Gases and Heat

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Gases and Heat
Chapter 18
Boltzmann’s Version of pV=nRT
pV = NkBT
N = Number of molecules
kB = Boltzmann’s Constant (1.38 X 10-23 J/K)
Mean Free Path
• Average distance between collisions
• Also used to describe electrons moving
through wires
• Light passing through glass, water
l=
1
4\/2 p (N/V)r2
l = mean free path
r = 0.5 X 10-10 m (monoatomic gases)
1.0 X 10-10 m (diatomic gases)
Example
A nitrogen molecule (N2, diatomic) is at 1.0 atm
of pressure and 20oC
a. Convert the temperature and pressure to the
proper units
b. Calculate the value N/V using Boltzmann’s
Equation
c. Calculate the mean free path of the
molecule.
Root-Mean Square Speed
• Velocity has a direction
• Overall, in a container of gases, the velocity is
zero
• Average speed is not zero
vrms = \/(v2)avg
vrms =
3kBT
m
m is the mass of one molecule (kg)
Example 1
Nitrogen molecules (N2) are at room
temperature, 20oC.
a. Calculate the mass of one molecule (kg)
b. Calculate the rms speed
Example 2
Cesium atoms can be cooled to 1.0 mK (m is 10-6)
a. Calculate the mass of 1 cesium atom
b. Calcuate the rms speed
Translational Energy
Eavg = 3/2 kBT
(for one molecule)
Eavg = 3/2NkBT = 3/2 nRT (for all molecules)
N = number of molecules
n = number of moles
Example
A balloon contains 2.00 grams of Helium at 25.0oC
a. Calculate the average energy of one atom of
Helium.
b. Calculate the average energy of the sample.
Second Law of Thermodynamics
2nd Law
– Natural processes tend to move toward a
state of greater disorder
– Heat goes hot to cold (Clausius)
– No device converts all heat to work (KelvinPlanck)
DS >0
Calculating Entropy
DS = Q
T
• Reversible process
• Constant Temperature
• T in Kelvin
Calculating Entropy: Example 1
How much entropy is produced when 30 grams
of ice melts at 0oC?
Q = mLfusion
Q = (0.030 kg)(3.33 X 105 J/kg)
Q = 9,990 J
DS = 9,990 J
273 K
=
36.6 J/K
Calculating Entropy: Example 2
How much entropy is produced when 500 g of
water boils and vaporizes at 100oC?
Q = mLvaporization
Q = (0.500 kg)(22.6 X 105 J/kg)
Q = 11.3 X 106 J
DS = 11.3 X 106 J=
3030 J/K
373 K
Calculating Entropy: Example 3
How much entropy is produced when 300 g of
lead melts at 327 oC? The latent heat of
fusion of lead is 2.45 X 104 J/kg
Q = mLvaporization
Q = (0.300 kg)(2.45 X 104 J/kg)
Q = 7.35 X 103 J
DS = 7.35 X 103 J
=
12.3 J/K
600 K
Entropy and the Universe
• Natural processes tend to move towards a
state of greater disorder
• DSuniverse > 0
• Only local order can be produced
• Cleaning your room example
• Eventually the entire universe will be the same
temperature (no heat engines possible). Heat
death.
Thermal Pollution
• Thermally polluting power plants
– Coal plants
– Oil plants
– Nuclear plants
• Non-thermally polluting power plants
– Hydroelectric
– Tidal energy
– Wind
– Solar