1. Gas Laws
Edward Wen, PhD

2. Properties of Gases expand to completely fill their container
take the shape of their container
low density
much less than solid or liquid state
compressible when pressure is changed.
mixtures of gases are always homogeneous (common air)
fluid

3. Kinetic Molecular Theory

4. Properties of Gas: Indefinite Shape & Volume
Gas molecules have enough kinetic energy (~500 m/s) and little attractions among each other
 keep moving around and spreading out
fill the container of whatever shape
Hit the surface of container

5. Pressure: Gases Pushing Surface
as Gas molecules strike a surface, they push on that surface
push = force
Pressure of gas = Hitting force per Hitted area
Pressure depends on:
number of gas particles in a given volume
volume of the container
average speed of the gas particles

6. Measuring Air Pressure
Barometer: Height of column of mercury supported by air pressure
Force of the air on the surface of the mercury = Gravity on the column of mercury
Standard pressure:
exactly 1 atmosphere = 760 mmHg
 14.7 psi (pound per square inch)
(“mmHg” read as “millimeter mercury”)
gravity

7. The Effect of Gas Pressure
Wind: whenever there is a Pressure difference, a gas will flow from area of High pressure  area of Low pressure
the bigger the difference in pressure, the stronger the flow of the gas
Vacuuming & Drinking through straw: if there is something in the gas’ path, the gas will try to push it along as the gas flows

8. Atmospheric Pressure & Altitude
Altitude↑ Atmospheric pressure↓
At the surface, P = 14.7 psi,
At 10,000 ft altitude (Big Bear Lakes, Ca) P = 10.0 psi
Rapid changes in atmospheric pressure may cause your ears to “pop”  an imbalance in pressure on either side of your ear drum (driving or flying)
Demo: Can you make a piece of paper uphold a bottle of water?

9. Common Units of Pressure
Average Air Pressure at Sea Level
pascal (Pa)
101,325
kilopascal (kPa)
atmosphere (atm)
1 (exactly)
millimeters of mercury (mmHg)
760 (exactly)
torr (torr)
pounds per square inch (psi, lbs./in2)
14.7

10. Practice: Convert Pressure between units
735.0 mmHg = ? atm
35. psi = ? torr
Ans: atm
Ans: 1.8 × 103 torr

11. Boyle’s Law For the gas contained
Pressure of a gas is inversely proportional to its volume:
P /V
constant T and amount of gas
P x V = constant
P1 x V1 = P2 x V2

12. When you double the pressure on a gas,
the volume reduces to one half, (as long as the
temperature and amount of gas do not change)

13. Boyle’s Law Demonstrations
Puffing Mushmellow
Ballon (long video)
Why unfinished drinking water bottle collapsed after coming home from Big Bear Lakes?

14. Example of Boyle’s law: A cylinder equipped with a moveable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm?
Information
Given: P1 = 4.0 atm V1 = 6.0 L
P2 = 1.0 atm
Find: V2 = ? L
Answer: 24 L

15. We’re losing altitude.
Quick Professor, give your
lecture on Charles’ Law!

16. Charles’ Law For the gas contained and at constant Pressure:
Volume is directly proportional to temperature (Kelvin): V  T
constant P and amount of gas
graph of V vs T is straight line
as T increases, V also increases
Kelvin K = °C + 273
V = constant x T
if T measured in Kelvin

17. Example of Charles’ Law: A gas has a volume of 2
Example of Charles’ Law: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0°C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in celsius?
Information
Given: V1 = 2.80 L
V2 = 2.57 L t2 = 0°C
Find: temp1 in K and °C
Answer: T = 297 K = 24°C

19. Avogadro’s Law
Volume directly proportional to the number of gas molecules
V = constant x n
constant P and T
more gas molecules = larger volume
count number of gas molecules by moles
Equal Volumes of gases contain Equal numbers of molecules
the gas doesn’t matter

20. Avogadro’s Law: At constant P and T, more gas  more volume

21. Combined Gas Law Boyle’s Law : Pressure and Volume
at constant temperature
Charles’ Law : Volume and absolute Temperature
at constant pressure
 Volume of a sample of gas when both the Pressure and Temperature change

22. Apply Gas laws to solve real world problems
Which Gas Law to use?
Boyle’s law: study P and V at constant T
Charles’ law: study V and T at constant P
Combined Gas Law for problems where two of P/V/T changes

23. Convert temperature to Kelvin:
Example of Combined Gas Law: A sample of gas has a volume of 158 mL at a pressure of 755 mmHg and a temperature of 34°C. The gas is compressed to a volume of 108 mL and heated to 85°C, what is the final pressure in mmHg?
Information
Given: V1 = 158 mL, P1 = 755 mmHg,
t1 = 34°C
V2 = 108 mL, t2 = 85°C
Find: P2, mmHg
Convert temperature to Kelvin:
T1 = 307 K, T2 = 358 K,

