 3RT ∴ v = = m µ
8
3.the most probable speed
f (v)
df (v) ∵ =0 dv
f (v)
2kT 2 RT ∴v p = = m µ
O
v vp
v
9
Example:
Try to calculate the percent of the molecules that the speed is in the interval between vp- vp/100 and vp+vp/100. 100.
5
三、Three speeds:
They are: 1.average speed
v
（平均速率） 平均速率）
2.the root-mean-square speed v 2（方均根速率） 方均根速率） 3.the most probable speed 最可几速率） 最可几速率 vP (最可几速率）.