1000 Solved Problems in Modern Physics

264 4 Thermodynamics and Statistical Physics

P−=

dAdt

2 λ

∫∞

0

νdn

∫∞

0

e−r/λdr

∫π/ 2

0

sinθcosθ

(

mu+rcosθ

dmu

dz

)

dθ

The factore−r/λis included to ensure that the molecule in traversing the

distancertoward dAdoes not get scattered and prevented from reaching dA.

Similarly, transport of momentum upward, from molecules in the lower

hemisphere through dAin time dtis

P+=

dAdt

2 λ

∫∞

0

νdn

∫∞

0

e−r/λdr

∫π/ 2

0

sinθcosθ

(

mu−rcosθ

dmu

dz

)

dθ

Hence net momentum transfer to the reference plane through an area dAin

time dtis

P=P−−P+=

dAdt

λ

mdu

dz

∫∞

0

νdn

∫∞

0

re−r/λdr

∫π/ 2

0

cos^2 θsinθdθ

=

dAdt

λ

m

du

dz

n<ν>

λ^2

3

=

m

3

dAdt

du

dz

λn<ν>

(the first integral givesn<ν>, the second oneλ^2 and the third one a factor

1/3)

Momentum transported per second is force

F=

λ

3

dAn<ν>m

du

dz

The viscous force is

ηdA

du

dz

=

λ

3

dAn<ν>m

du

dz

or η=

1

3

mn<ν>λ=

1

3

ρ<ν>λ

wheremn=ρ=density of molecules.

4.16λ=

1

√

2 πnσ^2

=

1

√

2 π× 3 × 1025 ×(2. 5 × 10 −^10 )^2

= 1. 2 × 10 −^7 m

f=

ν

λ

=

1 , 000

1. 2 × 10 −^7

= 8. 33 × 109 s−^1

4.17 T 1 V 1 γ−^1 =T 2 V 2 γ−^1 ,

(

V 2

V 1

)γ− 1

=

T 1

T 2

or 2γ−^1 = 1. 32 ,γ= 1. 4

Number of degrees of freedom,

f=

2

γ− 1

=

2

1. 4 − 1

= 5