Gases
Define pressure, give units of pressure, and
describe how pressure is measured.
State the standard conditions of
temperature and pressure and convert units
of pressure.
Use Dalton’s law of partial pressures to
calculate partial pressures and total
pressures.
Pressure (P): the force per unit area on a surface.
What causes pressure?
 collisions of the gas molecules with each other and with
surfaces with which they come into contact.
 depends on volume, temperature, and the number of
molecules present.
Pressure = Force
Area
where P = Pressure, F = Force & A = Area

The SI unit for force is the Newton, (N):
the force that will increase the speed of a one-kilogram mass by
one meter per second each second that the force is applied.
Consider a person with a mass of 51 kg. At Earth’s
surface, gravity has an acceleration of 9.8 m/s2.
What is the value of force?
Force = mass x acceleration
Force = 51 kg × 9.8 m/s2
= 500 kg • m/s2 = 500 N
 Pressure is force per unit area, so the
pressure of a 500 N person on an area of the
floor that is 325 cm2 is:
500 N ÷ 325 cm2 = 1.5 N/cm2
 The greater the force on a given area, the greater the
pressure.
 The smaller the area is on which a given force acts, the
greater the pressure.

barometer: device used to measure
atmospheric pressure
 millimeters of mercury (mm Hg)
 A pressure of 1 mm Hg is also called 1 torr in honor of
Torricelli for his invention of the barometer.





torr
atmosphere of pressure (atm)
bar
pounds per square inch (psi)
Pascal (Pa) – SI Unit pressure exerted by a force
of 1 N acting on an area of one square meter
 kiloPascal (kPa)
1 atm = 101.3 kPa = 760 mmHg = 760 Torr
The average atmospheric pressure in Denver,
Colorado is 0.830 atm. Express this pressure in:
a. millimeters of mercury (mm Hg) and
b. kilopascals (kPa)
Given: atmospheric pressure = 0.830 atm
Unknown: a. pressure in mm Hg
b. pressure in kPa
A)
760 mm Hg
atm  mm Hg; atm 
 mm Hg
atm
760 mm Hg
0.830 atm 
 631 mm Hg
atm
B)
101.325 kPa
atm  kPa; atm 
 kPa
atm
101.325 kPa
0.830 atm 
 84.1 kPa
atm
 The pressure of each gas in a mixture is
called the partial pressure.
 John Dalton discovered that the pressure
exerted by each gas in a mixture is
independent of that exerted by other gases
present.
 Dalton’s law of partial pressures: the
total pressure of a gas mixture is the sum of
the partial pressures of each gas.

Dalton derived the following equation:
PT = P1 + P2 + P3 + …
Total Pressure = sum of pressures of each
individual gas
Water molecules at the liquid surface evaporate and
mix with the gas molecules. Water vapor, like other
gases, exerts a pressure known as vapor pressure.
Gases produced in the laboratory are often collected
over water. The gas produced by the reaction
displaces the water in the reaction bottle.
Step 1: Raise bottle until water level inside matches the
water level outside. (Ptot = Patm)
Step 2: Dalton’s Law of Partial Pressures states:
Patm = Pgas + PH2O
To get Patm, record atmospheric pressure.
Step 3: look up the value of PH2O at the temperature of
the experiment in a table, you can then calculate Pgas.
KClO3 decomposes and the oxygen gas was collected
by water displacement. The barometric pressure and
the temperature during the experiment were 731.0
torr and 20.0°C. respectively. What was the partial
pressure of the oxygen collected?
Given:
PT
= Patm = 731.0 torr
PH2O = 17.5 torr (vapor pressure of water at
20.0°C, from table A-8 in
your book)
Patm = PO2 + PH2O
Unknown: PO2 in torr

Solution:
Patm = PO2 + PH2O
PO2 = Patm - PH2O

substitute the given values of Patm and
into the equation:
PO2 =731.0 torr – 17.5 torr = 713.5 torr
mole fraction: ratio of the number of moles of
one component of a mixture to the total
number of moles
Mole fraction of a gas(XA) =
Moles of gas A (nA)
Total number of moles of a gas
(ntot)
Go To: Page 2 of Packet
Partial pressures can be determined from mole fractions
using the following equation:
PA = XA PT
Go To: Page 3 of Packet
Use the kinetic-molecular theory to explain the
relationships between gas volume, temperature
and pressure.
Use Boyle’s law to calculate volume-pressure
changes at constant temperature.
Use Charles’s law to calculate volumetemperature changes at constant pressure.
Use Gay-Lussac’s law to calculate pressuretemperature changes at constant volume.
Use the combined gas law to calculate volumetemperature-pressure changes.



