Ideal Gas Law = a special equation of state - a function relating P, r and T
ASSUMPTIONS: The Ideal Gas Law assumes that the molecules are negligibly small and the collisions are like those of hard spheres - no excitation of molecules, no dissociation, no dissipation, no vibration,....
All the versions we are interested in have Pressure proportional to Temperature. The proportionality is a Density times a constant. The constant depends on what kind of density you want. - number of molecules per volume (n) , mass/volume (r), moles/volume (N) ...
P=n kT or P=r/M RT or P=N RT
In atmospheric science we usually keep the volume V constant - so the version PV=NRT (where N= number of molecules) is not often used. We are usually interested in a unit volume for which n=N/V = ( number of molecules / volume ) is more useful.
P=n kT : n= #molecules/volume, k = Boltzmann constant
P=N RT : N = #moles/volume, R = k NA where NA is # molecules in a mole
P=r/M RT : r= mass/volume, M=mass /mole.
Example 1- At the surface of Earth (P= 1 bar and T = 270K, M = 0.028 kg/mole) what is r?
Picking one of these densities - say n - we can then use the ideal gas law to translate the baratropic law we derived in class 1 from pressure to density...
P = P0 exp (-z/H) => n = n0 exp (-z/H)or for mass density r= r0 exp (-z/H)
where in both cases H = scale height = kT/mg
m= mass of average molecule in atmosphere = average mass number <A> x mass of proton mp
Unit check: kT/mg - ( kg m2 s-2 K-1)(K) (kg)-1 (m s-2)-1 = m
For Earth, T~270 K, <A>=28.96 amu, g=9.8 m s-2 making H = (1.38 x 10-23)(270) / ((28.96 x 1.6 x 10-27)(9.8)) = 8000m
... Homework 2 involves adding H to your table of planetary atmospheres....
Back to the density profile.... n = n0 exp (-z/H) or for mass density r= r0 exp (-z/H)
Let's integrate this from the surface to infinity to get the TOTAL MASS in a column of unit area (1 m2 or cm2)
Here we are using our force balance relation (see class 1) dP = - rg dz
Psfc = P0 = surface pressure, g = surface gravity.
Taking the case of the Earth we have Mc = 1 bar/ g = 105 Pa / 9.8 m s-2 = 104 kg m-2 - that's 10 tons of atmosphere above every square meter!
Just for completeness - let's multiply by the area of the Earth 4 p R2 => 5 x 1018 kg total mass. Seems a lot? Divide by the mass of the Earth (6 x 1024 kg) - the atmosphere is only 0.00008% of the mass - nada.
Now do the same for Titan Example 2 ..... P0 = 1.6 bar, g= 1.36 m s-2 - what's Mc?
If we take this total column mass Mc and spread it in a uniform layer of height H - what would be the density <r>?
Mc = <r> x H - so, <r> = Mc / H = 104 kg m-2 / 8000 m = 10/8 = 1.25 kg m-3
Wait a minute!!! That was what we got for r0 - the surface density of an exponential atmosphere.... (Can you prove they are equivalent?)
Example 3 - what's the density of the atmosphere at the surface of Titan?
We have assumed so far that T = constant and that M = constant (or <A> = constant - the composition is constant). Realistically....
or... expanding out farther...
Troposphere-from the Greek tropo (turning)-The lowest region of the
Earth’s atmosphere just above the surface where convection and steep temperature
gradients occur. Contains 90% of the mass of Earth’s atmosphere.
Tropopause-the altitude at which the troposphere pauses-the height of the tropopause varies with latitude, but is near 10 km.
Stratosphere-from the Latin strato (stratified)-a region of the upper atmosphere in which the temperature increase with altitude making the atmosphere relatively stable against vertical motions. It contains about 9.5% of the mass of Earth’s atmosphere, including the ozone layer.
Stratopause-end of the stratosphere at about 50km.
Mesosphere-A region of the Earth’s atmosphere in which the temperature declines with altitude. It is the location where small shooting stars burn up, and contains the lowest lying, heavily ionized portions of the atmosphere. The mesosphere contains about 0.5% of the mass of the atmosphere.
Mesopause-end of the mesosphere at about 85 km
Thermosphere-A low density region in which heating by x-rays and other high energy radiation causes the temperature to increase with altitude. This portion of the atmosphere is highly ionized which is useful for long range radio communication.
Ignorosphere-Region that few people understand and studies are under-funded.
Fundopause-where the funding stops.
And for Titan we have
Remember "the mole" from chemistry? A useful (confusing?) term for a fixed quantity of a chemical.
1 mole = Avogadro's number of atoms or molecules = NA = 6.02 x 1023 mol-1
Mass of a mole M = A x mp x NA where A=atomic or molecular mass number (e.g. 28 for N2) and mpis the mass of a proton. So, M(N2) = 28 x 1.66 x 10-27 x 6.02 x 1023 = 0.028 kg mol-1 = 28 g mol-1
Ideal Gas Law (also see Homework 1 answers)
Version 1: P=n kT
P=pressure (pascal or Pa = N m-2 = kg m-1s-2)
n=number density of gas molecules (# m-3)
k=Boltzman constant = 1.38 x 10-23 (J K-1 or kg m2s-2 K-1)
T= atmospheric gas temperature (Kelvin or K)
units check: m-3 J K-1 K = m-3 J = m-3 kg m2 s-2 = kg m-1 s-2= Pa
Version 2: P=r/( <A> mp)
r=atmospheric gas mass density (kg m-3 )
<A> = average molecular mass number of atmospheric gas (e.g. 28 for pure N2, 28.964 for the Earth's atmosphere on average, at sea level)
mp = mass of the proton = 1.67 x 10-27 kg
units check: are the units of r/( <A> mp) the same as n? kg m-3 / kg = m-3
Version 3: P=r/M RT
M = molecular weight of the atmospheric gas (e.g. =28.964 g mol-1 = 0.0289 kg mol-1 for Earth)
R = gas constant = NA k = 8.31 J K-1 mol-1
NA = Avogadro's number = the number of molecules in a mole = 6.02 x 1023 mol-1
units check: (kg m-3 )/(kg mol-1) ( J K-1 mol-1)(K)=J m-3= kg m2 s-2 m-3= kg m-1 s-2= Pa
Version 4: P=N RT
N = n/ NA = density of moles per volume (mol m-3)
units check: (mol m-3 )( J K-1 mol-1)(K)= m-3 J= m-3 kg m2 s-2 = kg m-1 s-2= Pa