Pressure
Pressure: Force applied per unit area.
Barometer: A device that measures atmospheric
pressure.
Manometer: A device for measuring the pressure
of a gas in a container.
Pressure
Units of Pressure
Pascal: (abbrev. Pa) The SI unit for pressure.
1 standard atmosphere = 1.000 atm = 760.0 mm Hg = 760.0 torr
1 standard atmosphere = 101,325 Pa = 101.325 kPa
1.000 atm = 14.69 psi
Pressure and Volume:
Boyle’s Law
Boyle’s Law: Pressure times Volume equals a constant. PV=k where
k is a constant at a specific temperature for a given amount of gas.
If we know the volume of a gas at a given pressure, we can predict the
new volume if the pressure is changed, provided that neither the
temperature nor the amount of gas is changed.
Pressure and Volume:
Boyle’s Law
Example
A sample has a volume of 1.51 L at a
pressure of 635 torr. Calculate
the final volume of the gas if the
final pressure is 785 torr.
PV
1 1  PV
2 2
V2 
PV
(635torr )(1.51L)
1 1

 1.22 L
P2
(785torr )
Volume and Temperature:
Charles’s Law
Charles’s Law: Proportionality constant times temperature is equal to
volume. V = bT where T is in Kelvins and b is the proportionality
constant.
Charles’s Law implies that the amount of gas (moles) and pressure are
constant. The volume of the gas is directly proportional to
temperature on the Kelvin scale.
Volume and Temperature:
Charles’s Law
Example
A sample has a temperature of 28oC
and a volume of 23 cm3 at 1 atm.
The final temperature was found
to be 18oC, assuming no change
in pressure. Calculate the final
volume.
28  273.15 K  301K
18  273.15 K  291K
V1 V2

T1 T2
V1
23cm3
V2  T2 ( )  (291K )(
)  22cm3
T1
301K
Volume and Moles:
Avogadro’s Law
Avogadro’s Law: For a gas at constant temperature and
pressure, the volume is directly proportional to the
number of moles of gas.
V = an or
V = a
n
where V is the volume of the gas
N is the number of moles
a is the proportionality constant.
Volume and Moles:
Avogadro’s Law
Example
3H2(g) + N2(g)  2NH3(g)
If one has 15.0 L of H2(g), what
volume of N2(g) is required for a
complete reaction, given that
1mol N 2
both gases are at the same
15.0L H 2 
 5.00 L N 2
3mol H 2
temperature and pressure?
The combined gas law
Combined Gas Law: The following equation is called the combined
gas law. It holds when the amount of gas (moles) is held constant.
P1V1
T1
=
P2V2
T2
The combined gas law
Example
A sample has a volume of 11.0 L at a
temperature of 13oC and a
PV
PV
pressure of 0.747 atm. The
1 1
2 2

sample is heated to 56oC at a final T1
T2
pressure of 1.18 atm. Calculate
(0.747 atm)(11.0 L)
the final volume.

(13  273.15 K )
V2  8.01L
(1.18atm)V2
(56  273.15 K )
Standard (STD) molar volume
Lets define the volume occupied by 1 mol of a gas under specified
conditions. For 1 mol of an ideal gas at 273.15 K and 1.0 atm, the
volume of the gas is 22.414 L, regardless of gas.
0oC (273.15 K) and 1.0 atm = standard temperature and pressure
(STP)
The Ideal Gas Law
Ideal gas Law: The equation for the ideal gas law is PV=nRT where
R=0.08206 L atm/mol K (universal gas constant).
Derived from STP and standard molar volume
The Ideal Gas Law
Example
A 1.5 mol of a sample of gas has a
volume of 21.0 L at 33oC. What
is the pressure of the gas.
PV  nRT
P(21.0 L)  (1.5mol )(.08206 L  atm
P  1.8atm
mol  K
)(306 K )
Density of Gases





 = m/V
PV = nRT
n = m/MM
 PV = (m/MM)RT
m/V = PMM/RT = 
Dalton’s Law of Partial
Pressures
Partial Pressure: The pressure that the gas exerts if it were above in
the container. For a mixture of gases in a container, the total
pressure exerted is the sum of the partial pressures of the gases
present. This can be expressed as Ptotal = P1 + P2 + P3 + . . . . .
where the subscripts refer to the individual gases. The pressures
P1, P2, and P3 are the partial pressures.
Important Points
1. The volume of the individual gas particles must not be very
important.
2. The forces among the particles must not be very important.
.
Dalton’s Law of Partial
Pressures
For a mixture of ideal gases, it is the total number of moles of particles
that is important, not the identity of the individual gas particles. We
can calculate the partial pressure of each gas from the ideal gas law.
Dalton’s Law of Partial
Pressures
Example
A 2.0 L flask contains a mixture of
N2 and O2 gas 25oC. The total
pressure of the mixture is 0.91
atm. The mixture is known to
contain 0.050 mol of N2.
Calculate the partial pressure of
O2 and the number of moles of O2
present.
PV  nRT
PN2 (2.0 L)  (0.050mol )(0.08206 L  atm
mol  K
)(298 K )
mol  K
)(298 K )
PN2  0.61atm
0.91atm  0.61atm  0.30atm O2
PV  nRT
(0.30atm O2 )(2.0 L)  n(0.08206 L  atm
n  0.25molO2
Dalton’s Law of Partial
Pressures



Pressure of gas in mixture of gases =
product of its mole fraction and total
pressure of mixture
PA = XAPtot
Ex: Xnitrogen gas = 0.78. What’s its partial
pressure at STP?
The Kinetic Molecular
Theory of Gases
Kinetic Molecular Theory: The behavior of individual
particles (atoms or molecules) in a gas.
1.
2.
3.
4.
Postulates of the Kinetic Molecular Theory of Gases
Gases consist of tiny particles (atoms or molecules).
These particles are continually in rapid and random
motion.
The particles are assumed not to attract or to repel
each other.
All gases, regardless of MM, have = average KE @ =
temp.
The Implications of the
Kinetic Molecular Theory
Gas speed
u2 = (3RT/MM)
Maxwell’s equation
= gas molecule speeds  temperature
Same av. KE @ same temp
BUT, different av. Speeds!
Smaller MM molecules go faster
The Implications of the
Kinetic Molecular Theory





Diffusion = random mixing of gases
Effusion = gas movt through a small
opening in a container into another
container with lower pressure
Graham’s Law:
Effusion rate1/effusion rate2 =
(MM2/MM1)
Gas effusion used initially for 235UF6/238UF6
separation
Gas Stoichiometry
Example
Calculate the volume of H2 produced at 1.50 atm and 19oC by the
reaction of 26.5 g of Zn with excess HCl.
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
Solution
mol Zn mol H 2
26.5gZn 

 0.405mol H 2
65.39g mol Zn
PV=nRT
(1.50atm)V=(0.405mol)(0.08206 L  atm
)(292 K )
mol  K
V  6.47 L