446
10 Heat and Matter
Now, for a small increase in height dy, the pressure decreases by d p and
is given by
d p = −ρg dy = −
P
P 0
ρ 0 g dy
(3)
where we have used (2). The negative sign shows that the pressure
decreases as the height increases.
Integrating
h
0
dy = −
P 0
ρ 0 g
P
P 0
d p
P
∴ h = −
p 0
ρ 0 g
ln
p
P 0
∴ p = p 0 exp
−
ρ 0 gh
P 0
(4)
Now, P 0 V 0 = μRT
(5)
since the temperature is assumed to be constant and μ is in moles. Furthermore
ρ 0 =
μM
V 0
(6)
where M is the molecular weight. Combining (5) and (6)
ρ 0
P 0
=
M
RT
(7)
Substituting (7) in (4)
p = p 0 exp
−
Mgh
RT
(8)
(b) If n is the number density, that is, the number of molecules per unit
volume and m 0 the mass of each molecule then
ρ = m 0 n
and ρ 0 = m 0 n 0
∴
ρ
ρ 0
=
n
n 0
(9)
10 Heat and Matter
Now, for a small increase in height dy, the pressure decreases by d p and
is given by
d p = −ρg dy = −
P
P 0
ρ 0 g dy
(3)
where we have used (2). The negative sign shows that the pressure
decreases as the height increases.
Integrating
h
0
dy = −
P 0
ρ 0 g
P
P 0
d p
P
∴ h = −
p 0
ρ 0 g
ln
p
P 0
∴ p = p 0 exp
−
ρ 0 gh
P 0
(4)
Now, P 0 V 0 = μRT
(5)
since the temperature is assumed to be constant and μ is in moles. Furthermore
ρ 0 =
μM
V 0
(6)
where M is the molecular weight. Combining (5) and (6)
ρ 0
P 0
=
M
RT
(7)
Substituting (7) in (4)
p = p 0 exp
−
Mgh
RT
(8)
(b) If n is the number density, that is, the number of molecules per unit
volume and m 0 the mass of each molecule then
ρ = m 0 n
and ρ 0 = m 0 n 0
∴
ρ
ρ 0
=
n
n 0
(9)