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Chapter 11
Gas Pressure
 Pressure is the force per unit area.
 Example: pounds per square inch
 Gas pressure is related to volume, temperature and the
number of gas particles.
Pressure vs Number of Particles
 Increasing the number of gas particles in a container
increases the pressure. More particles means more
collisions with the walls and, therefore, greater
pressure.
 Example: pumping air into a car or bike tire
Pressure vs Temperature
 Increasing the temperature of a gas increases the
pressure. Increasing temperature gives particles more
KE.
 They move faster and collide with walls more often and
with greater force, therefore causing greater pressure.
Standard Temperature and Pressure (STP)
 Standard Temperature
0ºC
273 K
 Standard Pressure
760 mmHg
101.3 kPa
1 atm
14.7 psi
29.92 inHg
760 torr
Temperature Conversions
 All temperatures in gas problems must be converted to
Kelvin.
K = Cº + 273
Cº = K – 273
(Remember: There are no negative Kelvin temperatures)
Example #1: Convert -15ºC to Kelvin
K = -15ºC + 273 = 258 K
Example #2: Convert 355 K to ºC
Cº = 355 – 273 = 82ºC
Pressure Conversions
Use standard pressures for pressure conversions.
Manometers
 Manometer
 device used to measure the pressure of a gas.
 Barometer
 Device used to measure air pressure.
 Made by Torricelli in mid-1600s
 Air pushes Hg up a vacuum tube (see diagram).
 Measured in mmHg, inHg (weather forecasts), kPa.
 https://www.youtube.com/watch?v=EkDhlzA-lwI
Manometers- closed
x and y points in the picture
are at same level, thus
pressures acting on these
points are equal.
Pressure at point x is the
pressure of gas and pressure
at point y is the pressure of
mercury at h height.
In this CLOSED system
pressure of gas is equal to;
Pgas=h
Manometers- open
I: Since height is
greater on AIR
side, gas pressure
is greater than atm
pressure
Pgas = Po +h
II: Air and gas
pressure are
exactly equal
Pgas = Po
III: Since height is
less on AIR side,
gas pressure is
less than atm
pressure
Pgas = Po -h
 Sphygmomanometer
 Device used to measure blood pressure.
 How it works
 A cuff is wrapped around the arm and inflated until the cuff
stops the blood flow in your artery. A valve lets some of the air
out of the cuff which allows blood to start flowing. A stethoscope
is used to listen to the blood flowing/rushing back through the
artery. The first thumping sound is called the systolic blood
pressure. When the sound is no longer heard, it is called the
diastolic blood pressure.
 All pressures are measured in mmHg but are usually
reported as numbers.
 Example – “120 over 80”
Ideal Gas Behavior
 An ideal gas has two properties:
1.
2.
It has mass but no volume (called a point mass).
It has no mutual attraction for other particles.
Real Gas Behavior
 All gases are “real” gases. However, most behave like
ideal gases (no significant volume and little attraction
for other molecules).
 The gases behave near ideal conditions until they are
at:
1. extremely low temperatures
2. extremely high pressures
 For 1st year chemistry, we will assume that all gases
behave like ideal gases.
Boyle’s Law
 The pressure and volume of a gas at constant
temperature are inversely proportional.
 Assuming temperature is held constant:
 If pressure increases, volume decreases.
 If pressure decreases, volume increases.

i.e., pressure and volume move in opposite directions.
Boyle’s Law
 Boyle’s law equation:
PV = k
 If temperature, k, is held constant…we can manipulate
the equation to:
P1V1 = P2V2
P = pressure (kPa, mmHg, psi, atm, …)
V = volume (L, mL, dm3, cm3)
Example #1:
A gas is compressed at constant temperature from 25.0 L
to 4.0 L. If the initial pressure was 0.50 atm, what is the
new pressure?
Example #1:
A gas is compressed at constant temperature from 25.0 L
to 4.0 L. If the initial pressure was 0.50 atm, what is the
new pressure?
P1V1 = P2V2
(0.50 atm)(25.0 L) = (P2)(4.0 L)
P2 = 3.1 atm
Example #2:
A sample of argon gas at STP has a volume of 450. mL.
If the temperature remains constant and the pressure
changes to 685 mmHg, what is the new volume?
*Remember, STP = standard temperature and pressure
Example #2:
A sample of argon gas at STP has a volume of 450. mL. If
the temperature remains constant and the pressure changes
to 685 mmHg, what is the new volume?
*Remember, STP = standard temperature and pressure
P1V1 = P2V2
(760 mmHg)(450. mL) = (685 mmHg)(V2)
V2 = 499 mL
Example #3:
A 24.5 L sample of hydrogen gas exerts 125 kPa of
pressure at 18ºC. What is the pressure of the hydrogen
if expanded to 65.0 L at 18ºC? What is that pressure
equal to in psi?
Example #3:
A 24.5 L sample of hydrogen gas exerts 125 kPa of
pressure at 18ºC. What is the pressure of the hydrogen
if expanded to 65.0 L at 18ºC? What is that pressure
equal to in psi?
P1V1 = P2V2
(125 kPa)(24.5 L) = (P2)(65.0 L)
P2 = 47.1 kPa
Dalton’s Law of Partial Pressures
 The total pressure of a gas mixture is the sum of the
partial pressures of the component gases
Ptotal = P1 + P2 + P3 +…
Example #1
 A mixture of oxygen, hydrogen, and
nitrogen gases exerts a total pressure of
278 kPa. If the partial pressures of
oxygen and hydrogen are 112 kPa and
101 kPa respectively, what would be the
partial pressure exerted by the
nitrogen?
