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Gases  Regardless of their chemical identity, gases tend to exhibit similar physical behaviors  Gas particles can be monatomic (Ne), diatomic (N 2.

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Presentation on theme: "Gases  Regardless of their chemical identity, gases tend to exhibit similar physical behaviors  Gas particles can be monatomic (Ne), diatomic (N 2."— Presentation transcript:

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2 Gases

3  Regardless of their chemical identity, gases tend to exhibit similar physical behaviors  Gas particles can be monatomic (Ne), diatomic (N 2 ), or polyatomic (CH 4 ) – but they all have some common characteristics: The Nature of Gases 1.Gases have mass. 2.Gases are compressible. 3.Gases fill their containers. 4.Gases diffuse. 5.Gases exert pressure. 6.Pressure is related to temperature

4 Kinetic Molecular Theory  Theory used to explain the behaviors and experimental characteristics of ideal gases – The theory states that the tiny particles in all forms of matter are in continuous motion.The theory states that the tiny particles in all forms of matter are in continuous motion.  There are 3 basic assumptions of the KMT as it applies to ideal gases.

5 KMT Assumption #1  A gas is composed of small particles.  The particles have an insignificant volume and are relatively far apart from one another.  There is empty space between particles.  No attractive or repulsive forces between particles.

6  The particles in a gas move in constant random motion.  Particles move in straight paths and are completely independent of each other  Particles path is only changed by colliding with another particle or the sides of its container. KMT Assumption #2

7  All collisions a gas particle undergoes are perfectly elastic.  They exert a pressure but don’t lose any energy during the collisions. KMT Assumption #3

8 Gases have mass.  Gases are classified as matter, therefore, they must have mass.

9 Gases are squeezable  The gas particles empty space can be compressed by added pressure giving the gas particles less room to bounce around thus decreasing the overall volume.

10 Gases are squeezable  There are a huge number of applications Storm door closersStorm door closers Pneumatic tube delivery devicesPneumatic tube delivery devices TiresTires Air tanksAir tanks

11 Gases fill their containers  Gases expand until they take up as much room as they possibly can.

12 Gases fill their containers  The random bouncing motion of gases allows for the mixing up and spreading of the particles until they are uniform throughout the entire container.

13 Gases diffuse  Gases can move through each other rapidly. The movement of one substance through another is called diffusion.The movement of one substance through another is called diffusion.  Because of all of the empty space between gas molecules, gas molecules can pass between each other until the gases mix uniformly.

14 Gases diffuse

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17  This doesn’t happen at the same speeds for all gases though. Some gases diffuse more rapidly then other gases based on their size and their energy.Some gases diffuse more rapidly then other gases based on their size and their energy.  Diffusion explains why gases are able to spread out to fill their containers.  It’s why we can all breathe oxygen anywhere in the room.  It also helps us avoid potential odoriferous problems.

18 Gases exert pressure  Gas particles exert pressure by colliding with objects in their path.  The definition of pressure is the force per unit area – so the total of all of the tiny collisions makes up the pressure exer- ted by the gas

19 Gases exert pressure It’s the pressure exerted by the gases that hold the walls of a container out The pressure of gases is what keeps our tires inflated, makes our basketballs bounce, makes hairspray come out of the can, helps our lungs inflate, allow vacuum cleaners to work, etc.

20 Pressure depends on Temp  Temperature measures the average kinetic energy of the particles in an object. Therefore, the higher the temperature the more energy the gas particle has.Therefore, the higher the temperature the more energy the gas particle has.  So the collisions are more often and with a higher force. Think about the pressure of a set of tires on a car.Think about the pressure of a set of tires on a car.

