# ppi.bioinfo.asia.edu.twklngCh12_temp_gas_law.ppt

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```					            Chapter 12
Temperture, Kinetic Theory, and the
Gas Law
12.1 Temperature
Operational - measured by thermometer,
using physical properties of materials, such as
volume change, resistance change and color change
Relation among Fahrenheit, Celsius, and Kelvin temperture scales.
Logarithmic scale of tremendous range of
tempertures in nature.
12.2 Thermal expansion of solids and liquids

p.299 Table 12.1, coefficient of linear and volume(~3*linear)
expansion
Objects expand in all directions as temperature incresass.
(a) Area increases
(b) Size of the hole increases
(c) Volume increase
This crack in a concrete sidewalk was created by thermal stress,
an indication of how great such stress can be.
12.3 The Ideal Gas Law
Ideal gas  no attraction among gas molecules

PV = NkT

P = pressure of gas , V = volume of gas
T =temperature (K)
N= number of atoms or molecules in the gas
k = 1.38*10-23 J/K

P ↑ as T ↑
V ↓ as P ↑
- indepdent of the type of gas        Gas – atoms or molecules
are widely separate
When air is pumped into a deflated tire,
its volume increased without pressure change

To a certain point, the tire
wall resist further expansion
and P ↑ with more air.

Once the tire is inflated,
P ↑ as T ↑
Moles and Avogardro’s Number, NA
A mole = amount of a substance that contains as many atoms or
molecules as there are atoms in exactly 0.012kg of carbon-12.
NA = 6.02*1023 /mol

A more precise value  wait until Einstein’s theory used to
determine the size and masses of atoms

Example 12.5 How many atoms and molecules
are there in a volume of gas at STP?
Solutions
Given STP  P=1 atm, V=1 m3, T=00C,
N=?
N=PV/kT = 1.01*105*1/(1.38*10-23*273)
= 2.68*1025
NA = 6.02*1023 /mol

Macroscopic  like this mole of table tennis balls covering the
Earth to a depth of about 35 km !
The Ideal Gas Law Restated using Mole

PV  NkT
N
PV     N A kT
NA
R  N Ak    R = Universal gas constant
PV  nRT      = 6.02*1023 *1.38*10-23
= 8.31 J/(mol K)
= 1.99 cal/(mol K)
= 0.0821 L atm/(mol K)
n = number of moles
12.4 Kinetic Theroy: Molecular explanation of pressure and temperture

-Kinetic theoryexplain pressure and temperture on a submicroscopic view
-Assume elastic collision of gas molecules with the wall of a container
Force on the wall (rate of change of momentum)
Number of molecules ↑, P ↑
Average velocity ↑, P ↑
1
PV  Nmv 2        See p.306 for derviation
3
v 2  v X  vY  vZ
2    2     2
PV  NkT
1
mv 2  NkT
3
1       3
K E  mv  kT
2

2       2
3kT
vrms  v 
m
12.4 Kinetic Theroy: Molecular explanation of pressure and temperature
Example 12.8 Energy and Speed of a gas molecule
(a) Average KE of a gas molecule at 200C = ?
(b) rms speed of N2 molecule at 200C = ?
Solutions
(a) K E  1 mv 2  3 kT
2        2       =1.5*1.38*10-23*293 = 6.07*10-21 J
(b)         3kT   3(1.38 *1023 * 293
vrms                    26    = 511 m/s
m        4.65 *10

Molecules bounce furiously-       Individual molecules do not move very
Billions of collisions per second far- sound waves are transmiited
at speeds related to the molecular speed
Distribution of Molecular Speeds, p.308
Maxwell-Boltzmann distribution of molecular speeds in an ideal gas

- vp = the most likely speed < vrms
- Only a tiny fraction of molecules have
very high speeds
Distribution of Molecular Speeds, p.308

vp is shifted to higher speeds
temperature.

