1000 Solved Problems in Modern Physics

262 4 Thermodynamics and Statistical Physics

f= 4 π

(

32 × 1. 67 × 10 −^27

2 π× 1. 38 × 10 −^23 × 300

) 3 / 2

(200)^2

exp

(

−

32 × 1. 67 × 10 −^27 × 2002

2 × 1. 38 × 10 −^23 × 300

)

×(2)

= 2. 29 × 10 −^3

4.11<ν^2 >^1 /^2 =

(

3 kT

m

) 1 / 2

νrms(600 K)=[νrms(300 K)](600/300)^1 /^2

= 1270 ×

√

2 = 1 ,796 m/s

4.12 Relative velocityνrelof one molecule and another making an angleθis

νrel=(ν^2 +ν^2 −2(ν)(ν) cosθ)^1 /^2 = 2 νsin(θ/2)

Now, all the direction of velocitiesvare equally probable. The probability

f(θ) thatvlies within an element of solid angle betweenθandθ+dθis given

by

f(θ)= 2 πsinθdθ/ 4 π=

1

2

sinθdθ

νrelis obtained by integrating overf(θ) in the angular interval 0 toπ.

<νrel>=

∫π

0

νrelf(θ)=

∫π

0

2 νsin

(

θ

2

)(

1

2

sinθdθ

)

= 2 ν

∫π

0

sin^2

(

θ

2

)

cos

θ

2

dθ= 4 ν

∫π

0

sin^2

θ

2

d

(

sin

θ

2

)

= 4 ν/ 3

4.13νe=(2gR)^1 /^2 ;νrms=(3kT/m)^1 /^2

νrms=νe

T=

2 mgR

3 k

=

2 ×(2× 23. 24 × 10 −^27 )(9.8)(6. 37 × 106 )

3 × 1. 38 × 10 −^23

= 1. 4 × 105 K

4.14 Fraction of gas molecules that do not undergo collisions after path lengthx
is exp(−x/λ). Therefore the fraction of molecules that has free path values
betweenλto 2λis
f=exp(−λ/λ)−exp(− 2 λ/λ)
=exp(−1)−exp(−2)
= 0. 37 − 0. 14 = 0. 23

4.15 Consider a volume element dV = 2 πr^2 sinθdθdrlocated on a layer at a
heightz=rcosθ.Ifmuis the momentum of a molecule at the XY-plane
atz=0, then its value at dVwill bemu+

(d

dzmu

)

rcosθ(Fig. 4.2). At an

identical layer below the reference plane dA, the momentum would be