21 Pressure depends on Temp Pressure Gauge Pressure Gauge Today’s temp: 35°F

22 Today’s temp: 85°F Pressure depends on Temp Pressure Gauge Pressure Gauge

23 Measuring Gases  Variables that are very important to studying the behavior of gases:  Volume: generally in Liters (1L = 1000 mL )  Temperature :given in Celsius but must be converted to Kelvin for gas law problems Kelvin = °C  Pressure  1 atm=760 mmHg=760 Torr = 14.7 psi = kPa  amount generally given in moles

24 S T P  Since the behavior of a gas is dependent on temperature and pressure, it is convenient to designate a set of standard conditions, called STP in order to study gas behavior. Standard Temperature = 0°C or 273KStandard Temperature = 0°C or 273K Standard Pressure = 1atm or 760mmHg or 101.3kPa (depending on the method of measure)Standard Pressure = 1atm or 760mmHg or 101.3kPa (depending on the method of measure)

25 Atmospheric Pressure  The gases in the air are exerting a pressure called atmospheric pressure  Atmospheric pressure is a result of the fact that air has mass and is colliding with everything.  Atmospheric pressure is measured with a barometer.

26  Atmospheric pressure varies with altitude The lower the altitude, the longer and heavier is the column of air above an area of the earth.  Example: Recipes like the back of a cake box that describes how to cook a cake at a higher altitude Atmospheric Pressure

27 Boyle’s Law

28 Boyle’s Mathematical Law: since PV equals a constant P 1 V 1 = P 2 V 2 If we have a given amount of a gas at a starting pressure and volume, what would happen to the pressure if we changed the volume? Or to the volume if we changed the pressure?

29 Boyle’s Mathematical Law: List the variables or clues given:  P 1 = 2 atm  V 1 = 3.0 L  P 2 = 4 atm  V 2 = ? P 1 V 1 = V 2 P 2 Plug in the variables & calculate: Plug in the variables & calculate: (2 atm) (3.0 L) = (4 atm) (V 2 ) Ex: A gas has a volume of 3.0 L at 2 atm. What will its volume be at 4 atm? Ex: A gas has a volume of 3.0 L at 2 atm. What will its volume be at 4 atm?

30 Charles’ Law

31 Charles’s Mathematical Law: since V/T = k = V 1 V 2 T 1 T 2 If we have a given amount of a gas at a starting volume and temperature, what would happen to the volume if we changed the temperature? Or to the temperature if we changed the volume?

32 Charles’s Mathematical Law:  T 1 = 400K  V 1 = 3.0 L  T 2 = 500K  V 2 = ? List the variables or clues given: Plug in the variables & calculate: 3.0L400K 500K X L X L= Ex: A gas has a volume of 3.0 L at 400K. What is its volume at 500K

33 Gay-Lussac’s Law

34 since P/T = k P 1 P 2 T 1 T 2 = If we have a given amount of a gas at a starting temperature and pressure, what would happen to the pressure if we changed the temperature? Or to the temp. if we changed the pressure? Gay-Lussac’s Mathematical Law:

35  T 1 = 400K  P 1 = 3.0 atm  T 2 = 500K  P 2 = ? List the variables or clues given: Plug in the variables & calculate: 3.0atm 400K 500K X atm X atm = = Ex: A gas has a pressure of 3.0atm at 400K. What is its pressure at 500K? Ex: A gas has a pressure of 3.0atm at 400K. What is its pressure at 500K?

36 Combined Gas Law

37 Combined and ideal gas laws

38  There is a lesser known law called Avogadro’s Law which relates Volume & moles (n).  It turns out that they are directly related to each other.  As number of moles increases then Volume increases. V/n = k Avogadro’s Law

39  If we combine all of the laws together including Avogadro’s Law we get: Where R is the universal gas constant Normally written as Ideal Gas Law

40  Because of the different pressure units there are 3 possibilities for our ideal gas constant R =.0821 LatmmolK If pressure is given in mmHg or torrIf pressure is given in mmHg or torr R=62.4LmmHgmolK If pressure is given in kPaIf pressure is given in kPa R=8.314LkPamolK If pressure is given in atmIf pressure is given in atm Ideal Gas Constant

41 Practice 1. Use the Ideal Gas Law to complete the following table for ammonia gas (NH 3 ). PressureVolumeTempMolesGrams 2.50 atm 0C0C mmHg6.0 L 100  C

42 Variations of the Ideal Gas Law — We need to know that the unit mole is equal to mass divided by molar mass. n = m/MM PV = nRT

43 Variations of the Ideal Gas Law We can then use the MM equation to derive a version that solves for the density of a gas. We can then use the MM equation to derive a version that solves for the density of a gas. — Remember that D = m/V MW = dRT orMW = mRT P VP P VP Kitty cats say meow and they put dirt on their pee!!

