1. Gases
amount: n = number of moles; m = mass (g(; MM = molar mass

2. KMT – Kinetic Molecular Theory of Gases
Background for understand physical properties of gases
Assumptions:
gases are mostly empty space
molecules are in constant and chaotic motion
collisions are complete elastic
pressure caused by molecules colliding with container walls (KE = ½ mv2)
Air is mostly molecules (N2 (78%); O2 (21%); Ar (0.09%); CO2 (0.03%); ...) so particles is not generally used
amount: n = number of moles; m = mass (g); MM = molar mass

3. KMT can be used for all three states of matter.
Motion:
translational
rotational
vibrational
State
Gas ✔
Liquid
(✔)
Solid
(explained in detail next slide)
amount: n = number of moles; m = mass (g); MM = molar mass

5. Measurements
Volume
Amount
Temperature
Pressure
1 L
(1 cm3 = 1 mL)
(1 m)3 = (102 cm)3
n = m / MM
oC = K
1 atm
760 mmHg (torr)
kPa
14.7 psi
(1.013 bar)
Standard Temperature & Pressure (STP) for Gas Pbs.
T = 0oC (273 K) P = 1 atm
Standard molar volume = 22.4 L
Volume: converting vol (L) cubic length (m3): (cubed 1m = cubed 102 cm)
amount: n = number of moles; m = mass (g); MM = molar mass
AP Reference

6. Measurements – Measuring Devices
Barometer Manometer Sphygmomanometer
(atmospheric) (a gas)

7. Measurements - Manometer
Pgas = Patm P gas > Patm Pgas < Patm
Pgas is greater than Patm: Pgas is less than Patm
(Pgas = Patm + Δh) (Pgas = Patm – Δh)
Patm = 760mm Hg Patm = 760
Δh = 20 mm Hg Δh = 20 mm Hg
Pgas = 780 mm Hg Pgas = 740 mm Hg
Pgas = Patm
Pgas = Patm + DHg
Pgas = Patm – DHg

8. Combination of Four Gas Laws
Combined Gas Law
Combination of Four Gas Laws
Boyle (P & V)
Charles (V & T)
Gay-Lussac (P&T)
Avogadro
(at a fixed T & P, V is directly related to number of moles)

9. Graphically (Ideal Gases)
Boyle’s Law
Charles’s Law
Gay-Lussac’s Law

10. Combined Gas Law
A gas-filled balloon contains 1.1 mol in a 26.5 L at 25oC and 101 kPa of pressure. How many moles of gas are added or removed to produce 27.0 L at 52oC and 1050 mmHg?
determine which law to use (there’s other gas laws).
assign values to the variables – organizes: less confusion in initial & final conditions
convert T to K; units on both sides of equation are the same
rearrange equation
substitute values
solve; check

11. Combined Gas Law
A deodorant can has a volume of 175 mL and a pressure of 3.8 atm at 22oC. What would the pressure be if the can was heated to 100.oC?

12. PV = nRT IDEAL GAS LAW Units: P = atm V= L n = mol
R = Latm/ Kmol
T = K
Static conditions * (combined = change of conditions)
Ideal gas law does depends only on initial & final conditions (State Function)

13. PV = nRT
A 47.3-L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temperature?
P = 1.85 atm
V = 47.3 L
n = 1.62 mol
R = L-atm/K-mol
T = ?
T = (1.85 atm)(47.3 L)
(1.62 mol)( L-atm/K-mol)
Ideal gas law does depends only on initial & final conditions (State Function)
T = 658K (=385oC)

14. Ideal Gas Law – MOLAR MASS
rearrange for n: n = PV RT
Let M = molar mass (g/mol) (m=mass of sample) (n=number of moles)
M = m/n => M = mRT PV

15. Ideal Gas Law – DENSITY
M = mRT PV
M = m RT
V P
M = dRT P

16. IDEAL GAS LAW
conventional form: PV = nRT
molar mass MM = mRT PV
density MM = dRT P

17. Using Gas Laws to Determine a Balanced Equation
A piston expands after a gaseous reaction that is carried out under constant pressure & volume. (Assume isolated system)
Which of the following balanced chemical equations describes the reaction?
A2(g) + B2(g)  2AB(g) (2) 2AB(g) + B2(g)  2AB2(g)
(3) A(g) + B2(g)  AB2(g) (4) 2AB2(g)  A2(g) + 2B2(g)

18. Vinit & Tinit (from picture)
Using Gas Laws to Determine a Balanced Equation
Vinit & Tinit (from picture)
Pinit = Pinit (from problem)
Combined gas law: P1 V1 = P2 V2 n1 T n2 T2
From Avogadro’s law, if T doubles, n must be halved. (P & V are constant)
number of moles
A2(g) + B2(g)  2AB(g)
 2
(2) 2AB(g) + B2(g)  2AB2(g)
 2
(3) A(g) + B2(g)  AB2(g)
 1
(4) 2AB2(g)  A2(g) + 2B2(g)
2 

19. NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
Problem
Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25oC and 2.00 atm.
NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
How many moles of CO2(g) are produced?

20. NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
Problem
Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25oC and 2.00 atm.
NaHCO3(s) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
Stoichiometry: 1 CO2 : 1 NaHCO3
Determine mass of mole NaHCO3

21. Gas Mixtures (Dalton’s Law of Partial Pressures)
PT = Pi + Pii ... mole fraction (Xi) = Pi/PT
so, Pi = Xi PT
When 1 mole of methane (CH4) is heated with 4 moles of O2, 1 mol CO2 and 2 mol H2O are formed. What are the partial pressures if PT = 1.26 atm.
1CO2 + 2H2O  1CO2 + 2H2O
CO2: mole fraction: XCO2 = 1.00 / = 0.200
partial pressure: * 1.26 atm = atm
(all of the methane is consumed so n = 5)