1. General Physics: Ideal Gas Law and Kinetic Theory1

2. Temperature and Heat
Temperature: determined by the average KE of molecules.
Heat: a way to transfer energy other than work.
Heat flows from high temperature to low temperature.
Internal Energy: total energy of the molecules of a body.

3. Temperature and Heat
Total Internal Energy= Kinetic Energy of Molecules + Potential Energy of Molecules

4. Temperature Scales
Measuring temperature. Thermometer: using expansion of the volume of a column of liquid. Scale: set of numbers associated to the value of the temperature.

5. Scales of Temperature Celsius Kelvin 100 373 Boiling point of water
273
Freezing point of water
-273
Absolute zero

6. Temperature Scales Celsius-Kelvin
K = C

7. Temperature Scales Fahrenheit
C = (5/9)(F-32)

8. Aside: The Periodic Table05 1

9. The Periodic Table Explained ?
Look carefully
proton
neutron
electron 05 1

10. Energy vs Mass E = mc2 He (m=4.0026 u) O (M=15.9995 u)
4 x He = u
Mass difference = 0.01 u = binding energy
So energy is the same as mass somehow ??
E = mc2 06 1

11. The Mole, Avogadro's Number, and Molecular Mass

12. Atomic Mass Unit, U
By international agreement, the reference element is chosen to be the most abundant type of carbon, called carbon-12, and its atomic mass is defined to be exactly twelve atomic mass units, or 12 u.

13. Molecular Mass
The molecular mass of a molecule is the sum of the atomic masses of its atoms.
For instance, hydrogen and oxygen have atomic masses of u and u, respectively.
The molecular mass of a water molecule (H2O) is:
2( u) u = u.

14. Avogadro's Number NA
The number of atoms per mole is known as Avogadro's number NA, after the Italian scientist Amedeo Avogadro (1776–1856):
Molar Volume Vμ = l: Volume one mole of ideal gas will occupy at STP (To=273.15K, po = 1atm)

15. Number of Moles, n
The number of moles n contained in any sample is the number of particles N in the sample divided by the number of particles per mole NA (Avogadro's number):
The number of moles contained in a sample can also be found from its mass.

16. Molecular Picture of Gas
Gas is made up of many individual molecules
Number density is number of molecules/volume:
N/V = r/m
r is the mass density
m is the mass for one molecule
Number of moles: n = N / NA
NA = Avogadro’s Number = 6.022x1023 mole-1
1 mole of a substance = molecular mass in  in grams
e.g., 1 mole of N2 has mass of 2x14=28 grams

17. Atomic Act I Which contains the most molecules ?
1. A mole of water (H2O)
2. A mole of oxygen gas (O2)
3. Same
H2O O2
correct 12

18. Atomic Act II Which contains the most atoms ? 1. A mole of water (H2O)
2. A mole of oxygen gas (O2)
3. Same
H2O (3 atoms)
O2 (2 atoms)
correct 15

19. Atomic Act III Which weighs the most ? 1. A mole of water (H2O)
2. A mole of oxygen gas (O2)
3. Same
correct
H2O (M = )
O2 (M = ) 17

20. Thermodynamic variables. Pressure.
Pressure is an important thermodynamic variable.
Pressure is defined as the force per unit area: p = F/A
The SI unit is pressure is the Pascal: 1 Pa = 1 N/m2. Another common unit is the atm (atmospheric pressure) which is the pressure exerted by the atmosphere on us (1 atm = 1.013 x 105 N/m2).
A pressure of 1 atm will push a mercury column up by 76 cm.

