1
THE GASEOUS STATE
Kinetic Theory of Gases – an excellent model of gases
1. Gas molecules have negligible volume compared with space between molecules.
- molecules are spread out
2. Gas molecules move. When they hit side of container they contribute to pressure.
3. All collisions are elastic.
- molecules behave like billiard balls, not soggy softballs
4. Gas molecules have no attraction or repulsion for each other.
- gas molecules don’t know other molecules are around
5. The speed of the gas molecules is proportional to the temperature.
- kinetic energy of gas is related to speed
Kinetic theory of gases is able to explain many properties of gases.
1. Gases are compressible.
2. Gases have low densities.
3. Gases mix completely.
. Gases fill container uniformly.
5. Gases exert pressure on side of container.
Kinetic theory explains diffusion of gases very well.
- Diffusion is the travel of gases through space so that they mix completely.
- Gasoline molecules from open gas can diffuse through room.
PRESSURE
Definition
Force
Area
A square meter column of air weights 101,325 N.
- 22,730 lbs.
101325 N
N
Atmospheric Pr essure 
 101325 2
2
1m
m
SI Unit of pressure is Pascal – Pa
N
1 Pa  1 2
m
N
Atmo. Pr ess.  101325 2  101325 Pa  101325
.
kPa
m
Pr essure 
2
Pressure is measured with a barometer.
vacuum (no air)
pressure of Hg column
atmospheric pressure
Hg pool
- Pressure of mercury column on surface of Hg pool balances atmospheric pressure pushing
on Hg pool.
- Height of column indicates atmospheric pressure.
- At sea level, column of Hg is 760 mm (29.92 in.)
- At sea level, column of H2O is 33.7 ft!!
- see exhibit of atrium of DSC
UNITS OF PRESSURE
1.) Pascals (Pa)
2.) Atmospheres (atm)
3.) Torr or mmHg
Must be able to convert between atmospheres and Torr (mmHg).
1atm  760 Torr  760 mmHg
Example: How many mmHg is 0.913 atm?
- aside: Lowest pressure of Hurricane Katrina (2005)
Example: How many atmospheres is 3.8 x 108 Torr?
- aside: Pressure at the center of the earth
3
BOYLE’S LAW
Pressure and volume of a gas are inversely proportional to each other, all other factors being
constant (such as temperature and amount).
k
1
P
 PV  k  P1V1  P2 V2
P
V
V
Consider cylinder with piston
Pa
- gas molecules inside cylinder hit piston and
cylinder walls creating pressure
P
- initially pressure inside cylinder is
atmospheric pressure, Pa
a
Now apply additional external pressure, Pex.
- I.e., push down on piston.
P
P
ext
- pressure inside is greater
- gas molecules hit walls more often
a
- note volume is smaller
P
P
a
ext
Now pull out piston, to create lower pressure inside.
Pext
Pa
Pa Pext
- total pressue (pressure inside) is less than
atmospheric pressure
- volume of gas is larger
Restating Boyle’s Law
- as pressure increases, volume decreases
- as pressure decreases, volume increases
Example: A gas sample originally has a pressure of 900 Torr and a volume of 125 mL. If
the temperature is constant, what is the pressure of the gas when it occupies 500 mL?
P1V1  P2 V2
 P2 
P1V1 900 Torr 125 mL

 225Torr
V2
500 mL
4
CHARLES’ LAW
Volume and temperature are proportional to each other, all other factors being constant.
V1 V2
V

V kT 
k 
VT
T1 T2
T
As temperature decreases, volume decreases
As temperature increases, volume increases
**Must use temperature on Kelvin scale.**
Consider a foil helium balloon
- pressure inside is approximately atmospheric pressure
- taking balloon from room temperature to cold outside temperature (say 20 F),
“deflates” balloon
- temperature of gas decreases, therefore volume decreases
- taking balloon back inside increases temperature, therefore volume increases
Example: A 4.00 L balloon is purchased at a gift shop at 23 C. If the balloon shrinks to
3.45 L when taken outside, what is the outside temperature?
Must use Kelvin temperatures.
T1 = 23 C + 273 = 296 K
T2 = 255 K – 273 = -18 C (0 F)
5
AVOGADRO’S LAW
As the amount (in moles) increases, volume increases.
As the amount (in moles) decreases, volume decreases.
V
V k n 
k
Vn