44 Dalton’s Law of Partial Pressure States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.States that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. P T =P 1 +P 2 +P 3 +… What that means is that each gas involved in a mixture exerts an independent pressure on its container’s wallsWhat that means is that each gas involved in a mixture exerts an independent pressure on its container’s walls

45 Three of the primary components of air are CO 2, N 2, and O 2. In a sample containing a mixture of these gases at exactly 760 mmHg, the partial pressures of CO 2 and N 2 are given as P CO 2 = 0.285mmHg and P N 2 = mmHg. What is the partial pressure of O 2 ?Three of the primary components of air are CO 2, N 2, and O 2. In a sample containing a mixture of these gases at exactly 760 mmHg, the partial pressures of CO 2 and N 2 are given as P CO 2 = 0.285mmHg and P N 2 = mmHg. What is the partial pressure of O 2 ? Simple Dalton’s Law Calculation

46 P T = P CO2 + P N2 + P O2 Simple Dalton’s Law Calculation 760mmHg =.285mmHg mmHg + P O2 P O2 = 167mmHg

47 Partial pressures are also important when a gas is collected through water.Partial pressures are also important when a gas is collected through water. —Any time a gas is collected through water the gas is “contaminated” with water vapor. —You can determine the pressure of the dry gas by subtracting out the water vapor Dalton’s Law of Partial Pressure

48 P tot = P atmospheric pressure = P gas + P H 2 O Atmospheric Pressure Atmospheric Pressure —The water’s vapor pressure can be determined from a list and subtract- ed from the atmospheric pressure

49 WATER VAPOR PRESSURES Temp (°C) (mmHg)(kPa)

50 Determine the partial pressure of oxygen collected by water displace- ment if the water temperature is 20.0°C and the total pressure of the gases in the collection bottle is 730 mmHg.Determine the partial pressure of oxygen collected by water displace- ment if the water temperature is 20.0°C and the total pressure of the gases in the collection bottle is 730 mmHg. Simple Dalton’s Law Calculation P H2O at 20.0°C= 17.5 mmHg

51 P T = P H2O + P O2 Simple Dalton’s Law Calculation 730mmHg = P O2 P O2 = mmHg P H2O = 17.5 mmHg P T = 730 mmHg

52 Mole Fraction Moles of gas x x P T = P x Total moles A mixture of 4.00 moles of O 2 and 3.00 moles of H 2 exert a total pressure of 760 torr. What is the partial pressure of each gas? 4.00 moles of O 2 x 760 torr = 434 torr 7.00 total moles 3.00 moles of H 2 x 760 torr = 326 torr 7.00 total moles

53 Graham’s Law Thomas Graham studied the effusion and diffusion of gases.Thomas Graham studied the effusion and diffusion of gases. –Diffusion is the mixing of gases through each other. –Effusion is the process whereby the molecules of a gas escape from its container through a tiny hole

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55 Graham’s Law states that the rates of effusion and diffusion of gases at the same temperature and pressure is dependent on the size of the molecule.Graham’s Law states that the rates of effusion and diffusion of gases at the same temperature and pressure is dependent on the size of the molecule. –The bigger the molecule the slower it moves the slower it mixes and escapes. Graham’s Law

56 The velocities of two different gases are inversely proportional to the square roots of their molar masses.The velocities of two different gases are inversely proportional to the square roots of their molar masses. Rate of effusion of A == Rate of effusion of B MM B MM A

57 If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster? Graham’s Law Example Calc. Rate of effusion of A == Rate of effusion of B MBMBMBMB MAMAMAMA

58 Graham’s Law Example Calc. Rate of effusion of He = = Rate of effusion of Ar 40 g 4 g Helium is 3.16 times faster than Argon.


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