....

**Ideal Gas Law****Mass of atmospheres****Realistic atmospheres - layers**

*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 : **

**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 = P

_{0}exp (-z/H) => n = n_{0}exp (-z/H)or for mass density r= r_{0}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 m_{p}

*Unit check: kT/mg - ( kg m ^{2} 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 = n_{0} exp (-z/H) or for mass density
r= r_{0} exp (-z/H)

Let's integrate this from the surface to infinity to get the TOTAL MASS in
a column of unit area (1 m^{2} or cm^{2})

Here we are using our force balance relation (see class 1) dP = - rg dz

Psfc = P_{0} = surface pressure, g = surface gravity.

Taking the case of the Earth we have Mc = 1 bar/ g = 10^{5} Pa / 9.8
m s^{-2} = 10^{4} 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 R^{2} => 5 x 10^{18} kg total mass. Seems a lot?
Divide by the mass of the Earth (6 x 10^{24} kg) - the atmosphere is
only 0.00008% of the mass - nada.

Now do the same for **Titan** *Example 2 .*....
P_{0} = 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 = 10^{4} kg m^{-2 } / 8000 m = 10/8 = 1.25 kg m^{-3}

Wait a minute!!! That was what we got for r_{0}
- 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...

Some definitions:

**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

**The Mole**

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 = N_{A} = 6.02 x 10^{23}
mol^{-1}

Mass of a mole M = A x m_{p} x N_{A} where A=atomic or molecular
mass number (e.g. 28 for N_{2}) and m_{p}is the mass of a proton.
So, M(N_{2}) = 28 x 1.66 x 10^{-27} x 6.02 x 10^{23}
= 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^{-1}s^{-2})

n=number density of gas molecules (# m^{-3})

k=Boltzman constant = 1.38 x 10^{-23} (J K^{-1} or kg m^{2}s^{-2}
K^{-1})

T= atmospheric gas temperature (Kelvin or K)

*units check: m ^{-3} J K^{-1} K = m^{-3} J = m^{-3}
kg m^{2} s^{-2} = kg m^{-1} s^{-2}= Pa*

**Version 2: **P=r/( <A> m_{p})
kT

r=atmospheric gas mass density (kg m^{-3}
)

<A> = average molecular mass number of atmospheric gas (e.g. 28 for pure
N_{2}, 28.964 for the Earth's atmosphere on average, at sea level)

m_{p} = mass of the proton = 1.67 x 10^{-27} kg

*units check: are the units of r/( <A> m _{p})
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 = N_{A} k = 8.31 J K^{-1} mol^{-1}

N_{A} = Avogadro's number = the number of molecules in a mole = 6.02
x 10^{23} mol^{-1}

*units check: (kg m ^{-3} )/(kg mol^{-1}) ( J K^{-1}
mol^{-1})(K)=J m^{-3}= kg m^{2} s^{-2} m^{-3}=
kg m^{-1} s^{-2}= Pa *

**Version 4: P= N RT**

*units check: (mol m ^{-3} )( J K^{-1} mol^{-1})(K)=
m^{-3} J= m^{-3} kg m^{2} s^{-2 }= kg m^{-1}
s^{-2}= Pa*