n
V1 V2

n1 n 2
Consider cylinder with piston and air pump
air
pump
Pumping air into cylinder increases volume.
air
pump
Pumping air out of cylinder decreases volume.
air
pump
IDEAL GAS EQUATION
Three quantities have been found to be proportional to the volume.
Boyle’s Law
Charles’ Law
Avogadro’s Law
1
P
VT
V
Vn
All three of these proportionalities can be combined into one proportionality.
V
nT
P
Including a proportionality constant, the proportionality becomes an equation.
VR
nT
P
6
Rearranging equation into its most common form yields the Ideal Gas Equation.
PV = nRT
R is called the Ideal Gas Constant.
R = 0.08206 Latm/molK
Ideal gas constant is always the same.
Note ideal gas equation has 4 variables. If we know 3 of the variables, we can solve for the
fourth variable.
Example: How many moles of gas are in a 2.0 L helium balloon at 25 C and at atmospheric
pressure?
PV = nRT
Remember T must be in Kelvin.
T = 25 C + 273 = 298 K
Rearrange ideal gas equation to solve for n.
PV
10
. atm  2.0 L
n

 0.0818 mol
L  atm
RT
0.08206
 298K
mol  K
Example: What is the volume of one mole of gas at standard temperature and pressure
(STP)?
Standard temperature = 0 C = 273 K
Standard pressure = 1 atm
Rearrange ideal gas equation to solve for V.
nR T
V

P
L  atm
 273K
mol  K
 22.4 L
1.0 atm
1mol  0.08206
Ideal gas equation can be used to find ratios between two sets of conditions.
- Use ideal gas equation to spot proportionality relationships.
PV = nRT
- Variables on opposite sides of equation are directly proportional.
P1 P2
PT 

T1 T2
- Variables on same side of equation are inversely proportional.
1
n
 n1T1  n 2 T2
T
Example: A constant amount of gas starts at 384 K, 63.2 L and 1.26 atm. What is the new
pressure when the volume becomes 43.4 L and the temperature becomes 298 K?
PV
P1V1 P2 V2
PV nRT 
 nR  k


T
T1
T2
V T
63.2 L 298 K
 P2  P1 1 2  126
. atm
 142
. atm
V2 T1
43.4 L 384 K
7
APPLICATIONS IN STOICHIOMETRY
Since we can relate number of moles of gas to other properties of gases, we can relate ideal
gas equation to stoichiometric properties.
Example: What volume of SO2 (g) is produced at STP, when 10.0 g of sulfur burns with an
excess of oxygen according to the reaction? S8 (s) + 8 O2 (g)  8 SO2 (g)
** Always compare moles to moles! **
- how many moles of SO2 are produced?
Now use ideal gas equation to convert moles to volume.
Example: How much NaN3 is needed to inflate a 50.0 L air bag to 1.15 atm given the
following chemical reaction? 2 NaN3 (s)  2 Na (s) + 3 N2 (g). Assume the
temperature is 25 C.
First we need to know how many moles of N2 are in an inflated air bag. Use ideal gas
equation.
Now use stoichiometry to find number of NaN3 grams.
8
DALTON’S LAW OF PARTIAL PRESSURE
The total pressure of a system equals the sum of the pressures of the component gases that
each would exert if they were alone.
PT = P1 + P2 + P3 + …
P1, P2, P3, - partial pressures
- Gases behave as if they are not affected by other gas molecules.
- This follows from the kinetic theory of gases.
For air
+
P(N2) =
0.781 atm
P(O2) =
0.209 atm
P(Ar) =
0.009 atm
P(CO2) =
0.001 atm
P(total) = 1.000 atm
total pressure equals sum of
partial pressures
Example: 0.73 mol O2, 1.92 mol N2, 0.41 mol CO2 are inside a 12.5 L container at 470 K.
Calculate the partial pressure of each gas and the total pressure.
Calculate partial pressures
PO 2
PN 2
PCO2
L  atm
n R T 0.73 mol  0.08206 mol  K  470 K


 2.25 atm
V
12.5 L
L  atm
. mol  0.08206
 470 K
n R T 192
mol  K


 5.92 atm
V
12.5 L
L  atm
n R T 0.41 mol  0.08206 mol  K  470 K


 127
. atm
V
12.5 L
P(total) = P(O2) + P(N2) + P(CO2) = 2.25 atm + 5.92 atm + 1.27 atm = 9.44 atm
Note if we sum of the moles of ideal gas first, then use the ideal gas equation to find the total
pressure.
n(total) = n(O2) + n(N2) + n(CO2) = 0.73 mol + 1.92 mol + 0.41 mol = 3.06 mol
Ptotal
L  atm
n R T 3.06 mol  0.08206 mol  K  470 K


 9.44 atm
V
12.5 L
- note same answer as before
- this is not a coincidence!