Transcript Gases - Research at UVU

Gas Pressure
Air Pressure
Pressure Units
• Units of pressure: atmosphere (atm)
Pa (N/m2, 101,325 Pa = 1 atm)
Torr (760 Torr = 1 atm)
bar (1.01325 bar = 1 atm)
mm Hg (760 mm Hg = 1 atm)
lb/in2 (14.696 lb/in2 = 1 atm)
in Hg (29.921 in Hg = 1 atm)
Universal Gas Behavior
• Unlike solids and liquids, gas behavior is
generally independent of chemical identity.
• Depends on four things only:
– Absolute temperature
– Pressure
– Volume
– Amount (moles)
Kinetic Molecular Theory
• This theory presents physical properties of gases in terms of the
motion of individual molecules.
• Kinetic Theory (in this class) will be based upon six assumptions:
• Average Kinetic Energy  Kelvin Temperature
• Gas molecules are points separated by a great distance
• Particle volume is negligible compared to gas volume
• Gas molecules are in rapid random motion
• Gas collisions are perfectly elastic
• Gas molecules experience no attraction or repulsion
Gas Behavior:
Gases in a Box
• Insert 1 mole of gas into a fixed volume container.
Then:
1. Gas expands to fill the container. Why?
2. The pressure becomes whatever value the gas laws
dictate for that volume, mole, and temperature
combination.
Gas Behavior:
Gases in a Piston
• Insert 1 mole of gas into a piston. Then:
1. Gas fills the piston. Why?
2. The piston changes volume until the pressure
inside is equal to the pressure outside. Why?
Understanding the Gas Laws
• Two keys to understanding the gas laws:
– Understand which parameters are changing
– Understand which are NOT changing
Boyle’s Law
• Pressure–Volume Law (Boyle’s Law):
Boyle’s Law
• Pressure–Volume Law (Boyle’s Law):
1
Volume 
Pressure
• The volume of a fixed amount of gas maintained at constant temperature is
inversely proportional to the gas pressure.
1 1
VP  X
Charles’ Law
• Temperature–Volume Law (Charles’ Law):
Charles’ Law
• Temperature–Volume Law (Charles’ Law):
V  T
• The volume of a fixed amount of gas at constant pressure is directly V
1
proportional to the Kelvin temperature of the gas.
T1
X
Avogadro’s Law
• The Volume–Amount Law (Avogadro’s Law):
Avogadro’s Law
• The Volume–Amount Law (Avogadro’s Law):
V n
• At constant pressure and temperature, the volume of a gas is directly
proportional to the number of moles of the gas present.
V1
X
n1
Collecting the Gas Laws
• Mathematically one can combine all of the
statements we’ve made about gases.
• Two equivalent equations come from this:
– Combined gas law
– Ideal gas law
Combined Gas Law
• Combining the law gives:
P1 V1
 X
n1  T1
• But if it equals a constant, then after any change it will still
be equal to the constant:
P1 V1
P2 V2 P3 V3 P4 V4
 X


n1  T1
n2  T2 n3  T3 n4  T4
• We write it this way:
P1  V1
P2  V2

n1  T1
n2  T2
• Nothing needs to be held constant now
• Remember that anything that does stay constant can be
cancelled.
Ideal Gas Law
• This constant “X” is just a number.
• Units of (pressure * volume) / (moles * temp)
• That is, L·atm·K–1·mol–1
• Numerically, this constant has a value of
R = 0.08206 L·atm·K–1·mol–1
Ideal Gas Law
• The equation then becomes
P V
 R
n T
We usually write it this way instead:
PV = nRT
STP
• Standard temperature: 273.15 K
• Standard pressure: 1 atm
Ideal gas law vs. combined gas law
• Ideal gas law
– Under unchanging conditions
• Combined gas law
– Under changing conditions
What is the volume of one mole
of helium gas at STP?
22.4 L
What is the volume of one
mole of argon gas at STP?
22.4 L
What is the volume of one
mole of radon gas at STP?
22.4 L
What is the density of one mole
of helium gas at STP?
4.003 g / 22.4 L = 0.179 g/L
What is the volume of one
mole of argon gas at STP?
39.948 g / 22.4 L = 1.78 g/L
What is the volume of one
mole of radon gas at STP?
222 g / 22.4 L = 9.91 g/L
What information would you need to
calculate the molar mass of a gas?
• Mass / moles (m / n)
• Enough information to get mass
• P,V,T to use ideal gas law to get n
• What is the molar mass of a gas with a density of
1.342 g/L–1 at STP?
 1.342 g  22.4 Lat STP 
g


  30.06
1 m ole
m ole
 1 L 

Funky questions
• At what temperature do you have 0.1 moles/atm of
helium in a 1 L pure helium sample?
T
PV
1 atm  1L

 121.9 K
nR 0.1 m oles 0.08206 L atm
m ol K
• In one mole of chlorine gas at STP, how many Kelvins
are there per liter?
T
P
1 atm
K


