1
PRESSURE
Force
Area
A square meter column of air weights 101,325 N.
- 22,730 lbs.
101325 N
Atmo. Pr ess. 
 101325 Pa  101.325 kPa
1m 2
SI Unit of pressure is Pascal – Pa
Definition:
Pr essure 
1 Pa  1
N
m2
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
Relative pressures are measured with manometers.
1. Open-end manometer
Mercury (or another dense fluid) is used to fill the tube. The end of the tube is
exposed to atmospheric (laboratory) pressure.
System pressure is less than
atmospheric pressure.
System pressure is greater
than atmospheric pressure.
psys  p Hg  p atm
psys  p atm  p Hg
2
2. Closed-end manometer
The end of the tube is sealed from the atmosphere and pump to a vacuum.
The end of the tube is sealed from the atmosphere and pump to a vacuum.
vacuum
System pressure cannot be
directly compared to
atmospheric pressure.
psys  p Hg
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).
1
k
 P1 V1  P2 V2
P
P
 PV  k
V
V
Consider cylinder with piston
Pa
- gas molecules inside cylinder hit piston and
cylinder walls creating pressure
- initially pressure inside cylinder is
atmospheric pressure, Pa
Pa
Now apply additional external pressure, Pex.
- I.e., push down on piston.
Pext
- pressure inside is greater
- gas molecules hit walls more often
Pa
- note volume is smaller
Pa Pext
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?
4
CHARLES’ LAW
Volume and temperature are proportional to each other, all other factors being constant.
V
V1 V2
V kT 
k
VT


T
T1 T2
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.
5
AVOGADRO’S LAW
As the amount (in moles) increases, volume increases.
As the amount (in moles) decreases, volume decreases.
V
V1 V2
V  n V k n 
k


n
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
V
1
P
Charles’ Law
VT
Avogadro’s Law 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
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.
6
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.
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.
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.
PT 
P1 P2

T1 T2
- Variables on same side of equation are inversely proportional.
n
1
T
 n1T1  n 2 T2
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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
P1 V1 P2 V2
 nR  k
PV nRT 


T
T1
T2
V T
63.2 L 298 K
 P2  P1 1 2  1.26 atm
 1.42 atm
V2 T1
43.4 L 384 K
Ideal gas equation can also be modified to calculate the molar mass of a gas.
Reconsider density
d
m
V
Mass can be rewritten in terms of molar mass and number of moles. m  n M
Thus density can be rewritten
d
nM
V
Rearrange to solve for molar mass and substitute ideal gas equation.
Md
RT
V
d
P
n
or M 
mRT
PV
However, the easiest method to understand is to apply the definition of molar mass
directly. That is, to calculate the molar mass, find the mass of the gas and the number
of moles of gas and divide moles into mass.
m
M
n
Example: 255 mL flask is filled with an unknown gas to a pressure of 745 Torr. The
evacuated flask weighs 238.120 g. The filled flask weighs 239.092 g. The
lab temperature is 22.0 C. Find the molar mass of the gas.
mass of gas 239.092 g – 238.120 g = 0.972 g
temperature 22.0 C + 273.15 = 295.2 K
pressure P  745 Torr
volume
V  255 mL
1atm
 0.980 atm
760 Torr
1L
 0.255 L
1000 mL
8
moles of gas n 
PV
0.980 atm  0.255 L

 0.0103 mol
RT 0.08206 L  atm  295.2K
mol  K
M
m
0.972 g

 94.4 g / mol
n 0.0103 mol
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.
Mass of
reactant
(g)
Mass of
product
(g)
M
Molar
mass
M
Molar
mass
pressure
of reactant
(atm)
PV = nRT
Moles of
reactant
(mol)
M
Molarity
Volume of
reactant
(L)
Balanced
equation
Moles of
product
(mol)
PV = nRT
M
Molarity
Volume of
product
(L)
pressure
of product
(atm)
9
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 temperature to be 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.
10
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) =
P(O2) =
P(Ar) =
+ P(CO2) =
P(total) =
0.781 atm
0.209 atm
0.009 atm
0.001 atm
1.000 atm
total pressure equals sum
of partial pressures
Example: 0.73 mol O2, 1.92 mol N2, 0.41 mol C2H6 are inside a 12.5 L container at
470 K. Calculate the partial pressure of each gas and the total pressure.
Calculate partial pressures
L  atm
 470 K
nRT

mol
K
 2.25 atm
PO 2 

12.5 L
V
L  atm
n R T 1.92 mol  0.08206 mol  K  470 K
 5.92 atm
PN 2 

V
12.5 L
L  atm
mol
.
.
 470 K
0
41
0
08206

nRT
mol  K
 1.27 atm
PC 2 H 6 

V
12.5 L
0.73 mol  0.08206
P(total) = P(O2) + P(N2) + P(C2H6) = 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(C2H6) = 0.73 mol + 1.92 mol + 0.41 mol = 3.06 mol
L  atm
n R T 3.06 mol  0.08206 mol  K  470 K
Ptotal 

