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