24. Ideal Gas Law Combined Gas Law + Avgadro’s Law  Ideal Gas Law
R is called the Gas Constant
the value of R depends on the units of P and V
R = atm · L/K · mol
convert P to atm and V to L
Application of Ideal Gas law: when T, P, V of a gas all changes

25. First, convert P or V or T to standard units: atm, Liter, Kelvin
Example of Ideal Gas Law : Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25°C
Information
Given: V = 3.2 L, P = 24.2 psi,
t = 25°C
Find: n, mol
Eq’n: PV = nRT
SM: P,V,T,R → n
First, convert P or V or T to standard units: atm, Liter, Kelvin
T = 298 K
P = 24.2 psi x ________ = atm
n = 0.22 mol

26. Applying Ideal Gas Law: Molar Mass of a Gas
Determination of the molar mass of an unknown substance:
Heat a weighed sample into gaseous state
Measure temperature T, pressure P and volume V of the gas
Use the Ideal Gas Law: #mole n = PV/RT

27. Example: Find Molar Mass A sample of a gas has a mass of 0. 311 g
Example: Find Molar Mass A sample of a gas has a mass of g. Its volume is L at a temperature of 55°C and a pressure of 886 mmHg. Find its molar mass.
Information
Given: V = L, P =,
t = 328 K, m = g
Find: molar mass, (g/mol)
Eq’n: PV = nRT; MM = mass/moles
SM: P,V,T,R → n & mass → mol. mass
Molar Mass = mass/moles Mass: given
To find mole of gas? PV = nRT
P = 886 mmHg x _____________
Mole of gas n = ____________

28. Air: Mixtures of Gases Air is a mixture (N2 , O2)
Each gas in the mixture behaves independently of the other gases
though all gases in the mixture have the same volume and temperature
all gases completely occupy the container, so all gases in the mixture have the volume of the container
Gas
% in Air, by volume
nitrogen, N2 78
argon, Ar
0.9
oxygen, O2 21
carbon dioxide, CO2
0.03

29. Collecting gas over water
Zn metal reacts
with HCl(aq) to
produce H2(g).
The gas flows
through the tube
and bubbles into
the jar, where it
displaces the
water in the jar.
Pgas = PH2O + PH2
Because water
evaporates, some
water vapor gets
mixed in with
the H2.

30. Reactions Involving Gases
Reaction stoichiometry (Chapter 8) can be combined with the Gas Laws for reactions involving gases
in reactions of gases, the amount of a gas is often given as a Volume
instead of moles
as we’ve seen, must state pressure and temperature
Ideal Gas Law: from Volume of the gas + P & T  Moles; then we can use the coefficients in the equation as a mole ratio

31. Example: Gases in Chemical Reactions

32. gram of KClO3  mol KClO3  mol O2 (gas)  volume of O2
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 308 K
Information
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 308 K
Find: VO2, L
Eq’n: PV=nRT
Gram A  mole A  mole A  gram B
Plan:
gram of KClO3  mol KClO3  mol O2 (gas)  volume of O2
Molar mass KClO3 = g/mol n = 3.60 mol O2 P = atm
V = 90.7 L

33. Standard Conditions (STP)
Common reference points for comparing
Standard Temperature & Pressure
Standard Pressure = 1.00 atm
Standard Temperature = 0°C = 273 K

34. Molar Volume of a Gas at STP
Definition: The volume of 1 (exact) mole gas at STP
Use the Ideal Gas Law: PV = nRT
1 mole of any gas at STP will occupy 22.4 L
==> Molar volume
can be used as a conversion factor
as long as you work at STP
1 mol gas  _______ L

35. Molar Volume So much empty space between molecules in the gas state,
 the volume of the
gas is not effected
by the size of the
molecules, (under
ideal conditions).

36. Density of Gas at STP
Since every exactly one mole of any gas has a volume of 22.4 L, whereas the mass of such gas would be as the molar mass in grams
Density of Gas = Molar mass / Molar Volume
Example:
Density of Oxygen gas at STP
= g/mol / 22.4 L/mol = 1.43 g/L

37. Density of Common Gases
At STP, the density of common gases (in g/L) as: H He CH N
Air O CO Cl
Which one, hydrogen gas or helium gas, is better in blimps in providing lift?
Why carbon dioxide is used in putting out fire? What if its density is less than the air?

