AOS 101 Discussion 303 TA Brian Miretzky

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AOS 101 Discussion 303TA- Brian Miretzky
Ideal Gases February 11th, 2008
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REVIEW

• COUNTOURING- YOU LOVE IT!
• Helps gain a better sense of location and
  strength of certain past or present weather
  features
• Time to put someone on the spot

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For temperature contouring

• Gradient the spatial rate of change of a given
  field (i.e. how close are the lines together)
• Lines closely packed steep gradient
• For isotherms, closely packed lines (temperature
  gradients) front.

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For Pressure Contouring

• For isobars, closely packed lines strong winds.
• Also, winds blow nearly parallel to isobars
• Counterclockwise around lows cyclonic
• Clockwise around highs anticyclonic

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Station Model Homework

• Generally very good
• Couple of points to make
• Visibility- what is when it is clear?
• Winds
• Weather and cloud cover relation to different
  pressures
• Examples/ Current Weather

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Changing gears

• Robert Boyle
• Was the first to analyze the behaviors of gases
  scientifically in the 17th century
• He discovered that PV constant if temperature
  is also held constant
• Thus if P increases then V must decrease if
  temperature is constant
• This is now known as Boyles Law

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Jacque Charles

• A French chemist who several years after Boyle
  came to another important conclusion
• At a constant pressure, the volume of any gas is
  directly proportional to the temperature
• V/T constant
• i.e., the ratio of the volume to the temperature
  will remain constant if the pressure remains
  constant
• Thus, if we increase the temperature of a gas,
  yet keep the pressure the same, the volume will
  also increase.

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One more French guy- Joseph Louis Gay-Lussac
- In 1802, another French chemist, Joseph Louis
Gay-Lussac discovered yet another important law
of gases. - At a constant volume, the pressure of
any gas is directly proportional to the
temperature (in degrees Kelvin!) P/T constant
(with constant volume!!) - i.e., the ratio of
the volume to the temperature will remain
constant if the pressure remains constant - Thus,
if we increase the temperature of a gas, yet keep
the pressure the same, the volume will also
increase.
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Combining these laws gives

• The Ideal Gas Law
• PV nRT
• Or
• P ?RT (because ? m/V) This is what is used in
  meteorology because it makes for easier
  comparison by combining two variables into one

T constant As P increases, ? increases P
constant As T increases, ? decreases ?
constant As T increases, P increases

• But, as Prof. Martin said, dont just memorize
  this! First, remember the definitions of each of
  the three variables

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Variable Definitions

• P, Pressure Force of the molecules that make up
  the gas, exerted on the surface the gas is making
  contact with (per unit area) P Force/Area.
  Units1 mb 1 hPa
• 1 hPa 100 Pascals (Standard Unit)
• T, Temperature Average kinetic energy of the
  molecules that make up the gas. KE 1/2mv2
  scale (K)
• ?, Density Mass per unit volume (of the gas
  analyzed). ? m/V. The more molecules in a
  specific volume, the greater the density. (kg/m3)
• R, The gas constant for dry air 287 J/kg K

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Examples to think about

• 1) Whipped cream in a sealed bell jar.
• 2) Pressure is decreased inside the jar.
• What will happen to the whipped cream?
• More precisely, what will happen to the tiny air
  bubbles that are inside the whipped cream?
• In this case, temperature is constant, and
  pressure and density (volume) are allowed to
  fluctuate

First Step Always figure out which variable is
held constant!
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Result

• The whipped cream appears to have expanded!
• What happened?
• As pressure decreased in the system and
  temperature remains constant, the density of the
  air bubbles inside the whipped cream must also
  decrease.

Because density mass/volume, the volume of the
air inside the whipped cream actually increases
as pressure decreases
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Lets try a calculation

• If the temperature of an air parcel is 252.5 K,
  and its density is 0.690 kg/m3, what is the
  pressure of the air parcel?
• From the gas law, p ? R T
• T 252.5 K
• ? 0.690 kg/m3
• R 287 J/kg K
• So, p 0.690 252.5 287 50000 Pa
• In millibars, p 500 mb
• Again look to my website for links to other
  websites or tutorials that deal with this weeks
  discussion