21. Physical Characteristics of Gases Physical Characteristics
Typical Units
Volume, V
liters (L)
Pressure, P
atmosphere
(1 atm = 1.015x105 N/m2)
Temperature, T
Kelvin (K)
Number of atoms or molecules, n
mole (1 mol = 6.022x1023 atoms or molecules)

22. “Father of Modern Chemistry” Chemist & Natural Philosopher
Boyle’s Law
Pressure and volume are inversely related at constant temperature.
PV = constant
As one goes up, the other goes down.
P1V1 = P2V2
“Father of Modern Chemistry”
Robert Boyle
Chemist & Natural Philosopher
Listmore, Ireland
January 25, 1627 – December 30, 1690

23. Boyle’s Law: P1V1 = P2V2

24. Boyle’s Law: P1V1 = P2V2

25. Gas Laws, Absolute Temperature, and the Kelvin Temperature Scale
When the temperature and mass of an ideal gas is held constant, pV is constant, Boyle’s law:

26. Calculations with Boyle’s Law

27. Calculation with Boyle’s Law
Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mm Hg after its pressure is changed to 2200 mm Hg at constant T?
1. Set up a data table:
Conditions 1 Conditions 2
P1 = mm Hg P = mm Hg
V1 = L V2 = ?

28. Calculation with Boyle’s Law (Continued)
2. When pressure increases, volume decreases.
Solve Boyle’s Law for V2:
P1V1 = P2V2
V = V1 x P1 P2
V = L x 550 mm Hg = L
2200 mm Hg
pressure ratio
decreases volume

29. Gas Law Problems BOYLE’S LAW P V
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa P V
WORK:
P1V1 = P2V2
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL

30. Learning Check 1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg
A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant).
1) mm Hg
2) mm Hg
3) 1200 mm Hg

31. Solution 1) 200. mm Hg Data table Conditions 1 Conditions 2
P1 = mm Hg P2 = ???
V1 = L V2 = L
P2 = P1 x V1 V2
600. mm Hg x L = mm Hg L

32. PV Calculation (Boyle’s Law)
A quantity of gas has a volume of 120 dm3 when confined
under a pressure of 93.3 kPa at a temperature of 20 oC.
At what pressure will the volume of the gas be 30 dm3 at
20 oC?
P1 x V1 = P2 x V2
(93.3 kPa) x (120 dm3) = (P2) x (30 dm3)
P2 = kPa

33. Jacques-Alexandre Charles Mathematician, Physicist, Inventor
Charles’ Law
Volume of a gas varies directly with the absolute temperature at constant pressure.
V = KT
V1 / T1 = V2 / T2
Jacques-Alexandre Charles
Mathematician, Physicist, Inventor
Beaugency, France
November 12, 1746 – April 7, 1823

34. Charles’s Law

35. Charles’ Law: V1/T1 = V2/T2

36. Charles’ Law: V1/T1 = V2/T2

37. Gas Laws, Absolute Temperature, and the Kelvin Temperature Scale
When the pressure is held constant, V is proportional to T, Charles’ Law

38. EXAMPLE 6.15
Charles’s Law: Temperature-Volume Relationships
A balloon indoors, where the temperature is 27 °C, has a volume of 2.00 L. What would its volume be (a) in a hot room where the temperature is 47 °C, and (b) outdoors, where the temperature is –23 ºC? (Assume no change in pressure in either case.)

39. Gas Law Problems CHARLES’ LAW T V
A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
CHARLES’ LAW
GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K T V
WORK:
P1V1T2 = P2V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3

40. Charles’s Law: Temperature-Volume Relationships continued
A sample of oxygen gas occupies a volume of 2.10 L at 25 °C. What volume will this sample occupy at 150 °C? (Assume no change in pressure.)
b. A sample of hydrogen occupies 692 L at 602 °C. If the pressure is held constant, what volume will the gas occupy after being cooled to 23 °C?
At what Celsius temperature will the initial volume of oxygen in Exercise 6.15A occupy L? (Assume no change in pressure.)

