Thermal Velocities and Maxwell-Boltzmann Distribution Homework Assignment - Prof. Pradeep , Assignments of Thermal Physics

Solutions to problems related to thermal velocities and the maxwell-boltzmann distribution. Topics include defining variables, calculating the number of particles with certain speeds, and determining the average density, mean free path, and collision frequency.

Typology: Assignments

Pre 2010

Uploaded on 09/17/2009

koofers-user-xar-1
koofers-user-xar-1 🇺🇸

10 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
HOMEWORK ASSIGNMENT 11
Problems Chapter 11
11-7
The Maxwell speed distribution (number of particles with speed in the range
dv) is
N(v)dv = 4πN m
2πkT 3/2
v2emv2
2kT dv
a) Define
x=v
vm
, dx =dv
vm
, vm=2kT
m1/2
Then
N(x)dx =N(xvm)d(xvm) = 4N
πx2ex2dx
b) The total number of particles with v < v0is
P(v0) = Zv0
0
dvN (v) = Zx0
0
dxN(x)
=4N
πZx0
0
x2ex2dx
=2N
πZx0
0
xdex2
dx dx
=2N
πZx0
0
dxex2
dx d(x)
dx ex2
dx
=2N
πx0ex2
0Zx0
0
ex2dx
c) For v0=vm, x0= 1 and
P(vm)
N=2
πe1Z1
0
ex2dx= 0.428
11-8 The definitions are
vm=2kT
m1/2
,v=8kT
πm 1/2
, vrms =3kT
m1/2
For the Oxygen molecule
m= 32 1.67 1027
1
pf2

Partial preview of the text

Download Thermal Velocities and Maxwell-Boltzmann Distribution Homework Assignment - Prof. Pradeep and more Assignments Thermal Physics in PDF only on Docsity!

HOMEWORK ASSIGNMENT 11

Problems Chapter 11

The Maxwell speed distribution (number of particles with speed in the range

dv) is

N (v)dv = 4πN

m

2 πkT

v

2 e

− mv

2 2 kT (^) dv

a) Define

x =

v

vm

, dx =

dv

vm

, vm =

2 kT

m

Then

N

∗ (x)dx = N (xvm)d (xvm) =

4 N

π

x

2 e

−x^2 dx

b) The total number of particles with v < v 0 is

P(v 0 ) =

∫ (^) v 0

0

dvN (v) =

∫ (^) x 0

0

dxN

∗ (x)

4 N

π

∫ (^) x 0

0

x

2 e

−x^2 dx

2 N

π

∫ (^) x 0

0

x

de

−x^2

dx

dx

2 N

π

∫ (^) x 0

0

d

xe−x

2

dx

d (x)

dx

e

−x^2

 (^) dx

2 N

π

x 0 e

−x^20 −

∫ (^) x 0

0

e

−x^2 dx

c) For v 0 = vm, x 0 = 1 and

P(vm)

N

π

e

− 1 −

0

e

−x^2 dx

11-8 The definitions are

vm =

2 kT

m

, v =

8 kT

πm

, vrms =

3 kT

m

For the Oxygen molecule

m = 32 ∗ 1. 67 ∗ 10

− 27

a) For T = 300K

vm = 393. 8 , v = 444. 3 , vrms = 482. 3

b) For T = 1000K

vm = 2273, v = 2565, vrms = 2784

11-11 The average density is

N

V

NA

V

N

NA

NA

V

n =

NA

V

P V

RT

NAP

RT

The mean free path is

` =

π

N

V

σ =

π

NAP

RT

πd

2 = 3. 85 ∗ 10

− 8

and the ratio of the mean free path to the particle diameter is

`

d

Finally, the collision frequency is

νc =

v

`

π

kT

m

NAP

RT

πd

2 = 9. 9 ∗ 10

9