2. One of the most amazing things about gasses is that , despite wide differences in chemical properties, all gases more or less obey the gas law.
The gas law deal with how gases behave with respect pressure, volume, temperature and amount.

3. A Gas Uniformly fills any container. Easily compressed.
Mixes completely with any other gas.
Exerts pressure on its surroundings.
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4. Pressure SI units = Newton/meter2 = 1 Pascal (Pa)
1 standard atmosphere = 101,325 Pa
1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr
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5. Barometer Device used to measure atmospheric pressure.
Mercury flows out of the tube until the pressure of the column of mercury standing on the surface of the mercury in the dish is equal to the pressure of the air on the rest of the surface of the mercury in the dish.
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6. Manometer
Device used for measuring the pressure of a gas in a container.
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7. Pressure Conversions: An Example
The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals.
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9. Gas laws can be deduced from observations like these.
Mathematical relationships among the properties of a gas (Pressure, Volume, Temperature and Moles) can be discovered.
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10. Another way to describing it is that their product are constant.
Boyle’s Law
The volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant.
Another way to describing it is that their product are constant.
As one goes up, the other goes down
PV = k

11. Boyle’s Law: P1V1 = P2V2

12. Boyle’s Law
Pressure and volume are inversely related (constant T, temperature, and n, # of moles of gas).
PV = k (k is a constant for a given sample of air at a specific temperature)
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14. P1V1= P2V2 = (0.956atm)(12,4l)=(1.20)(V2)
EXERCISE!
A sample of helium gas occupies 12.4 L at 23°C and atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant?
P1V1= P2V2 = (0.956atm)(12,4l)=(1.20)(V2)
(0.956atm)(12.4L)/(1.20) = V2
9.88 L
(0.956 atm)(12.4 L) = (1.20 atm)(V2)
The new volume is 9.88 L.
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15. A sample of oxygen gas occupies a volume of 250. mL at 740
A sample of oxygen gas occupies a volume of 250.mL at 740. torr pressure. What volume will it occupy at 800. torr pressure?
(740.torr)(250mL)=(800.torr)(V2)
V2 = 231 mL

16. (125Kpa)(3,50L)=(p2)(2.00L) P2= 219Kpa
A sample of carbon dioxide occupies a volume of 3.50 liters at 125 kPa pressure. What pressure would the gas exert if the volume was decreased to 2.00 liters?
(125Kpa)(3,50L)=(p2)(2.00L)
P2= 219Kpa

17. A 2. 0 liter container of nitrogen had a pressure of 3. 2 atm
A 2.0 liter container of nitrogen had a pressure of 3.2 atm. What volume would be necessary to decrease the pressure to 1.0 atm.
 (3.2atm)(2,0L) = (1,0atm)(V2)
V2= 6.4 L

18. Charles’s Law

19. Charles’s Law
Volume and Temperature (in Kelvin) are directly related (constant P and n).
V=bT (b is a proportionality constant)
K = °C + 273
0 K is called absolute zero.
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21. Charles’ Law: V1/T1 = V2/T2

22. Charles’s Law To play movie you must be in Slide Show Mode
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23. EXERCISE!
Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)? Remember to convert the temperature to Kelvin (1.30)/( )= V2/( ) L
The new volume will become L. Remember to convert the temperatures to Kelvin.
(1.30 L) / ( ) = V2 / ( )
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25. 4. 40 L of a gas is collected at 50. 0 °C
4.40 L of a gas is collected at 50.0 °C. What will be its volume upon cooling to 25.0 °C?
4.06 L

26. 5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure (as required by Charles' Law)?
127 ◦C

27. Calculate the decrease in temperature when 2. 00 L at 20
. Calculate the decrease in temperature when 2.00 L at 20.0 °C is compressed to 1.00 L.
146.5 K

28. 600.0 mL of air is at 20.0 °C. What is the volume at 60.0 °C?

29. Avogadro’s Law
This law states that equal volume of gases at constant temperature and pressure contains the same number of particle
V = an V1/n1 = V2/n2

30. Avogadro’s Law
Volume and number of moles are directly related (constant T and P).
V = an (a is a proportionality constant)
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31. (2.45mol)/(89.0l) = (2.10)/ (V2) (89.0)(2.10)/(2.45) 76.3 L
EXERCISE!
If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure?
(2.45mol)/(89.0l) = (2.10)/ (V2)
(89.0)(2.10)/(2.45)
76.3 L
The new volume is 76.3 L.
(2.45 mol) / (89.0 L) = (2.10 mol) / (V2)
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32. Avogadro’s Law
If 0.50 moles of oxygen occupies a volume of 7.20 L at STP, what volume does 1.5 moles of O2 occupy at STP
21.6 L

33. If 3. 25 moles of argon gas occupies a volume of 1
If 3.25 moles of argon gas occupies a volume of 1.00L at STP, what volume does mole of argon occupy
4.35 L

34. If 12. 40 moles of CO2 occupies a volume of 96
If moles of CO2 occupies a volume of 96.8 L, how many moles occupy 72.6 L.
9.30 moles

35. We can bring all of these laws together into one comprehensive law:
V = bT (constant P and n)
V = an (constant T and P)
V = (constant T and n)
PV = nRT
(where R = L·atm/mol·K, the universal gas constant)
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36. EXERCISE!
An automobile tire at 23°C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire. PV = nRT (3.18atm)(25.0L) = n( )(23+273) 3.27 mol
The number of moles of air is 3.27 mol.
(3.18 atm)(25.0 L) = n( )(23+273)
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37. EXERCISE!
What is the pressure in a L tank that contains kg of helium at 25°C? Convert Kg of helium to moles of helium (p)(304.0L) =([5.670×1000]/4.003)(0.0821)(25+273) 114 atm
The pressure is 114 atm. Remember to convert kg of helium into moles of helium before using the ideal gas law.
(P)(304.0) = ([5.670×1000]/4.003)(0.0821)(25+273)
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38. EXERCISE!
At what temperature (in °C) does 121 mL of CO2 at 27°C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm? P1V1/T1 = P2V2/T2 (1.05atm)(121mL)/(27+273) = (1.40atm)(293mL)/(T2+273) 696°C
The new temperature is 696°C.
(1.05 atm)(121 mL) / (27+273) = (1.40 atm)(293 mL) / (T )
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39. A sample of hydrogen gas(H2) has a volume of 8
A sample of hydrogen gas(H2) has a volume of 8.56L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the moles of H2molecukes present in this gas sample

40. Molar Volume of an Ideal Gas
For 1 mole of an ideal gas at 0°C and 1 atm, the volume of the gas is L.
STP = standard temperature and pressure
0°C and 1 atm
Therefore, the molar volume is L at STP.
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41. What is the volume (in liters) occupied by 49.8 g of HCl at STP?

42. EXERCISE!
A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O2 are present? Pv= nRT (1.00atm)(2.40L)= n( )(273) N= mol O2 ×32.00g/mol 3.57 g
3.57 g of O2 are present.
(1.00 atm)(2.50 L) = n( )(273)
n = mol O2 × g/mol = 3.57 g O2
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43. Molar Mass of a Gas d = density of gas T = temperature in Kelvin
P = pressure of gas
R = universal gas constant
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44. EXERCISE!
What is the density of F2 at STP (in g/L)? D = (MM)/(R)(T) D = (19.00g/mol×2)(1.00atm)/( )(273K) 1.70 g/L
d = (MM)(P) / (R)(T)
d = (19.00g/mol × 2)(1.00 atm) / ( )(273 K)
d = 1.70 g/L
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45. A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/l at 36 c and 2.88 atm. Calculate the molar mass of the compound

46. For a mixture of gases in a container, PTotal = P1 + P2 + P3 + . . .
The total pressure exerted by a mixture of gases is the sum of the partial pressure of each individual gas that is present.
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47. After the valve is opened, what is the pressure of the helium gas?
EXERCISE!
Consider the following apparatus containing helium in both sides at 45°C. Initially the valve is closed.
After the valve is opened, what is the pressure of the helium gas?
The pressure is 2.25 atm. There are two ways to solve the problem:
1) Find the new pressure of the gas on the left side after the valve is opened (P1V1=P2V2). The pressure becomes 1.50 atm.
(2.00 atm)(9.00 L) = (P2)(9.00 L L)
P2 = 1.50 atm
Find the new pressure of the gas on the right side after the valve is opened (P1V1=P2V2). The pressure becomes 0.75 atm.
(3.00 atm)(3.00 L) = (P2)(9.00 L L)
P2 = 0.75 atm
The total pressure is therefore = 2.25 atm.
2) Find the number of moles in each chamber separately (using PV = nRT), add the number of moles and volumes on each side, then use PV = nRT to solve for the new pressure.
3.00 atm
3.00 L
2.00 atm
9.00 L
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48. Calculate the new partial pressure of oxygen.P1V1=P2V2 6.13 atm
EXERCISE!
27.4 L of oxygen gas at 25.0°C and 1.30 atm, and 8.50 L of helium gas at 25.0°C and 2.00 atm were pumped into a tank with a volume of 5.81 L at 25°C.
Calculate the new partial pressure of oxygen.P1V1=P2V2
6.13 atm
Calculate the new partial pressure of helium.
2.93 atm
Calculate the new total pressure of both gases =
9.06 atm
For oxygen:
P1V1=P2V2
(1.30 atm)(27.4 L) = (P2)(5.81 L)
P2 = 6.13 atm
For helium:
(2.00 atm)(8.50 L) = (P2)(5.81 L)
P2 = 2.93 atm
The total pressure is therefore = 9.06 atm.
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49. Mixture of helium and oxygen can be used in scuba diving tanks to help prevent “the bends” for a particular drive, 46 L He and 1.0 atm and 12 L O2 at 25C and 1.0 atm were pumped into a tank with 5.0 L. calculate the partial pressure of each gas and total pressure in the thank at 25C.

51. Postulates of the Kinetic Molecular Theory
1) The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero).
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52. Postulates of the Kinetic Molecular Theory
2) The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas.
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53. Postulates of the Kinetic Molecular Theory
3) The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other.
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54. Postulates of the Kinetic Molecular Theory
4) The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.
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55. Root-Mean- Square speed
One way to estimate molecular speed it to calculate the root-mean-square speed

56. Root Mean Square Velocity
R = J/K·mol
(J = joule = kg·m2/s2)
T = temperature of gas (in K)
M = mass of a mole of gas in kg
Final units are in m/s.
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57. Diffusion – the mixing of gases.
Effusion – describes the passage of a gas through a tiny orifice into an evacuated chamber.
Rate of effusion measures the speed at which the gas is transferred into the chamber.
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58. Graham’s Law of Effusion
M1 and M2 represent the molar masses of the gases.
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