Constant: temperature, amount of gas
If you decrease the volume, what happens
to the pressure?
If you increase the volume, what happens
the pressure?
Pressure and volume are _____________
related.
P1V1 = P2V2
A sample of oxygen gas has a volume of
150.0 mL when its pressure is 0.947 atm.
What will the volume of the gas be at a
pressure of 0.987 atm if the temperature
remains constant?
P1 = 0.947 atm
V1 = 150.0 mL
P2 = 0.987 atm
V2 = ?
PV
(0.947 atm)(150.0 mL O2 )
1 1
V2 

 144 mL O2
P2
0.987 atm
Constant: pressure, amount of gas
 If you increase the temperature of a gas,
what will happen to the volume?
 If you decrease the temperature of gas,
what will happen to the volume?
 Volume and temperature are ______________
related.
 Units: Farenheit, Celsius, and Kelvin
 absolute zero: when all motion stops
 K = 273 + °C.
V1 V2

T1 T2
A sample of neon gas occupies a volume of
752 mL at 25°C. What volume will the gas
occupy at 50°C if the pressure remains
constant?
Temperature must be in KELVIN!!!
V1 = 752 mL
T1 = 25°C
V2 = ?
T2 = 50°C
V1T2
V2 
T1
V1T2 (752 mL Ne)(323 K)
V2 

 815 mL Ne
T1
298 K
Constant: volume, amount of gas
 If you increase the temperature of a gas
what will happen to the pressure?
If you decrease the temperature of gas
what will happen to the pressure?
 Pressure and temperature are _____________
related.
P1 P2

T1 T2
The gas in a container is at a pressure of
3.00 atm at 25°C. Directions on the
container warn the user not to keep it in a
place where the temperature exceeds
52°C. What would the gas pressure in the
container be at 52°C?
Temperature must be in KELVIN!!!
P1 = 3.00 atm
T1 = 25°C
P2 = ?
T2 = 52°C
PT
1 2
P2 
T1
P2 = P1T2 = (3.00 atm) (325 K) = 3.27 atm
T1
298 K
Constant: amount of gas
 combined gas law: used when pressure,
temperature, and volume change within a
system
PV
P2V2
1 1

T1
T2
NOTE: P & V are directly related to
T, while P is inversely related to V
A helium-filled balloon has a volume of 50.0
L at 25.0°C and 1.08 atm. What volume
will it have at 0.855 atm and 10.0°C?
Temperature must be in KELVIN!!
P1 = 1.08 atm
V1 = 50.0 L
T1 = 25.0°C
P2 = 0.855 atm
V2 = ?
T2 = 10.0°C
PV
1 1T2
V2 
P2T1
PV
(1.08 atm)(50.0 L He)(283 K)
1 1T2
V2 

