Ch. 12
Behavior of Gases
Gases
• Gases expand to fill its container, unlike solids
or liquids
• Easily compressible: measure of how much
the volume of matter decreases under
pressure
Variables that describe a gas
• Pressure (P)
– Measured in kilopascals, kPa
– Pressure and number of molecules are directly
related
 increase molecules = increase pressure
– Gases naturally move from areas of high pressure
to low pressure, due to the available space to
move into
Gas Pressure- collision of gas molecules
with the walls of the container
Variables that describe a gas
• Volume (V)
– Measured in Liters, L
– Volume and pressure are inversely related
• As volume decreases, the pressure increases
• Smaller container = less room for movement, therefore
molecules hit sides of container more often
II. Factors Affecting Gas Pressure
B. Volume
Reduce volume - ↑ pressure
Increase volume - ↓ pressure
Ex: piston in a car
Variables that describe a gas
• Temperature (T)
– Measured in Kelvin, K
– The temperature and pressure are directly related
• Increase in temp = increase in pressure
• Volume must be held constant
• Molecules hit the walls harder (due to increase in K.E.)
and more frequently.
Think about a tire in hot weather…
Variables that describe a gas
• Amount
– Measured in moles, mol
– Moles and pressure are directly related
• Increase in # of moles = increase in pressure
Ex: Inflating a balloon is adding more molecules.
• Temperature must remain constant
II. Factors Affecting Gas Pressure
A. Amount of Gas
Add gas - ↑ pressure
Remove gas - ↓ pressure
Ex: pumping up a tire
adding air to a balloon
aerosol cans
Gas Pressure (Cont.)
-- 3 ways to measure pressure:
»atm (atmosphere)
»mm Hg
»kPa (kilopascals)
U-tube Manometer
III. Variables that describe a gas
Variables
Units
Pressure (P) –
kPa, mm Hg, atm
Volume (V) –
L , mL , cm3
Temp (T) –
°C , K
(convert to Kelvin)
K = °C + 273
Mole (n) -
mol
How pressure units are related:
1 atm = 760 mm Hg = 101.3 kPa
How can we make these into
conversion factors?
1 atm
760 mm Hg
101.3 kPa
1 atm
Guided Problem:
1. Convert 385 mm Hg to kPa
385 mm Hg
x 101.3 kPa
760 mm Hg
= 51.3 kPa
2. Convert 33.7 kPa to atm
33.7 kPa
x
1 atm
101.3 kPa
= .33 atm
STP
Standard Temperature and Pressure
Standard pressure – 1 atm, 760 mmHg,
or 101.3 kPa
Standard temp. – 0° C or 273K
Gas Laws
• Describe how gases behave
• Change can be calculated
• Know the math and the theory!!
Boyle’s Law (1662)
• Gas pressure is inversely related to volume
(as volume increases, pressure decreases)
• Temperature is constant
P1V1= P2V2
Example Problems
pg 335 # 10 &11
10. The pressure on 2.50 L of anesthetic gas changes from
105 kPa to 40.5 kPa. What will be the new volume if the
temp remains constant?
P1 = 105 kPa
V1 = 2.5 L
P2 = 40.5 kPa
V2 = ?
P1 × V1 = P2 × V2
(105) (2.5) = (40.5)(V2)
262.5 = 40.5 (V2)
6.48 L = V2
Example Problems
pg 335 # 10 &11
11. A gas with a volume of 4.00L at a pressure of 205 kPa is
allowed to expand to a volume of 12.0L. What is the
pressure in the container if the temp remains constant?
P1 = 205 kPa
P2 = ?
V1 = 4.0 L
V2 = 12.0 L
P1 × V1 = P2 × V2
(205) (4.0) = (P2)(12)
820 = (P2) 12
68.3 L = P2
Charles’s Law (1787)
• Volume is directly proportional to temp.
(increase volume, increase temp)
• Pressure is constant
𝑉1 𝑉2
=
𝑇1 𝑇2
Example Problems pg. 337 # 12 & 13
12. If a sample of gas occupies 6.80 L at 325°C, what
will be its volume at 25°C if the pressure does not
change?
V1= 6.8L
V2 = ?
T1 = 325°C = 598 K T2 = 25°C = 298 K
6.8 = V2
598 298
598 × V2 = 2026.4
598
598
V2 = 3.39 L
Example Problems pg. 337 # 12 & 13
13. Exactly 5.00 L of air at -50.0°C is warmed to
100.0°C. What is the new volume if the pressure
remains constant?
V1= 5.0L
T1 = -50°C = 223 K
5 = V2
223
373
(223) V2 = 1865
223
223
V2 = 8.36 L
V2 = ?
T2 = 100°C = 373 K
Gay-Lussac’s Law (1802)
• Pressure and temperature are directly related
(Increase pressure= Increase temperature)
• Volume is constant!
𝑃1
𝑇1
=
𝑃2
𝑇2
Example Problems
1. The gas left in a used aerosol can is at a pressure
of 103 kPa at 25°C. If this can is thrown onto a fire,
what is the pressure of the gas when its
temperature reaches 928°C?
P1= 103 kPa
T1 = 25°C = 298 K
103 = P2
298 1201
298 × P2 = 123,703
P2 = 415 kPa
P2 = ?
T2 = 928°C = 1201 K
Example Problem pg. 338 # 14
14. A gas has a pressure of 6.58 kPa at 539 K. What
will be the pressure at 211 K if the volume does
not change?
P1= 6.58 kPa
T1 = 539 K
6.58 = P2
539 211
539 × P2 = 1388
539
539
P2 = 2.58 kPa
P2 = ?
T2 = 211 K
Combined Gas Law
• Combines 3 gas laws: Boyle’s, Charles’, and GayLussac’s
• Used when it is difficult to hold any one variable (P,
V, or T) constant
𝑃1𝑉1
𝑇1
•
=
𝑃2𝑉2
𝑇2
Can take away any variable that is constant
– Take temp away = Boyle’s
– Take Pressure away = Charle’s
– Take Volume away = Gay-Lussac’s
How to remember each Law!
Cartesian Divers
Boyles
P
V
Gay-Lussac
Charles
Fizz Keepers
T
Balloon and flask Demo
Ex: 3.0 L of Hydrogen gas has a pressure of
1.5 atm at 20oC. What would the volume be
if the pressure increased to 2.5 atm at 30oC?
Ideal Gas Law
• Used for gases that behave “ideally”
• Allows you to solve for # of moles of a contained gas
when P, V, and T are known.
• Use constant
(𝐿∙𝑘𝑃𝑎)
R=8.31
(𝑚𝑜𝑙 ∙𝐾)
𝑃𝑉 = 𝑛𝑅𝑇
P(pressure)- must be in kPa
V (volume)- must be in L
n (# of moles)- muse be in moles of gas
R- gas constant
T (Temperature)- Must be in Kelvin (oC + 273= K)
Ideal Gas Law
• A gas behaves “ideally” if it conforms to the gas laws
– Gases do not usually do this
– Real gases only behave this way at:
1. High temps (molecules move fast)
2. Low pressure (molecules are far apart)
• This is because gases will stay a gas under these conditions
– Molecules are not next to each other very long so attractive forces can’t
play a role b/c molecules are moving too fast
– Ideal Gases do no exist because:
1.
2.
Molecules do take up space
There are attractive forces between molecules otherwise no
liquid would form.
(Molecules slow down to become liquids)
E. Ideal Gas Law
•You can calculate the # of n of gas at
standard values for P, V, and T
PV
=R
Tn
(1 atm)(22.4L)
(273K)(1 mol)
UNIVERSAL GAS CONSTANT
R= 0.0821 atm∙L/mol∙K
You don’t need to memorize this value!
=R
E. Ideal Gas Law
PV=nRT
P= pressure in atm
V = volume in liters
n = number of moles
R= 0.0821 atm∙L/mol∙K
T = temperature in Kelvin
E. Example Problems
1. At what temperature will 5.00g of Cl2 exert a
pressure of 900 mm Hg at a volume of 750 mL?
2. Find the number of grams of CO2 that exert a
pressure of 785 mm Hg at a volume of 32.5 L and
a temperature of 32 degrees Celsius.
3. What volume will 454 g of H2 occupy at 1.05 atm
and 25°C.
Ex: What volume will 2.0 mol of N2
occupy at 720 torr and 20oC?
Dalton’s Law of Partial Pressures
• Used for mixture of gases in a container
• If you know the P exerted by each gas in a
mixture, you can calculate the total gas
pressure
• It is particularly useful in calculating pressure
of gases collected over water.
Ptotal = P1 + P2 + P3…
*P1 represents the “partial pressure” or the contribution by the gas
F. Dalton’s Partial Pressure Law
• The total pressure of a mixture of
gases equals the sum of the partial
pressures of the individual gases.
Ptotal = P1 + P2 + P3 + ...
F. Dalton’s Law
• Example problem:
1. Air contains oxygen, nitrogen, carbon dioxide,
and trace amounts of other gases. What is the
partial pressure of oxygen (PO2) if the total
pressure is 101.3 kPa. And the partial pressures
of nitrogen, carbon dioxide, and other gases are
79.10 kPa, 0.040 kPa, and 0.94 kPa.
PO2 = Ptotal – (PN2 + PCO2 + Pothers)
= 101.3 kPa – (79.10 kPa + 0.040 kPa + 0.94 kPa)
= 21.22 kPa
F. Dalton’s Law
2. A container holds three gases : oxygen , carbon
dioxide, and helium. The partial pressures of the
three gases are 2.00 atm, 3.00 atm, and 4.00 atm
respectively. What is the total pressure of the
container?
3. A gas mixture contains oxygen, nitrogen and carbon
dioxide. The total pressure is 50.0 kPa. If the carbon
dioxide has a partial pressure of 21 kPa and the
nitrogen has a partial pressure of 15 kPa, what is
the partial pressure of the oxygen?
4. A container contains two gases – helium and argon,
at a total pressure of 4.00 atm. Calculate the partial
pressure of helium if the partial pressure of the
argon is 1.5 atm.
Graham’s Law of Effusion
• Rate of effusion and diffusion are inversely proportional to
the square root of the mm of molecules
– Effusion: Gas escaping through tiny holes in a container
– Diffusion: movement from area of high concentration to low
concentration (ex: perfume spreading across a room)
(Both depend of the mm of the molecule, which determines speed)
𝑅𝑎𝑡𝑒 𝐴
𝑅𝑎𝑡𝑒 𝐵
=
𝑚𝑎𝑠𝑠 𝐵
𝑚𝑎𝑠𝑠 𝐴
• Type of Molecule is important
– Gases with lower mm effuse/diffuse faster
– Ex: Helium diffuses/effuses faster than Nitrogen from a balloon b/c
Helium moves faster due to lower mm.
Big = Slow
small = Fast
Ex:
𝑅𝑎𝑡𝑒 𝐻2
𝑅𝑎𝑡𝑒 𝑁2
A. Graham’s Law
• Diffusion
– The tendency of molecules to move toward
areas of lower concentration.
• Ex: air leaving tire when valve is opened
• Effusion
– Passing of gas molecules
through a tiny opening in
a container
A. Graham’s Law
Tiny opening
Diffusion
Effusion
Which one is Diffusion and which one is Effusion?