 12.2
V nR 1 m ol 0.08206 L atm
L
m ol K
Gas-phase stoichiometry
• We have a new route to moles PV=nRT
• But we need to know first how two different
gases behave when in the same space
Gas Mixtures
• Two gases in the same container have the
same volume—whatever the volume of the
container is.
• Two gases in the same container have the
same temperature—whatever the
temperature is inside the container.
Gas Mixtures
• Two gases in the same container do NOT have
the same pressure.
• They have whatever pressure they would have
if they were in the container alone.
• That is, solve PV=nRT for each gas in the
mixture separately.
Gas Mixtures
• The total pressure inside the container is
the sum of the pressures of the
individual gases.
Ptotal   Pi
i
• Dalton’s Law of Partial Pressures
New Density Unit: Mole Fraction
• For a two-component system, the moles of
components A and B can be represented by the
mole fractions (XA and XB).
nA
XA 
nA  nB
nB
XB 
nA  nB
XA  XB 1
Gas Stoichiometry
• In gas stoichiometry, for a constant temperature and pressure,
volume is proportional to moles.
• Assuming no change in temperature and pressure, calculate
the volume of O2 (in liters) required for the complete
combustion of 14.9 L of butane (C4H10):
2 C4H10(g) + 13 O2(g)  8 CO2(g) + 10 H2O(l)
Molecular Speed
• It can be shown that:
vrms
3RT

M
Molar mass
• So then for neon:
vrms
J
3  8.314
 298K
3RT
m iles
K m ol
m


 136000
 3000
sec
M
hr
4.00 g
m ol
Mean Molecular Speeds
Collisions
• It can be shown that:
 vmean P
z
kT
Collision frequency
kT

2P
Mean free path
• A room temp gas collides billions of times per
second
• The mean free path is less than 100 nm.
Maxwell speed distribution curves.
Same Behavior vs. Different Behavior
• Most gas behaviors are based upon
comparisons of their relative energies
(temperatures)
– Same temperature = same behavior
• Some gas behaviors are based upon
comparisons of their relative speeds
– Same speed = same behavior
Graham’s Law
• Diffusion is the
mixing of different
gases by random
molecular motion
and collision.
Graham’s Law
• Effusion is when gas
molecules escape
without collision,
through a tiny hole
into a vacuum.
Graham’s Law
• Graham’s Law: Rate of effusion is proportional
to its rms speed, vrms.
Rate  v rms
3RT

M
• For two gases at same temperature and
pressure: Rate1
M2
M2


Rate2
M1
M1
Behavior of Real Gases
• Test of
ideal gas
behavior.
Compressibility factor
• Z = PV/RT
This plot assumes
room temperature.
Real Gases
• All the assumptions of kinetic molecular
theory break down when explored in
sufficient detail.
• Two assumptions break down first:
– The volume of gas molecules is negligible
– There are no attractive or repulsive forces
between molecules
Non-negligible volumes
• The volume of molecules affects pressurevolume behavior more than temperaturepressure behavior.
• For a given small volume, the pressure will be
higher than the ideal gas suggests..
Behavior of Real Gases
• Test of
ideal gas
behavior.
Volume non-idealities
seen here!
Non-negligible interactions
• The long-range interactions of particles are
attractions, not repulsions.
• Thus a real gas sample takes up less space than the
ideal gas law suggests, when the molecules are not
crowded together.
• This effect fades as molecules move faster.
Behavior of Real Gases
• Test of
ideal gas
behavior.
Attractive force
non-idealities
seen here!
Behavior of Real Gases
• Corrections for non-ideality require a non-ideal gas
law. The van der Waals equation is one of them:

n 
 P  a 2   V – n  b   nRT
V 

2
Intermolecular
Attractions
Excluded
Volume
Van der Waals Constants
Gas
Helium (He)
Ammonia
(NH3)
Hydrogen (H2)
n-octane
Water
Carbon dioxide
a
(L2 atm / mole2)
0.03412
4.170
b
(L / mole)
0.02370
0.03707
0.2444
37.32
5.464
3.592
0.02661
0.2368
0.03049
0.04267
Other gas laws
• van der Waals:
• Peng-Robinson:
• Redlich-Kwong:
RT
a
P

V
V
 b RT
n
n
RT

P

V
V V

V


        
n
nn

n

RT
A
P

V
V V

B
T   B
n
nn

Unifying the Gas Laws
• Under normal temperatures you can liquefy a
gas simply by raising the pressure
• Above a certain critical temperature (Tc) you
cannot liquefy a gas under any pressure. The
pressure and volume of that “last” liquid are
Pc and Vc
Critical Constants
Species
Tc (K)
Pc (atm)
Vc (L)
Helium
5.195
2.2452
0.0578
Ammonia
405.3
109.84
0.0725
Water
647.126
217.66
0.05595
“Critical” adjustments
• Now we stop using temperature (and pressure
and volume) in the gas laws.
• Instead we write the reduced temperature (TR)
as a fraction of the critical temperature (Tc).
• That is TR = T / Tc
Compressibility
factor plots redone
Atmosphere
Smog (Inversions)
NO2  h  NO  O
O  O2  O3
Brownish haze
Acid Rain
S  O2  SO2
2 SO2  O2  2 SO3
SO3  H 2O  H 2 SO4
Global Warming