Problem 3 - (a) Chapter 20 problem
72 (b) Show that the avp
erage velocity satis…es hvi =vrms = 8=3 for a molecule in an
pideal
P=
gas.(c) Given that the velocity of sound in an ideal gas is vs =
where is a constant depending
on
the
nature
of
the
molecules
in
p
=3:
the gas, show that vs =vrms =
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 =vrms = (2=3)1=2 = :816
(b) The average velocity is
Z
Z 1
1 1
hvi =
vN (v) dv = A0
v3e
N 0
0
1
mv 2 =2kT
dv:
Integrating by parts by de…ning x = v 2 and dy = ve
kT
e
m
dx = 2vdv and y =
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=2
mv 2 =2kT
(kT )
16
=
m1=2 (2 )1=2
r
2kT
+
A0
m
1
=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
p
hvi =vrms = 8=3 = :921
(c) From the ideal gas law P V = N kT so that
P = N kT =V =
N m kT
M kT
=
= kT =m;
V m
V m
where M is the total mass of the gas. Hence the speed of sound is given by
vs2 =
P
= kT =m
vs = ( =3)1=2 vrms :
Problem 4 - Atmospheric pressure in Denver
(a) Given the average molecular weight of Earth’s atmosphere is m = 29:0
and = 1:4; show that speed of sound in the Earth’s atmosphere at T = 20 C
is vs = 343m=s: (b) Using the ideal gas law and dP=dz =
g; show that
P (z) = P0 e z=zo where zo = RT =mmol g: (c) Assuming that atmospheric
pressure at sea level is P0 = 1atm = 14:7lbs=in2 and the average temperature
is T = 14 C determine the atmospheric pressure in Denver (z = 1600m) in
lbs=in2 . Compare your result to the actual average barometric pressure of
12:1lbs=in2 :
2
(a) The speed of sound in Earth’s atmosphere is
p
p
p
vs =
kT =m =
RT =mmol = 1:4 8:314
vs = 343m=s
(273 + 20) =:029
(b) From the expression for the pressure gradient and P ( ) obtained from
the ideal gas law in part (a) we …nd
dP
=
dz
dP
=
P
P
ln
=
P0
mP
g=
kT
Z P
mmol g
dz !
RT
P0
g=
z=zo ! P = P0 e
mg
mmol g
P =
P
kT
RT
Z z
1
dP
=
dz
P
zo 0
z=zo
where zo = RT =mmol g:
(c) From the given data zo is given by
zo = 8:314 (287) = (:029
9:8) = 8396m:
Hence the atmospheric pressure in Denver is
P = 14:7e
1600=8396
= 12:15lbs=in2 :
This result is just slightly higher than the actual average of 12:1lbs=in2
3