 9.44 atm
V
12.5 L
- note same answer as before
- this is not a coincidence!
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DEFINITION: Mole fraction
n
n1
X1  1 
n n1  n 2  n 3 
For previous example
0.73 mol
0.73 mol

 0.24
0.73 mol  1.92 mol  0.42 mol 3.06 mol
1.92 mol
0.42 mol

 0.63
.
 013
X C2 H 6 
3.06 mol
3.06 mol
X O2 
XN2
Sum of all mole fractions always equals one.
X O 2  X N 2  X C 2 H 6  0.24  0.63  0.13  1.00
Note that for gases in constant volume container, partial pressures can be used to find
mole fractions.
X1 
n1 P1V / RT P1


n PV / RT P
 P1  X1P
- This equation is a restatement of Dalton’s Law !
Note for oxygen in above example
X O2 
2.25 atm
2.25 atm

 0.24
2.25 atm  5.92 atm  1.27 atm 9.44 atm
12
COLLECTING GASES OVER WATER
The pressure that a liquid exerts on the atmosphere is called the vapor pressure.
Molecules on surface of water have kinetic energy.
Some of the molecules get enough kinetic energy to escape into gas phase.
- pan of water at room temperature eventually evaporates.
When collecting gases over liquid, the vapor pressure of the liquid will contribute to the
total pressure of the gases.
volume
of gas
during reaction
reaction completed
Example: 4.84 g of CaCO3 decomposes in a dilute HCl solution. The CO2 gas is
collected over water into a volume of 1.16 L. The atmospheric pressure is
755 Torr. The temperature is 21 C. The vapor pressure of water is 18.65
Torr at 21 C.
a) Find the number of moles of CO2 collected.
Ptotal  PCO 2  PH 2O  755Torr PCO 2  755Torr  18.65Torr  736Torr
1atm
 116
. L
PV
760 Torr
n

 0.0466 mol
L  atm
RT
 294 K
0.08206
mol  K
736 Torr 
b) Calculate the percent yield
CaCO3 (s) + 2 HCl (aq)  CaCl2 (aq) + H2O (l) + CO2 (g)
4.84 g CaCO 3 
1 mol CaCO 3

1 mol CO 2
100.09 g CaCO 3 1 mol CaCO 3
% yield 
 0.0484 mol
0.0466 mol
 100%  96.2 %
0.0484 mol
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KINETIC THEORY 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
Higher temperature increases average speed
.55
0.5
298 K
0.4
398 K
f ( u  2  298)
0.3
f ( u  2  398)
2000 K
f ( u  2  2000)
0.2
0.1
0
0
0
2
4
0
6
u
8
10
12
12
velocity
- The average velocity of the gas molecules increases as the temperature increases.
- Note that a sample of gas molecules doesn’t have just one speed; the speeds are a
distribution.
14
Increased molar mass decreases average speed
4.5
SF6, 298 K
4
3
f ( u  2  298)
CO2, 298 K
f ( u  44  298)
f ( u  146  298)
2
SF6, 2000 K
f ( u  146  2000)
1
0
0
H2, 298 K
0
1
0
2
3
u
4
4.5
velocity
- The average velocity of the gas molecules decreases as the molar mass increases.
- Heavier molecules need more energy to move.
Breathing He makes your voice higher.
Breathing SF6 makes your voice lower.
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.
4. Gases fill container uniformly.
5. Gases exert pressure on side of container.
6. Gases flow easily (have low viscosity).
15
DIFFUSION AND EFFUSION
Diffusion is the transport of gases through space so that they mix completely.
- Gasoline molecules from open gas can diffuse through room.
- Diffusion rate of depends on speed on molecules.
- Diffusion rate also depends on density of gas.
- Higher density that molecules are closer together, smaller mean free path.
- Smaller mean free path means more collisions.
Effusion is the transport of gases thru a small pinhole.
- Principles of effusions nearly the same as diffusion.
- Kinetic theory explains diffusion of gases very well.
Graham’s Law of Effusion
The rate of effusion is inversely proportional to the square of the molar mass of the gas.

- Heavier gases move slower.
1
M

M2
1

2
M1