Example #1
 A mixture of oxygen, hydrogen, and nitrogen gases
exerts a total pressure of 278 kPa. If the partial
pressures of oxygen and hydrogen are 112 kPa and 101
kPa respectively, what would be the partial pressure
exerted by the nitrogen?
278 kPa = 112 kPa + 101 kPa + P3
P3 = 65 kPa
Example #2
752 cm3 of oxygen gas is collected
over water with a temperature of
32C. The total pressure of the gases
is 102.4 kPa. What is the partial
pressure of the dry gas?
 Vapor pressure of water at 32C = 4.8 kPa (appendix B-8)
Example #2
 752 cm3 of oxygen gas is collected over water with a
temperature of 32C. The total pressure of the gases is
102.4 kPa. What is the partial pressure of the dry gas?
 Vapor pressure of water at 32C = 4.8 kPa (appendix B-
8)
 102.4kPa = 4.8 kPa + Poxygen
 Poxygen = 97.6 kPa
Charles’s Law
 At constant pressure, the volume of a gas is
directly proportional to its Kelvin
temperature.
 Assuming pressure is held constant:
 If temperature increases, volume increases.
 If temperature decreases, volume decreases.

i.e., temperature and volume move in the same
direction.
Charles Law-- Equation
Charles’s Law equation:
V/T = k
V1 = V2
T1
T2
V = volume (L, mL, dm3, cm3)
T = temperature (K)
The equation could also be written without fractions as:
V1T2 = V2T1
Temperatures must be Kelvin temperatures. Convert from Celsius to Kelvin.
Example #1:
A balloon is filled with a 3.0 L of
helium at 310 K and 1 atm. The
balloon is placed in an oven where
the temperature reaches 450 K with
constant pressure. What is the new
volume of the balloon?
Example #1:
A balloon is filled with a 3.0 L of helium at 310 K and 1
atm. The balloon is placed in an oven where the
temperature reaches 450 K with constant pressure.
What is the new volume of the balloon?
V 1 / T1 = V 2 / T 2
3.0 L / 310 K = V2 / 450 K
V2 = 4.4 L
Example #2:
A flask contains 90.0 mL of
hydrogen gas at 695 torr and
22ºC. What is the new
temperature if the volume
changes to 70.0 mL at 695
torr?
Example #2:
A flask contains 90.0 mL of hydrogen gas at 695 torr
and 22ºC. What is the new temperature if the volume
changes to 70.0 mL at 695 torr?
V 1 / T1 = V 2 / T 2
22ºC = 295 K
90.0 mL / 295 K = 70.0 mL / T2
T2 = 229 K = -44ºC
Gay – Lussac’s Law
 P/T = k
 As the temperature of an enclosed gas increases, the
pressure increases if volume is constant
 Directly proportional, temp in K
 P1 = P2
T1 T2
Example #1:
A cylinder of gas has a pressure
of 4.40 atm at 25C. At what
temperature, in C, will it reach a
pressure of 6.50 atm?
Example #1:
A cylinder of gas has a pressure of 4.40 atm at 25C. At
what temperature, in C, will it reach a pressure of 6.50
atm?
Combined Gas Law
 PV = k
T
 P1V1 =P2 V2
T1
T2
Avogadro’s Law
 Equal volumes of gases at the same temperature and
pressure contain equal numbers of molecules.
 V = kn
 k = constant, n = amount of gas in moles
 Under STP conditions, one mole of any gas (6.02 x 1023)
takes up a volume of 22.4L
Example #1
 What volume does 0.0685 mol of gas occupy at STP?
Example #1
 What volume does 0.0685 mol of gas occupy at STP?
0.0686mol x 22.4L = 1.53L
1mol
Conversion
Factor
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.
 ie heavier gases travel slower
 **remember difference between effusion & diffusion?
Example #1
 Compare the rates of effusion of hydrogen and oxygen
at the same temperature and pressure.
Example #1
 Compare the rates of effusion of hydrogen and oxygen
at the same temperature and pressure.
Rate of H2 = √MO2 = √32.00 = 3.98
Rate of O2 √MH2 √2.02
Hydrogen effuses 3.98 x’s faster than oxygen
Ideal Gas Law
 The mathematical relationship between pressure,
volume, temperature and the number of moles of a gas
 PV = nRT
 R = the ideal gas constant
= 0.0821L•atm/(mol•K)
= 8.314L•kPa/(mol•K)
Example #1
 What is the pressure in atmospheres exerted by a
0.500 mol sample of nitrogen gas in a 10.0L container
at 298K?
Example #1
 What is the pressure in atmospheres exerted by a
0.500 mol sample of nitrogen gas in a 10.0L container
at 298K?
 P(10.0L) = (0.500mol)(0.0821L•atm)(298K)
(mol•K)
P = 1.22atm
Example #2
 Determine the mass of 5.60L of O2 at 1.75atm and
250.0K.
Example #2
 Determine the mass of 5.60L of O2 at 1.75atm and
250.0K.
(1.75atm)(5.60L) = n(0.0821L•atm)(250.0K)
(mol•K)
n = .477mol
.477mol O2 32.00g O2
1mol O2
= 15.3g O2
Gas Stoichiometry Under Nonstandard Conditions
 We know all there is to know about the gas laws, now
let’s add some stoichiometry…
Example #1
4.00mol C8H18 25molO2
2mol C8H18
?LO2
1molO2
How do we know what to put in the last conversion factor?
PV = nRT
V = RT **this will give us L/mol for oxygen
n P
V = (0.0821Latm)(308K) = 26.5L/molO2
n
(molK) (.953atm)
1320LO2
Example #2
PV = nRT
(1.3atm)(20.0L) = n(0.0821Latm)(263K) = 1.2molO2
(molK)
1.2molO2 18molH2O 18.02gH2O
25molO2 1molH2O
= 16gH2O