1. BEHAVIOR OF GASES Chapter 12
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2. Properties of Gases
Gas properties can be modeled using math. Model depends on—
V = volume of the gas (L)
T = temperature (Kelvin)
P = pressure (atmospheres)

3. Volume of Gas
In a smaller container, molecules have less room to move.
Hit the sides of the container more often.
As volume decreases, pressure increases. (think of a syringe)

4. Gas Laws
Here is the qualitative relationship between the pressure, temperature, and volume of a constant # of gas particles in a container:
(1) ___________ Law: At a constant temperature, as the volume of a container __________ the pressure of the container will ___________.
V___, P ___
*Example: Compressing the gas in a flexible container will _________ its volume.
Boyle’s
decreases
increase ↓ ↑
decrease
Pressure
Volume

5. Boyle’s Law If Temperature is constant, then P1V1 = P2V2
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If Temperature is constant, then
P1V1 = P2V2
This means that as Pressure goes up as Volume goes down. Pressure and Volume are indirectly related
Robert Boyle ( ). Son of Early of Cork, Ireland.

6. Boyle’s Law A bicycle pump is a good example of Boyle’s law.
As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

7. Boyle’s Law P1V1 = P2V2 P1 = initial pressure V1 = initial volume
P2 = final pressure
V2 = final volume
STP = standard temperature (273 Kelvin) and pressure (1 atmosphere)
Pressure can be measured in atmospheres (atm), torrs (torr), millimeters mercury (mmHg) or kilopascals (kpa)
Conversions:
1 atm = 760 torr = 760 mmHg = kpa
Boyle’s Law
P1V1 = P2V2

8. Boyle’s Law Practice Problems
If we have 4 L of methane gas at a pressure of
1.50 x 103 mmHg. What will the pressure of the gas be if the volume is decreased to 2.5 L?
P1 =
V1 =
P2 =
V2 =
1.50 x 103 mmHg =
Plug the #’s into the equation and solve for P2.
(1.05 x 103)
(4)
(P2)
(2.5)
4 liters ?
2.5 liters
(1.50 x 103) (4) = P2
2.5
2400 mmHg = P2

9. Boyle’s Law Practice Problems
2. You have a car with an internal volume of 12,000 L. If you drive your car into the river and it implodes, what will be the volume of the gas when the pressure goes from 1.0 atm to 1.4 atm?
P1 =
V1 =
P2 =
V2 =
1 atm =
Plug the #’s into the equation and solve for V2.
(1)
(12,000)
(1.4)
(V2)
12,000 L
1.4 atm ?
(1) (12,000) = V2
1.4
L = V2

10. Boyle’s Law Practice Problems
If P1 and P2 are given, they must both be in the same unit. If they are not, you must make a conversion. Usually the conversion is to atm’s since it is the measure of standard pressure. Make the following conversions using:
1 atm = 760 torr = 760 mmHg = kpa
Practice Problems:
(1) Convert 6.5 atm to torr
(2) Convert 98 torr to mmHg
(3) Convert 177 mmHg to atm
760 torr
6.5 atm x =
4940 torr
1 atm
760 mmHg
98 torr
98 mmHg x =
760 torr
1 atm
.23 atm
177 mmHg x =
760 mmHg

11. Gas Laws (continued) ↑ ↑
(2) ____________ Law: At a constant pressure, as the temperature of a container __________ the volume of the container will ___________.
T___, V ___
*Examples: Heating a balloon will cause it to ___________. Taking a balloon outside on a cold winter day will cause it to _____________.
If you could keep a gas from condensing, you could cool it off to absolute zero and the volume of the gas would be _________!
Charles’s
increases
increase ↑ ↑
inflate
shrink
zero
Volume
Temperature (K)

12. Charles’s Law If Pressure is constant, then
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If Pressure is constant, then
V1 T2 = V2 T1
This means as Volume goes up so does Temperature. This means Volume and Temperature are directly related.
Jacques Charles ( ). Isolated boron and studied gases. Balloonist.

13. Charles’s original balloon
Modern long-distance balloon

14. Charles’s Law

15. Kelvin = degrees Celsius + 273
V1 = initial volume
T2 = final temperature
V2 = final volume
T1 = initial temperature
STP = standard temperature (273 Kelvin) and pressure (1 atmosphere)
Conversions:
Kelvin = degrees Celsius + 273
Charles’s Law
V1T2 = V2T1

16. Charles’s Law Practice Problems
If we have 2 L of methane gas at a temperature of
40 degrees Celsius, what will the volume of the gas be if we heat the gas to 80 degrees Celsius?
V1 =
T2 =
V2 =
T1 =
2 liters
Plug the #’s into the equation and solve for V2.
(2)
(353) =
(V2)
(313)
80 ̊ C = 353 K ?
40 ̊ C = 313 K
(2) (353) = V2
313
2.26 L = V2