38. Volume H2  mol H2  mol H2O  gram H2O
Example of Using Molar Volume: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information
Given: 1.24 L H2
Find: g H2O
CF: 1 mol H2O = g
1 mol H2  22.4 L
STP? 1 mol gas = _____ L
Volume H2  mol H2  mol H2O  gram H2O
0.988 g H2O

39. Real Gases Ideal gas laws assume
No Attractions between gas molecules
No Volume: gas molecules do not take up space
based on the Kinetic-Molecular Theory
Real gases: often do not behave like Ideal gases at High pressure (“Squeezed”) or Low temperature (“Frozen”)

40. Ideal vs. Real

41. Puffing Mushmellow & Ballon Soda bottle Submarine
Pressure surge when bottle being squeezed
Volume of the air inside the dropper decreases
Water filling up the dropper
Increased density of dropper (Massdropper/Volume)
Dropper sinking!

42. Apply the Solution Map:
Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0°C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in celsius?
Information
Given: V1 = 2.80 L
V2 = 2.57 L t2 = 0°C
Find: temp1 in K and °C
Eq’n:
SM: V1, V2 T2 → T1
Apply the Solution Map:

43. Apply the Solution Map:
Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0°C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in celsius?
Information
Given: V1 = 2.80 L T1 = 297 K
V2 = 2.57 L t2 = 0°C, T2 = 273 K
Find: temp1 in K and °C
Eq’n:
SM: V1, V2 T2 → T1
Apply the Solution Map:
convert to celsius

44. Example: A sample of gas has a volume of 158 mL at a pressure of 755 mmHg and a temperature of 34°C. The gas is compressed to a volume of 108 mL and heated to 85°C, what is the final pressure in mmHg?
Information
Given: V1 = 158 mL, P1 = 755 mmHg,
t1 = 34°C
V2 = 108 mL, t2 = 85°C
Find: P2, mmHg
Eq’n:
SM: P1, V1, V2, T1, T2 → P2

45. Apply the Solution Map:
Example of Ideal Gas Law : Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25°C
Information
Given: V = 3.2 L, P = 24.2 psi,
t = 25°C
Find: n, mol
Eq’n: PV = nRT
SM: P,V,T,R → n
Apply the Solution Map:
convert the units

46. Apply the Solution Map:
Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25°C
Information
Given: V = 3.2 L, P = atm,
T = 298 K
Find: n, mol
Eq’n: PV = nRT
SM: P,V,T,R → n
Apply the Solution Map:

47. Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 308 K
Information
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 308 K
Find: VO2, L
Eq’n: PV=nRT
CF: 1 mole KClO3 = g
2 mole KClO3  3 moles O2
SM: g → mol KClO3 → mol O2 → L O2

48. Apply the Solution Map:
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 308 K
Information
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 308 K,
nO2 = 3.60 moles
Find: VO2, L
Eq’n: PV=nRT
CF: 1 mole KClO3 = g
2 mole KClO3  3 moles O2
SM: g → mol KClO3 → mol O2 → L O2
Apply the Solution Map:
convert the units

49. Apply the Solution Map:
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 308 K
Information
Given: 294 g KClO3
PO2 = mmHg, TO2 = 308 K,
nO2 = 3.60 moles
Find: VO2, L
Eq’n: PV=nRT
CF: 1 mole KClO3 = g
2 mole KClO3  3 moles O2
SM: g → mol KClO3 → mol O2 → L O2
Apply the Solution Map:

50. Example: How many grams of water will form when 1
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information
Given: 1.24 L H2
Find: g H2O
CF: 1 mol H2O = g
2 mol H2O  2 mol H2
1 mol H2  22.4 L
Write a Solution Map:
L H2
mol H2
mol H2O
g H2O

51. Apply the Solution Map:
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information
Given: 1.24 L H2
Find: g H2O
CF: 1 mol H2O = g
2 mol H2O  2 mol H2
1 mol H2  22.4 L
SM: L → mol H2 → mol H2O → g H2O
Apply the Solution Map:

52. Apply the Solution Map:
Example: Find Molar Mass A sample of a gas has a mass of g. Its volume is L at a temperature of 55°C and a pressure of 886 mmHg. Find its molar mass.
Information
Given: V = L, P = atm,
t = 328 K, m = g
Find: molar mass, (g/mol)
Eq’n: PV = nRT; MM = mass/moles
SM: P,V,T,R → n & mass → mol. mass
Apply the Solution Map:

53. Example of Using Molar Volume: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information
Given: 1.24 L H2
Find: g H2O
CF: 1 mol H2O = g
2 mol H2O  2 mol H2
1 mol H2  22.4 L
Write a Solution Map:
L H2
mol H2
mol H2O
g H2O

54. Apply the Solution Map:
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information
Given: 1.24 L H2
Find: g H2O
CF: 1 mol H2O = g
2 mol H2O  2 mol H2
1 mol H2  22.4 L
SM: L → mol H2 → mol H2O → g H2O
Apply the Solution Map: