Problem 3 - Chapter 20 problem 72
The number distribution for molecules in an ideal gas is
N (v) v = 4 N
m
2 kT
3=2
v2e
where
A0 = 4
mv 2 =2kT
m
2 kT
v = N A0 v 2 e
mv 2 =2kT
v
3=2
;
hence the probability distribution is
P (v) v =
N (v)
v = A0 v 2 e
N
mv 2 =2kT
v:
(a) The most probable molecular speed is found at the maximum of this
probability distribution. Taking the derivative of this distribution and setting
it equal to zero, we …nd
d
P (vo ) = A0 2vo
dv
kT
vo2 = 2 :
m
vo3 m=kT e
mv 2 =2kT
= 0;
Since the rms velocity is given by
2
v 2 = vrms
=3
kT
;
m
we see that the most probable speed is related to the rms velocity via
vo = (2=3)1=2 vrms = :816vrms
(b) The average velocity is
Z
Z 1
1 1
hvi =
vN (v) dv = A0
v3e
N 0
0
mv 2 =2kT
Integrating by parts by de…ning x = v 2 and dy = ve
dx = 2vdv and y =
1
kT
e
m
dv:
mv 2 =2kT
mv 2 =2kT
:
we …nd
Hence
2 kT
mv 2 =2kT
hvi =
A0 v
hvi =
2kT kT
A0
e
m
m
m
1=2
hvi =
e
1
0
1
mv 2 =2kT
(kT )
161=2
=
m1=2 (2 )1=2
r
2kT
+
A0
m
=2
0
Z
kT
m
1
ve
mv 2 =2kT
dv
0
2
A0 = 2
kT
m
2
4
m
2 kT
3=2
8 kT
m
From this the relation between the average speed and the rms velocity is
r
8
hvi =
vrms = :921vrms
3
(c) Note from the kinetic theory of an ideal gas the rms velocity is
v 2 = 3kT =m;
where m is the mass of an individual molecule. From the ideal gas law
P V = N kT so that
P = N kT =V =
M kT
N m kT
=
= kT =m;
V m
V m
where M is the total mass of the gas. The speed of sound is given by
vs2 =
P
= kT =m
vs = ( =3)1=2 vrms :
Since
5=3 for a monatomic gas and
7=5 for a diatomic gas, we see
that the speed of sound is also similar to the rms velocity of a molecule.
Problem 4 - Chapter 21 problem 66
(a) The bulk modulus is de…ned as
P
:
V =V
B=
When the bulk modulus is pressure dependent then a more general expression
is the derivative form
@P
B= V
:
@V
2
For an adiabatic expansion (expansion and compression in a sound wave is
adiabatic) then P V = const: Di¤erentiating this expression with respect to
the volume yields
@P
V + PV
@V
1
=0!
@P
=
@V
P
:
V
Substituting this result into the expression for the bulk modulus yields
B = P:
Thus the speed of sound is
vs =
p
B= =
p
P= :
(b) Since the mass density, ; of a gas (or any material) depends on it
mass divided the volume, = m=V; the bulk modulus can be expressed via
the chain rule as
B =
B =
m @P @
=
@ @V
@P
:
@
m @P
@
m
V2
The speed of sound can then be written as
s
p
@P
;
vs = B= =
@
which is the desired result.
3
2
=
@P
@