17. Charles’s Law Practice Problems
If you have 45 L of helium in a balloon at 25 ̊ C and you increase the temperature of the balloon to 55 ̊ C, what will the new volume of the balloon be?
V1 =
T2 =
V2 =
T1 =
45 liters
Plug the #’s into the equation and solve for V2.
55 ̊ C = 328 K
(45)
(328) =
(V2)
(298) ?
25 ̊ C = 298 K
(45) (328) = V2
298
49.53 L = V2

18. Gas Laws (continued) ↑ ↑
(3) ____________ Law: At a constant volume, as the temperature of a container __________ the pressure of the container will ___________.
P1T2 = P2T T___, P ___
*Example: Heating a rigid container causes the gas inside to move __________ which causes _________ pressure. Be careful! Too much heat will make it explode!
Guy-Lussac’s
increases
increase ↑ ↑
faster
more
Pressure
Temperature (K)

19. The Combined Gas Law

20. Combining the gas laws P1V1T2 P2V2 T1 = V1 T1 = V2 T2 P1 T1 = P2 T2
So far we have seen three gas laws:
Bachelor #3
Bachelor #1
Bachelor #2
I’m the hottest
pick me!
Check me out!
Pick me baby!
Robert Boyle
Jacques Charles
Joseph Louis Gay-Lussac V1 T1 = V2 T2 P1 T1 = P2 T2
P1V1 =
P2V2
These are all subsets of a more encompassing law: the combined gas law
P1V1T P2V2 T1 =

21. (initial conditions) = (final conditions)
The Combined Gas Law
This equation combines all of the previous three laws into one convenient form.
Boyles Law:
Pressure and volume
Guy-Lussac’s Law: Pressure and temperature
Combined Gas Law
Charles’s Law: Volume and temperature
Using the Combined Gas Law requires you to have the temperature in _____________ units. The pressure and volume units can be anything as long as the initial and final units are ______ __________.
P1 x V1 x T2
P2 x V2 x T1 =
Kelvin
(initial conditions) = (final conditions)
the same

22. Plug the #’s into the equation and solve for V2.
Standard Temperature and Pressure: (STP)
Often the volume of a gas is needed at “standard conditions.” For scientists, this means “STP”. Standard temperature is ______K, and standard pressure is ____________ atmosphere (atm)
1 atmosphere (atm) = 760 Torr = 760 mmHg = kPa = 14.7 lbs/in2 (psi)
Practice Problems:
1) mL of helium is in a balloon at 25˚C. What will the new volume of the balloon be if the temp. is raised to 100˚C?
273 1
(Since pressure is not mentioned, it can be assumed that it was constant. You can throw it out of our equation.)
P1 = ______
V1= ______
TK1= ______
P2 = ______
V2= ______
TK2= ______
Plug the #’s into the equation and solve for V2.
(373)
(298)
(80.0)
(V2)
80.0 mL
??? =
298 K
373 K
V2 = 100 mL

23. Practice Problems (continued):
2) A rigid steel container is filled with neon under a pressure of 760 mm Hg and a temperature of 325 K. If the temperature is reduced to standard temperature, what will the new pressure be?
P1 = ______
V1= ______
TK1= ______
760 mm
P2 = ______
V2= ______
TK2= ______
???
Plug the #’s into the equation and solve for P2.
(273)
(325)
(760)
(P2) =
325 K
273 K
P2 = 638 mm Hg
Volume is not mentioned, so assume it is constant.
3) A balloon at a pressure of 4.5 atmospheres, 300 K, and a volume of 35.0 liters is changed to STP conditions. What will the new volume of the balloon become?
P1 = ______
V1= ______
TK1= ______
4.5 atm
P2 = ______
V2= ______
TK2= ______
1 atm
Plug the #’s into the equation and solve for V2.
(273)
(4.5)(35.0) =
(300)
(1)(V2)
35.0 L
???
300 K
273 K
V2 = 143 L

24. Dalton's Law of Partial Pressures
Under Pressure
Dalton's Law of Partial Pressures

25. Dalton’s Law of Partial Pressure
What happens to the pressure of a gas as we mix different gases in the container?
The ______ of each individual gas pressure equals the _______ gas pressure of the container.
P(total)= P1+P2+P3…
total
sum

27. Dalton’s Law of Partial Pressures
1) A container has oxygen, nitrogen, and helium in it. The total pressure of the container is 2.4 atmospheres. If all the partial pressures are equal to one another, what ate the partial pressures of each gas?
Total number of gases = P =
2.4 atm
total 3
Pgas = 2.4 atm ÷ 3 = 0.8 atm