41. Gas Law Problems GAY-LUSSAC’S LAW P T
A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?
GAY-LUSSAC’S LAW
GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ? P T
WORK:
P1T2 = P2T1
(765 torr)T2 = (560. torr)(309K)
T2 = 226 K = -47°C

42. Ideal Gas Laws Combining gives the ideal gas law:
At STP p2= Pa, Vμ = l, V2 = nVμ, T2 = K
R=8.31 J/(mole x K) universal gas constant
PV/T = nR or

43. The Ideal Gas Law P V = n R T (Equation of State) P V = N kB T
n = number of moles
R = ideal gas constant 8.31 J/(molK)
P V = N kB T
P = pressure in N/m2 (or Pascals)
V = volume in m3
N = number of molecules
T = absolute temperature in K
k B = R/NA Boltzmann’s constant = 1.38 x J/K
Note: P V has units of N-m or J (energy!) 20

44. Ideal Gas Law ACT II PV = nRT
You inflate the tires of your car so the pressure is 3.0 barr when the air inside the tires is at 20 degrees C. After driving on the highway for a while, the air inside the tires heats up to 38 C. Which number is closest to the new air pressure (assume V = const.)?
1) 1.6 barr 2) 3.2 barr 3) 5.7 barr
Careful, you need to use the temperature in K
P = P0 (38+273)/(20+273) 23

45. Ideal Gas Law: ACT 1 pV = nRT
A piston has volume 20 ml, and pressure of 30 psi. If the volume is decreased to 10 ml, what is the new pressure? (Assume T is constant.)
1) ) ) 15
Change this to the pressure change with temperature, but give the temperature in C.
V=20
P=30
V=10
P=??
When n and T are constant, pV is constant (Boyle’s Law) 26

46. Balloon ACT 1
What happens to the pressure of the air inside a hot-air balloon when the air is heated? (Assume V is constant)
1) Increases 2) Same 3) Decreases
Balloon is still open to atmospheric pressure, so it stays at 1 atm 30

47. Balloon ACT 2
What happens to the buoyant force on the balloon when the air is heated? (Assume V remains constant)
1) Increases 2) Same 3) Decreases
FB = r V g
r is density of outside air! 32

48. Balloon ACT 3
What happens to the number of air molecules inside the balloon when the air is heated? (Assume V remains constant)
1) Increases 2) Same 3) Decreases
PV = NkT
P and V are constant. If T increases N decreases. 34

49. Gas Laws
In terms of the ideal gas law, explain briefly how a hot air balloon works.
Hot air has less mole density than cool air. So less hot air is required in order to achieve the same pressure as cool air. This makes the density of hot air less allowing it to float.
When temperature increases the volume of the gas increases, thus reducing the density of the gas making it lighter that then surrounding air, which causes the balloon to rise.
In regards to balloons, you should inhale helium again sometime, we all liked that.
Note! this is not a pressure effect, it is a density effect. As T increases, the density decreases the balloon then floats due to Archimedes principle. The pressure remains constant! 36

50. Gas Law Problems COMBINED GAS LAW P T V V1 = 7.84 cm3 P1 = 71.8 kPa
A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
COMBINED GAS LAW
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = kPa
T2 = 273 K
P T V
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=( kPa) V2 (298 K)
V2 = 5.09 cm3

51. Ideal Gas Law IDEAL GAS LAW P = ? atm n = 0.412 mol T = 16°C = 289 K
Calculate the pressure in atmospheres of mol of He at 16°C & occupying L.
IDEAL GAS LAW
GIVEN:
P = ? atm
n = mol
T = 16°C = 289 K
V = 3.25 L
R = Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289)
L mol Latm/molK K
P = 3.01 atm

52. Ideal Gas Law IDEAL GAS LAW V = ? n = 85 g T = 25°C = 298 K
Find the volume of 85 g of O2 at 25°C and kPa.
IDEAL GAS LAW
GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = kPa
R = dm3kPa/molK
WORK:
85 g 1 mol = 2.7 mol
32.00 g
= 2.7 mol
PV = nRT
(104.5)V=(2.7) (8.315) (298)
kPa mol dm3kPa/molK K
V = 64 dm3

53. Summary Ideal Gas Law PV = n R T Kinetic Theory of Monatomic Ideal Gas
P = pressure in N/m2 (or Pascals)
V = volume in m3
n = # moles
R = 8.31 J/ (K mole)
T = Temperature (K)
Kinetic Theory of Monatomic Ideal Gas
<Ktr> = 3/2 kB T 50

54. Summary Ideal gas law: Absolute zero is –273.15°C.
Celsius–Kelvin conversion: