Lecture 2
Ideal Gas Law
Stull: pg 4-9, Wallace & Hobbs: 63-67
Atmospheric Pressure
Atmospheric pressure change is approximately exponential with height, z.
If temperature were uniform with height (it’s not) then:
P  Poe
( a / T ) z
a= 0.0342 K/m, Po = scaled pressure = 101.325 kPa
P  Poe
z / H p
Hp = 7.29 Km = scale height for pressure, it is the
e-folding distance for pressure curve.
Atmospheric Pressure
P  Poe
z / H p
Because pressure decreases with height monotonically (one direction), P
can be used as a surrogate for a measure of altitude (Fig. 1.4, Stull).
(Example in Stull, pg. 6)
Density
  m /V
Density is defined as mass per unit volume.
• Density increases as the number and molecular weight of
molecules in a volume increase.
• Average density at sea level = 1.255 kg m-3 (Table 1.2, Stull)
Density
Because gases such as air are compressible, air density can
vary over a wide range. Density decreases roughly
exponentially with height in an atmosphere of uniform
temperature.
  o e
z / H p
ρo = 1.225 kg/m3, scale height, Hp = 8.55 km.
Temperature
When a group of molecules (microscopic) move in the same
direction, motion is called wind (macroscopic).
When they move in random directions, motion is associated
with temperature.
Higher temperatures, T are associated with greater average
molecular speeds, v.
T  a  mwv
2
a = 4 ·10-5 k m-2s2, scale height, mw= molecular weight of
most gases.
Temperature
Absolute units such as K (Kelvin) must be used for
temperature in all thermodynamic and radiative laws!
Standard average sea-level temperature is:
T= 15°C = 288 K = 59 °F
Virtual Temperature
For moist air, the gas constant changes because water vapor
is less dense than dry air. To simplify things, a virtual
temperature, Tv can be defined to include the effects of water
vapor on density:
Tv  T  (1  0.61 w)
where w is the water vapor mixing ratio (g water vapor / kg dry air) and all
temperatures are in K.
Moist air of temperature T behaves as dry air of temperature Tv.
Equation of State
Because pressure is caused by the movement of molecules,
we might expect the pressure P to be greater where there
are more molecules (i.e., greater density ρ), and where they
are moving faster (i.e., greater temperature T).
The relationship between pressure, density, and temperature
is called the Equation of State.
For dry air the gases in the atmosphere have a simple
equation of state known as the ideal gas law.
Ideal Gases
Molecules have zero volume
No intermolecular forces
Elastic collisions with walls of container
Works well for the Earth’s atmosphere
Molar Form
Variables
Pressure (p)
Volume (V)
Temperature (T)
Number of moles (n)
Ideal Gas Law
pV
*
 R (constant)
nT
Exercise: Determine units of R*
Universal Gas Constant
*
R
= 8.314
-1
-1
Jmol K
Ideal Gas Law: Molar Form
pV  nR T
*
Memorize this!
Sealed Container
cylinder
Gas
n fixed
Massless, Frictionless Piston
piston
Gas
Weights can be added or removed to change
pressure
Equilibrium  internal pressure = external pressure
Gas
Keep T fixed, Vary p
*
nR T constant
V

p
p
i.e., pressure and volume
are inversely proportional
(Boyle’s Law)
Increase Pressure
Increase Pressure
Increase Pressure
Ideal gas law
 V decreases
Increase Pressure
Ideal gas law
 V decreases
Increase Pressure
Ideal gas law
 V decreases
Decrease Pressure
Decrease Pressure
Decrease Pressure
Ideal gas law
 V increases
Decrease Pressure
Ideal gas law
 V increases
Decrease Pressure
Ideal gas law
 V increases
Keep p fixed, Vary T
nR T  nR * 
T
V
 
p
 p 
*
i.e., volume is directly
proportional to temperature
Increase T
Ideal gas law
 V increases
Apply Heat
Increase T
Ideal gas law
 V increases
Apply Heat
Increase T
Ideal gas law
 V increases
Apply Heat
Decrease T
Ideal gas law
 V decreases
Apply Cooling
Decrease T
Ideal gas law
 V decreases
Apply Cooling
Decrease T
Ideal gas law
 V decreases
Apply Cooling
Fix V
 nR
p  
 V
*

T

i.e., p is directly proportional to T
Increase T
Ideal gas law
 p increases
Piston doesn’t
move
Pressure
builds up
Apply Heat
(internal pressure 
external pressure)
Exercise
n = 1.00 mol
V = 1.00 m3
T = 300 K
Calculate p (in Pa)
Solution

1.00mol  8.314J  mol
p
1.00m
 2490J  m
3
3
 2490N  m  m
2
3
 2490N  m  2490Pa
1
K
1
 300K
Partial Pressure
Consider a mixture of k gases
Volume, V; temperature, T; pressure,
p; # moles = n
ni = # moles of gas i

n1 + n 2 +  + n k = n
Partial pressure, pi:
*
ni R T
pi 
V
Dalton’s Law
k
p

p
 i
i 1
Proof

RT
pi    ni 

i 1
 i 1  V
*
nR T

V
k
k
*
Ratios
pi ni

p n
Ratio of partial pressure to total pressure = mole fraction
Partial Volume
*
ni R T
Vi 
p
Volume Ratios
Vi ni

V
n
Explains why mole fractions are also
called volume fractions.
Ideal Gas Law – Mass Version
Start with molar version
pV  nR T
*
Let m = mass
Let M = molecular weight
(1)
Mass & Moles
m  n M

m
n
M
(2)
Substitute (2) into (1)
m *
pV 
R T
M
Rearrange
R
pV  m 
M
*
Define

T

*
R
R
M
Finally
pV  mRT
(3)
R is called the specific gas constant.
It depends on the molecular weight of the gas.
Gas Constant for Dry Air
Md = 28.97 gmol-1
1
1
1
1
8.314J  mol  K
Rd 
1
28.97g  mol
 0.2870J  g  K
Switch to SI Units
Replace g by kg
1000g
Rd  0.2870J  g  K 
kg
1
1
1
 287.0 J  kg  K
1
Alternate Forms
In meteorology, we are not dealing with
fixed volumes
Divide both sides of (3) by V
m
p  RT
V
What is m/V?
Density. Symbol: 
Density Form of Ideal Gas Law
p  RT
Exercise
p = 1.000 atm
= 1013.25 hPa
T = 15.00C
= 288.15K
Calculate the density of dry air
Solution
p
1013.25hPa


Rd T 287.0J  kg1  K 1  288.15K
 1013.25x102 

Pa


 
1
1
 287.0  288.15  J  kg  K  K 
 N  m 2
 1.225 
1
 J  kg



 N m

3
  1.225kg  m
 1.225
1 
 N  m  kg 
2
Yet Another Form of the Ideal Gas
Law
Divide both sides of (3) by m
V 
p   RT
m
V/m = specific volume
Symbol: 
Specific Volume Form of Ideal Gas
Law
p  RT
Don’t Memorize All the Forms!
Remember these


molar form: pV = nR*T
mass form: pV = mRT
Remember that  = m/V and  = V/m.
Note that  = 1/.
Gas Constant for Moist Air
Rair = R/Mair
Mair  Md/(1 + 0.61w)
Thus,
Rair
R*
R*


M air M d 1  0.61w 
*
R

 1  0.61w 
Md
 Rd  1  0.61w 
Gas Law for Moist Air
p  RairT
or
p  RairT
Density of Moist Air
p
p
1


RairT Rd T 1  0.61w

 dry
1  0.61w
For a given temperature and
pressure, the greater the mixing
ratio, the smaller the density.
Example
Given: T = 300K, p = 1000 hPa

(Take all zeros to be significant)
 dry
p

Rd T
1.000x105 Pa

287.0 J  kg1  K 1  300K
N  m 2
 1.16
J  kg1
N  m 2
 1.16
N  m  kg1
 1.16kg  m 3
Continued
Now, let w = 0.040 (high!)
3
1.16kg  m

1  0.61 0.040
3
1.16kg  m

1.02
3
 1.13kg  m
Conclusion
Density of moist air at most 3% less than
density of dry air
For most purposes, Rd is OK.
Virtual Temperature, Tv
Idea: Use Rd for moist air
Absorb information about moisture content
into T
p  RT  Rd (1  0.61w)T
p  RT  Rd (1  0.61w)T 
Tv
Ideal Gas Law with Tv
p  Rd Tv
Tv > T, but the difference is usually
small
T vs. Tv
Tv  T
 0.61w
T
Let w = 0.040

(About as large as w gets)
Tv  T
 0.61  0.040  .024
T
(A 2.4% difference)