 60.0 L He
P2T1
(0.855 atm)(298 K)
State the law of combining volumes.
State Avogadro’s law and explain its
significance.
Define standard molar volume of a gas and use
it to calculate gas masses and volumes.
State the ideal gas law.
Using the ideal gas law, calculate pressure,
volume, temperature, or amount of gas when
the other three quantities are known.
In the early 1800s, Joseph Gay-Lussac
observed that 2 L of hydrogen can react with
1 L of oxygen to form 2 L of water vapor.
hydrogen gas + oxygen gas → water vapor
2 L (2 volumes) 1 L (1 volume) 2 L (2 volumes)
The reaction shows a simple 2:1:2 ratio in
the volumes of reactants and products:
2 mL, 1 mL, and 2 mL
Avogadro’s law: states that equal volumes
of gases at constant temperature and
pressure contain equal numbers of
molecules.
According to Avogadro’s law, one mole of
any gas will occupy the same volume as one
mole of any other gas at the same
conditions, despite mass differences.
standard molar volume of a gas:
22.41410 L (rounded to 22.4 L)
Gay-Lussac’s law of combining volumes of
gases and Avogadro’s law can be applied in
calculating the stoichiometry of reactions
involving gases.
The coefficients in chemical equations of gas
reactions reflect not only molar ratios, but also
volume ratios (assuming conditions remain the
same).
 example—reaction of carbon dioxide formation:
2CO(g)
+
O2(g)
→
2CO2(g)
2 molecules
2 mole
2 volumes
1 molecule
1 mole
1 volume
2 molecules
2 mol
2 volumes
Number 1 on Practice Sheet
 What volume of nitrogen at STP would be
required to react with 0.100 mol of
hydrogen to produce ammonia?
N2 + 3 H2  2 NH3
0.100 mol H2 x 1 mol N2 x 22.4 L N2
3 mol H2
1 mol N2
= 0.747 L N2
Constant: pressure, temperature
 If you increase the amount of moles, what
happens to the volume?
 If you decrease the amount of moles what
happens to the volume?
 Amount of moles and volume are
____________ related.
V1 = V2
n1 n2
This equation is NOT in the book, it was
calculated during the Gas Simulation Lab
 ideal gas law: relates all variables –
pressure, volume, moles, temperature
PV = nRT
ideal gas constant, R:
 Its value depends on the units chosen for pressure,
volume, and temperature in the rest of the equation.
 Measured values of P, V, T, and n for a gas at near-ideal
conditions can be used to calculate R:
PV (1 atm)(22.414 10 L)
L  atm
R

 0.082 058
nT
(1 mol)(273.15 K)
mol  K
Usually rounded to 0.0821 (Latm/molK)
ALWAYS MATCH UP YOUR UNITS!!!!

What is the pressure in atmospheres
exerted by a 0.500 mol sample of nitrogen
gas in a 10.0 L container at 298 K?
P=?
V = 10.0 L
n = 0.500 mol
T = 298 K
nRT
P
V
(0.500 mol)(0.0821 L  atm)(298 K)
P
 1.22 atm
10.0 L
Number 2 on Practice Sheet
 What volume of nitrogen at 215OC and 715
mmHg would be required to react with
0.100 mol of hydrogen to produce
ammonia?
N2 + 3 H2  2 NH3
Note: This system is NOT at STP!!
0.100 mol H2 x 1 mol N2 = 0.033 mol N2
3 mol H2
P = 715 mmHg
V=?
n = 0.333 mol N2
R = 62.4 LmmHg/molK
T = 25OC + 273 = 488 K
Describe the process of diffusion.
State Graham’s law of effusion.
State the relationship between the average
molecular velocities of two gases and their
molar masses.
REMEMBER:
 DIFFUSION: the gradual mixing of two or
more gases due to their spontaneous,
random motion
 EFFUSION: process when the molecules of a
gas confined in a container randomly pass
through a tiny opening in the container
Graham’s law of effusion:
the rates of effusion of gases at the same
temperature and pressure are inversely
proportional to the square roots of
their molar masses.
MB
rate of effusion of A

rate of effusion of B
MA
Compare the rates of effusion of hydrogen
and oxygen at the same temperature and
pressure.
Given: identities of two gases, H2 and O2
Unknown: relative rates of effusion
Hydrogen = Compound A
Oxygen = Compound B
HINT: Always put the substance with the
larger molar mass on top as compound B.
1. Calculate:
MB
32.00 g/mol
rate of effusion of A
32.00 g/mol



 3.98
rate of effusion of B
2.02 g/mol
MA
2.02 g/mol
2. Rearrange the equation:
rate of effusion of A = 3.98 rate of effusion of B
3. Write a sentence:
Hydrogen diffuses 3.98 times faster than Oxygen