28. Dalton’s Law of Partial Pressures
2) Two flasks are connected with a rubber hose. The first flask contains N2 at a pressure of 0.75 atm., and the second flask contains O2 at a pressure of 1.25 atm. When the two flasks are opened and mix, what will the pressure be in the resulting mixture?
P1 =
0.75 atm
P2 =
1.25 atm
Ptotal = atm atm = 2.0 atm

29. Dalton’s Law of Partial Pressures
3) A mixture of neon and argon gases exerts a total pressure of 2.39 atm. The partial pressure of the neon alone is 1.84 atm. What is the partial pressure of the argon?
Ptotal =
2.39 atm
P1 =
1.84 atm
P2 = atm – 1.84 atm = 0.55 atm

30. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD
SAMPLE PROBLEM
In a laboratory, oxygen gas was collected by water displacement at an atmospheric pressure of 726 torr and a temperature of 22°C. Calculate the partial pressure of dry oxygen. Water vapor pressure is 2.6 kPa (from table in reading packet). Convert 2.6 kPa to torr to solve.

31. DALTON’S LAW OF PARTIAL PRESSURE: WATER DISPLACEMENT METHOD
SOLUTION to Sample Problem
PTOTAL = 726 torr
PWATER= 19.5 torr (22 °C)
POXYGEN = ?
PTOTAL = POXYGEN + PWATER
POXYGEN = PTOTAL – PWATER
= 726 torr – 19.5 torr
POXYGEN = torr

32. Dalton’s Law of Partial Pressures
3) A mixture of helium, neon and argon gases exerts a total pressure of 2.39 atm. What is the partial pressure of the argon if it makes up 40% of the mixutre?
Ppartial pressure = Ptotal x (%of mixture)
Ptotal =
2.39 atm
% =
0.40
Ppartial pressure = atm x 0.40 = atm

33. Avogadro’s Hypothesis
Avogadro’s hypothesis states that ________ volumes of gases (under the same temp. and pressure conditions) contain _______ number of particles.
If containers have the same ____, ____, and ___, then they will have the same ____ of particles regardless of the _________ of the gas particle.
You might think that a small gas molecule would take up ______ space than a large gas molecule, but it ___________ at the same _________________ and ______________!!
equal
equal
T P V #
size
less
doesn’t
temperature
pressure

34. AVOGADRO’S LAW The mathematical form of Avogadro’s law is: V1 = V2
n1 n2
SAMPLE PROBLEM 1
A sample of gas with a volume of 9.20 L is known to contain mol. If the amount of gas is increased to 2.85 mol, what new volume will result if the pressure and temperature remain constant?

35. AVOGADRO’S LAW SOLUTION Given: V1 = 9.20 L V2 = ?
n1 = mol n2 = 2.85 mol
Solution:
V2 = n2V1 = (2.85 mol) (9.20 L) = 21.4 L
n (1.225 mol)

36. PV=nRT The Ideal Gas Law
An equation used to calculate the __________ of gas in a container (in units of _________.)
PV=nRT
The units for T= __________, V = _________, n = # of moles
R = Ideal Gas Constant
The value of R changes depending on the unit of ____________ used in the equation:
R = 62.4 (mm Hg)(L)/(mole)(K) R = 8.31 (kPa)(L)/(mole)(K)
R = (atm.)(L)/(mole)(K) R = 2.45 (in. Hg)(L)/(mole)(K)
amount
moles
Kelvin
Liters
pressure

37. “Ideal” Gases
Real gases, (like nitrogen), will eventually ___________ into a liquid when the temperature gets too ____ or the pressure gets too _____.
If you want a gas to act more ideally, keep the temperature _____ and the pressure ______. That way, they will act more like an ideal gas and never have a chance of _______________.
The best real gas that acts like an ideal gas is __________. It doesn’t condense until the temperature gets to ______K.
condense
low
high
high
low
condensing
helium 4
Real Gas

38. Ideal Gases vs. Real Gases

40. Ideal Gas Law SAMPLE PROBLEM 1
What pressure will be exerted by mol of gas in a 5.00 L container at 17.0°C?
Given:
n = mol
V= 5.00 L
T= 17.0 °C = 290 K
Solution:
L  atm
P = nRT = (0.400 mol) ( mol  K) (290 K) = 1.9atm
V (5.00 L)

41. Diffusion vs. Effusion
The spreading out of a gas from _______ to _____ concentrations is called diffusion.
*Example: ___________ in a room spreads out
A gas escaping through a ______ _______ in a container is called effusion. As the size of a molecule _____________, the effusion speed and diffusion rate ______________...(inverse relationship.)
high
low
Perfume
tiny hole
increases
decrease
Effusion