Total probaility = 1
12.5 Phase Changes (相變)

Real gas – attraction among molecules
Condensing to liquid (Gas  Liquid)
freezing to a solid (Liquid  Solid)
Volume dramatic ↓

Absolute zero
12.5 Phase Changes (相變), PV diagram

Hyperbolic shape(雙曲線), isotherms(等溫線)

Condensate/Vapourize
雙曲線

Liquid
Gas
V↓ a little, as P↑
V↓ a lot, as P↑
12.5 Phase Changes (相變)
Critical point TC – above which liquid cannot exist
CO2 cannot be liquefied at T > 310C
Critical pressure – minimum pressure needed for liquid to exist at TC
12.5 Phase Changes (相變)
p.310, Table 12.2 (Critical temp.and pressure),Helium is last liquefied gas
PT graph – Phase diagram for water                 Liquid phase not exist
solid lines  phase equilibrium                    at any P

(1) liquid-vapour curve
- boiling point
- critical point
(2) solid-liquid curve
- melting point (00C at 1 atm)
- at fixed temp., (00C)                                 boiling point
ice  water (by↑P)
PT graph – Phase diagram for water

(3) solid-vapour curve
- interesting, at lower pressure,
there is no liquid phase
- exist either as gas or solid
- sublimation(昇華)
- for water this is true for
P > 0.0060 atm
Triple point  all three phases in
equilibrium, at 273.16 K (0.010C)

A more accurate calibration temp.
than melting point !

p.310, Table 12.3 Triple point
p.315 - Phase diagram of CO2
Equilibrium between liquid and gas at two different boiling points

Equilibrium at lower temperature
 Lower rate of condensation and
vaporization                     Equilibrium at higher temperatur
Higher rate of condensation and
vaporization

Dynamics
equilibrium
Vapor pressure, Partial pressure and Dalton’s Law

Vapor pressure = the gas pressure created by the liquid or solid phases
of a substance

Partial pressure = the gas pressure created if it alone occupied the total
volume available

Dalton’s Law of partial pressures

total  pressure           partial  pressure
i  gases
i

Total pressure = sum of partial pressures of the component gases,
Assume ideal gases and no chemical reactions
12.6 Humidity, Evaprotation, and Boiling

p.312, Table 12.4 - Saturated vapor density of water
T ↑  vapor pressure ↑
Relative humidity – how much water vapor is in the air compare with
the maximum possible.

Relative humidity is related to the partial pressure of water vapor in the air
-At 100% humidity (dew point)
partial pressure of water vapor = vapor pressure

- partial pressure < vapor pressure  evaporataion (humidity < 100%)
- partial pressure > vapor pressure  condensation

Relative humidity,
RH= (vapor density/saturation vapor density)*100%
Example 12.10 Density and vapor pressure

At T=200C, vapor pressure = 2.33*103N/m2, use ideal gas law to
Calculate the density of water vapor in g/cm3 that would create a
partial pressure = vapor pressure.
Solutions
PV = nRT  n/V = P/RT
P/RT = 2.33*103/8.31/293 = 0.953 mol/m3
The molecular mass of water = 18.0  18.0 g/mol
 mass = 18*(number of mole = n)
 density = mass/V = 0.953*18 = 17.2 g/m3
12.6 Humidity, Evaprotation, and Boiling

(a) Some water molecules can escape – Maxwell-Boltzmann distribution
(b) Sealed container – evaporation will continue until
evaporation = condensation
Vapor pressure = partial pressure of vapor  Saturation(飽和)
12.6 Humidity, Evaprotation, and Boiling
Example 12.11 Humidity and Dew point
(a) Calculate RH, T=250C, density of water vapor = 9.4 g/m3
(b) At what T=?, will this air reach 100% RH – dew point
(c) What is the humidity when T=25.00C and the dew point is -10.00C
Solutions
(a) RH = (9.40/23.0)*100% = 40.9% Table 12.4
(b) From Table 12.4, 9.4 g/m3  RH is 100% at 10.00C
(c) From Table 12.4, at -10.00C, saturated vapor density = 2.36 g/cm3
RH = (2.36/23.0)*100% = 10.3%
12.6 Humidity, Evaprotation, and Boiling

Why does air formed when water boils ?
Bubble started at 200C has P = 1 atm
As T↑ water vapor enters the bubble, vapor pressure ↑
Bubble expands to keep P = 1 atm
As T↑more water vapor enter the bubble  bubble expand
Buoyant force on it increase  bubble breaks  boiling
Homework, due next week

Ch. 11 11.21, 11.29, 11.31, 11.45, 11.53
Ch. 12 12.37, 12.47, 12.57

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